Nested Recursions, Simultaneous Parameters and Tree Superpositions
Abraham Isgur, Vitaly Kuznetsov, Mustazee Rahman, Stephen Tanny
NNESTED RECURSIONS, SIMULTANEOUS PARAMETERS AND TREESUPERPOSITIONS
ABRAHAM ISGUR, VITALY KUZNETSOV, MUSTAZEE RAHMAN, AND STEPHEN TANNY
Abstract.
We apply a tree-based methodology to solve new, very broadly defined fam-ilies of nested recursions of the general form R ( n ) = (cid:80) ki =1 R ( n − a i − (cid:80) pj =1 R ( n − b ij )),where a i are integers, b ij are natural numbers, and k, p are natural numbers that we use todenote “arity” and “order,” respectively, and with some specified initial conditions. Thekey idea of the tree-based solution method is to associate such recursions with infinitelabelled trees in a natural way so that the solution to the recursions solves a countingquestion relating to the corresponding trees. We characterize certain recursion familieswithin R ( n ) by introducing “simultaneous parameters” that appear both within the re-cursion itself and that also specify structural properties of the corresponding tree. First,we extend and unify recently discovered results concerning two families of arity k = 2,order p = 1 recursions. Next, we investigate the solution of nested recursion families bytaking linear combinations of solution sequence frequencies for simpler nested recursions,which correspond to superpositions of the associated trees; this leads us to identify andsolve two new recursion families for arity k = 2 and general order p . Finally, we extendthese results to general arity k >
2. We conclude with several related open problems. Introduction
In this paper, all values of the parameters and variables are integers.Loosely speaking, a nested recurrence relation (also called a meta-Fibonacci recursion) isany recursion where some argument contains a term of the recursion. In a series of recentpapers (see [1, 3, 9, 10, 11, 12]) infinite labelled trees are used to solve certain families ofnested recursions with the following general form: R ( n ) = k (cid:88) i =1 R ( n − a i − p (cid:88) j =1 R ( n − b ij )) , (1.1)where a i are integers, b ij , k , and p are natural numbers, and with some specified initialconditions. We call k and p the “arity” and “order,” respectively, of the recursion, andrefer to a recursion with arity k and order p as k -ary order p . Sometimes we refer to arecursion of the form (1 .
1) as a generalized Conolly-Hofstadter (CH) recursion, for reasonswhich will become clear below.
Date : November 3, 2018.2000
Mathematics Subject Classification.
Primary 11B37, 05C05; Secondary 05A15, 05A19.
Key words and phrases.
Nested recursion, meta-Fibonacci sequence, ( α, β )-Conolly sequence, simultane-ous parameter, slowly growing (or slow) sequence, frequency function, tree superposition.Mustazee Rahman’s research was supported by a NSERC CGS grant. Vitaly Kuznetsov was partiallysupported by an Ontario Graduate Scholarship and a NSERC PGS grant. a r X i v : . [ m a t h . C O ] J a n A. ISGUR, V. KUZNETSOV, M. RAHMAN, AND S. TANNY
A solution to (1 . R ( n ) or R to refer both to therecursion and its solution, if one exists .We say that a solution sequence is slowly growing or slow if it has the property thatsuccessive differences are either 0 or 1 and it tends to infinity. Any slowly growing sequence A ( n ) can be described by its frequency sequence φ A ( v ), which counts the number of timesthat v > A ( n ).It is evident that the nesting structure of (1 .
1) makes it impossible to apply the usual tech-niques used for solving (ordinary) difference equations, such as characteristic polynomialsand generating functions. Further, except in the simplest cases, there is no explicit orclosed form for the solution.As in [1, 3, 9, 10, 11, 12], we solve the recursion using our “tree-based” methodology.By this we mean that we show the existence of an infinite sequence that satisfies therecursion together with its initial conditions, where the n th term of the solution sequencehas a counting interpretation in terms of a labelled infinite tree. In this combinatorialinterpretation, we have an infinite tree, labelled with integers in preorder, and the solutionsequence to the nested recursion counts labels (or some analogue) on the leaves of this tree.It follows that this solution method will naturally identify a slow solution, which is why werestrict ourselves to such solutions in this paper. A fundamental contribution of [10, 11] has been to locate what we call here “simultaneousparameters.” These are parameters that both appear in the recursion and that also corre-spond to structural properties of the infinite tree used to derive and interpret its solution.For example, in [10], it is shown that the parameters s ≥ j ≥ C ( n ) = C ( n − C ( n − C ( n − − C ( n − , C (1) = 1 , C (2) = 2 (1.2)and the original H recursion (see [1]) H ( n ) = H ( n − H ( n − H ( n − − H ( n − , H (1) = H (2) = 1 , H (3) = 2 (1.3)to create the more general recursion families R s,j ( n ) = R s,j ( n − s − R s,j ( n − j )) + R s,j ( n − s − j − R s,j ( n − j )) (1.4)and H s,j ( n ) = H s,j ( n − s − H s,j ( n − j )) + H s,j ( n − s − j − H s,j ( n − j )) . (1.5) For example, we use Q ( n ) or Q to refer to Hofstadter’s nested recursion, which is defined in [4] by Q (1) = Q (2) = 1 and Q ( n ) = Q ( n − Q ( n − Q ( n − Q ( n − n > Q is a famous example whereit is not known whether or not a solution exists, although the first billion Q recursion values have beencomputed. In some cases (see [3, 11] for examples) one can derive generating functions, difference sequences, andfrequency sequences for a solution to the nested recursion, but only as a result of prior analysis of the natureof the solution sequence. See [7] where the tree-based methodology is modified to derive a combinatorial interpretation for asolution sequence with successive differences that are either 0 or d >
1. Also, in [8] some nested recursionswith slow solutions are studied that do not have a combinatorial interpretation. In these earlier papers we referred to these parameters as “shift” parameters. Our new terminologyemphasizes the greater generality of these parameters and the dual role they play in both the nestedrecursion and its corresponding infinite tree.
ESTED RECURSIONS, SIMULTANEOUS PARAMETERS AND TREE SUPERPOSITIONS 3
Further, and most importantly, it is shown how to alter the labelling of the infinite binarytrees corresponding to the solution sequence to C ( n ) and H ( n ) respectively, to create newlabelled infinite binary trees that correspond to the solution sequences for the more generalrecursion families including the additional parameters s and j . For example, to derive thesolution for (1.4), s labels are inserted in the previously empty nodes along the upper spineof the infinite binary tree, and j labels are inserted in each node rather than 1 label pernode (see [10] for a detailed explanation). The other simultaneous parameters k and p in(1.1) are discussed in [12] and [3].Identifying simultaneous parameters has proven to be a very powerful way of expandingthe range of nested recursions that we can solve. For this reason there is significant interestin finding more such parameters. Once one has been found, the tree methodology offers aneffective way to prove how the simultaneous parameter affects the solution to the nestedrecursion.The search for new families of nested recursions that can be defined by identifying simul-taneous parameters is the starting point for this paper. In Section 2 we introduce the newsimultaneous parameter m into (1 .
4) and use the tree methodology to solve the resultingorder 1 recursion, namely, R s,j,m ( n ) = R s,j,m ( n − s − R s,j,m ( n − j )) + R s,j,m ( n − s − j − m − R s,j,m ( n − j − m )) (1.6)with s a nonnegative integer, j a natural number, m an integer with 0 ≤ m ≤ j , and withappropriate initial conditions.Observe that when m = 0 (1 .
6) is identical to (1 .
4) while when m = j we have (1 . .
6) contains the above two previously known butseemingly unrelated recursion families as special cases, and also introduces all of the inter-mediate families of recursions lying “between” these two previously unconnected recursionfamilies. Thus, by solving (1.6) we are able to unify and extend the results in [10] in animportant way.In view of this beautiful and unexpected result, it is natural to ask if it is possible to extendother known results about families of nested recursions by combining the simultaneousparameters k, s, j, m and p in interesting ways. For example, recall that in [3] the so-called( α, β )-Conolly recursion of order p is defined as R ( n ) = R ( n − p (cid:88) i =1 R ( n − i + 1)) + R ( n − α − β − p (cid:88) i =1 R ( n − α − β − i + 1)) (1.7)with α even, β ≥ α + β ≥ p = α/ β . With appropriate initial conditions thisrecursion has a slow, Conolly-like solution sequence; that is, its frequency sequence is ofthe form α + βφ C ( m ), where C is the Conolly sequence (1 . H sequence (1 .
3) is the constant sequence 2, it followsthat the frequency sequence for the solution to the order p nested recursion (1.7) can bewritten as a linear combination of the frequency sequences to the two simple order 1 nestedrecurrences H ( n ) and C ( n ) defined above. In that sense we can view (1.7) as an order p extension of these two latter order 1 recursions.In Section 3 we show how to introduce simultaneous parameters s , j and m into (1 .
7) in anatural way. Subsequently, we use tree-based solution methods to solve the resulting nestedrecursion. We identify some interesting analogies between the solution to the extended order
A. ISGUR, V. KUZNETSOV, M. RAHMAN, AND S. TANNY p recursion and the order 1 recursion (1 .
6) that also contains these same simultaneousparameters.As it turns out, however, the solution to the more general order p nested recursion defined inSection 3 is not entirely analogous to that for the original ( α, β )-Conolly order 1 recursion.In particular, its frequency sequence fails to have the elegant property that it is a linearcombination of the frequency sequences of the solutions for (1 .
4) and (1 . s, j extensions to the original C and H recursions.We address this issue in Section 4, where we enhance the tree-based methodology via thenotion of tree superposition to derive a different 2-ary, order p nested recursion whosesolution does have the desired property that its frequency sequence is a linear combinationof the frequency sequences of the solutions for (1 .
4) and (1 . p recursion through an understandingof the structure of the labelled infinite tree that would be required to provide the desiredsolution property. This approach is a sort of “reverse engineering” of the analytical processwe have followed to this point, where we have used the tree methodology only to solve agiven nested recursion. Once the new recursion is identified in this way, we apply the treemethodology to derive its solution.In Section 5 we continue our study of nested recursions via the lens of simultaneous param-eters by extending our approach to certain k -ary, order p recursion families. In particularwe introduce the simultaneous parameter k ≥ m already introduced in Section 4, to yield a new k -aryorder p family containing m . We show that for appropriate choices of m and p the solutionof this k -ary recursion has frequency sequence γk + δφ C k , where C k is the solution of the k -ary Conolly recursion studied in [12].We conclude in Section 6 with some open questions and comments about future directionsfor this work.2. Unifying Two Seemingly Unrelated Recursion Families
This section concerns the process of finding and proving a combinatorial interpretation forthe recursion (1 . .
4) and (1 . m .Fix s ≥ j ≥
1, and m with 0 ≤ m ≤ j . For m < m > j , the recursions (1 .
6) seem toalways be undefined; we will see some heuristic justification for this in the combinatorialinterpretation to come. In general, for fixed s, j, m and where there is no confusion, weomit the subscripts and just write R ( n ).Define T = T s,j,m to be the following tree. First, draw a skeleton of an infinite binarytree (see Figure 2.1). We call the nodes on the extreme left except for the very first nodeon the bottom left supernodes (see the bold boxes in the diagram). All the nodes on thebottom level (including the bottom leftmost node) are called leaves , while all other nodesare regular nodes . By the left (right) subtree of a node X we mean the left (right) child of X ESTED RECURSIONS, SIMULTANEOUS PARAMETERS AND TREE SUPERPOSITIONS 5 together with all descendants of that child. The leaves that are left (right) children of theirparent are called left (right) leaves. Nodes on the second level of T are called penultimate nodes and are the parents of the leaves. Finally, subdivide each leaf into j cells. Figure 2.1.
Skeleton of an infinite binary tree with j = 3 cells in each leaf.For each n ≥ T ( n ) denote the infinite tree T with n labels, where these labels areinserted in the nodes of T in preorder as follows: Insert s labels into each supernode, j − m labels into each regular node, and j + m labels into each leaf, placing 1 label in every leafcell but the last, and 1 + m labels into the last cell of each leaf. Continue in this way untilwe have placed n labels in total in preorder. See Figure 2.2 for our running example in thissection, the tree T , , (31), with s = 1 , j = 3 , m = 1, and n = 31. Figure 2.2.
The tree T , , (31), corresponding to the value of R , , (31) =17 in the solution sequence to the recursion (1 . C T ( n ) to be the number of nonempty cells (that is,cells with at least one label) in the leaves of T ( n ). In the running example, C T (16) = 9.We say that a recursion R with corresponding tree T has initial conditions that follow thetree up to t if R ( n ) = C T ( n ) for 1 ≤ n ≤ t . For example, for the recursion R , , ( n ) above,the initial conditions 1,2,3,3,3,4,5,6,6 follow the tree T , , ( n ) up to t = 9, which coincideswith the last label in the second leaf.The key result in this section is that the leaf cell counting function satisfies (1 .
6) withsufficiently many initial conditions that follow the tree. More precisely:
A. ISGUR, V. KUZNETSOV, M. RAHMAN, AND S. TANNY
Theorem 2.1.
Suppose that the recursion (1 . has initial conditions R ( n ) = C T ( n ) for n ≤ j + 3 m + 2 s , that is, the initial conditions follow the tree until the right leaf of thesecond penultimate level node. Then for all n , R ( n ) = C T ( n ) , that is, C T ( n ) solves therecursion. Notice that this combinatorial interpretation for the solution of (1 .
6) suggests why wecannot allow m < m > j in recursion (1 . m < m = −
1) or a negative number of labels ( m < − m > j would lead to negative numbers of labels in the regular nodes.By definition, C T ( n ) is the sum of the number of nonempty cells in the left leaves of T ( n )and the number of nonempty cells in the right leaves of T ( n ). Observe that the number ofnonempty cells in the right leaves of T ( n ) equals the number of nonempty cells in the leftleaves of T ( n − j − m ): to see this, note that if l is a label on a left leaf other than the firstleaf in T ( n − j − m ), then l + m + j is a label on a right leaf in T ( n ). Conversely, if r isa label on a right leaf other than the second leaf of T ( n ), then r − m − j is a label on aleft leaf of T ( n − j − m ). Since the initial conditions require that we are beyond the firsttwo leaf nodes, which are full and thus have the same number of labels, the fact that thiscorrespondence doesn’t hold for the first leaf pair doesn’t matter. Thus it follows that thereis a one-to-one correspondence between nonempty cells in the left leaves of T ( n − j − m )and nonempty cells in the right leaves of T ( n ).Therefore, to prove Theorem 2.1 it is enough to show that for n ≥ j + 2 m + 2 s , the numberof nonempty cells in the left leaves of T ( n ) is C T ( n − s − C T ( n − j )). We can then applythis result to the tree T ( n − j − m ) and use the preceding correspondence to deduce that C T ( n − s − j − m − C T ( n − j − m )) counts the number of nonempty cells in the rightleaves of T ( n ), provided that n − j − m ≥ j + 2 m + 2 s , that is, n ≥ j + 3 m + 2 s . Addingthese cell counts together and combining with the given initial conditions yields the desiredsolution to the recursion.In order to prove that for n ≥ j + 2 m + 2 s the number of nonempty cells in the left leavesof T ( n ) is C T ( n − s − C T ( n − j )) we define the pruning operation for T ( n ). Note that when n ≥ j + 2 m + 2 s the left leaf of the second penultimate level node necessarily is full.See Figures 2 . , . , . , and 2 . initial correction step: : Delete the s labels in the first supernode (the leftmostpenultimate node). Then take the j − m largest labels n − ( j − m ) + 1 , . . . , n − , n and move them into the now-empty first supernode. deletion step: : For every cell in T ( n ), if it has at least one label less than or equalto n − j , delete the first label from that cell. This will delete precisely C T ( n − j )labels, by definition of the leaf cell counting function C T . At the end of the deletionstep, we have deleted s + C T ( n − j ) labels in total. lifting step: : In all nonempty leaves of our tree (except possibly the last), we willhave m labels in the last cell (the last nonempty leaf might have less than m labelsin the last cell); this is because we deleted one label from each cell, but the last cell This doesn’t hold for the first leaf since the labeling of the first supernode intervenes when s > ESTED RECURSIONS, SIMULTANEOUS PARAMETERS AND TREE SUPERPOSITIONS 7
Figure 2.3.
In the initial correction step for T , , (31), the label 5 (in red)is removed from the first supernode and labels 30 and 31 are moved into thefirst supernode. Figure 2.4.
In the deletion step for T , , (31), one label (indicated in red)is deleted from every cell that has a label less than or equal to 28.of each leaf had 1 + m labels, and will thus have m left. Move all remaining leaflabels into the parent of the leaf node they started in.At this point of the pruning operation, all penultimate nodes (including the firstsupernode) other than (perhaps) the last nonempty penultimate node have exactly j + m labels (the last nonempty penultimate node may have fewer labels). The firstsupernode had all of its original s labels removed, j − m labels added from the end,and m labels added from each of its two children. All the other penultimate nodes(except possibly the last nonempty one) started with j − m labels and gained m from each of its two children.As the last part of the lifting step, convert all the penultimate level nodes into leavesby dividing them into j cells, with one label in each cell but the last, and 1 + m labels in the last cell of each leaf. The last nonempty penultimate node may nothave the j + m labels needed to fill all of its cells, in which case simply fill as manycells as the number of labels on it allows. Finally, delete the bottom level nodesof the current tree (the original leaves), all of which are now empty. This processresults in a new tree with the same skeleton as T ( n ). A. ISGUR, V. KUZNETSOV, M. RAHMAN, AND S. TANNY
Figure 2.5.
In the lifting step for T , , (31), all the remaining leaf labelsmove up into their parent penultimate nodes, respectively. The now emptyleaves are deleted and the penultimate nodes become the new leaves, withcell divisions introduced and labels inserted according to the rules. Figure 2.6.
In the relabelling step for T , , (31), the 15 remaining labelsare replaced with 1 through 15, showing that T ∗ , , (31) = T , , (15) . relabelling step: : Renumber the labels of the new tree in preorder (so that 1 is thefirst label, 2 the second, and so on). It is readily seen that the new tree so labelled,which we denote by T ∗ ( n ), is identical to T ( n − s − C T ( n − j )), since it has thesame skeleton structure and has n − s − C T ( n − j ) labels.For convenience, we define C T ∗ ( n ) to be the number of nonempty leaf cells of T ∗ ( n ), thatis, we define C T ∗ ( n ) = C T ( n − s − C T ( n − j )).We can think of every node in T ∗ ( n ) as being part of T ( n ), that is, we can identify eachnode in T ∗ ( n ) with the node that it was in T ( n ). For example, we identify the penultimatenode with labels 24 and 25 in Figure 2.2 with the leaf node containing labels 13, 14 and 15in Figure 2 . T ( n ) with nonempty cells in the leaves of the pruned tree T ∗ ( n ). In this waywe prove that C T ( n − s − C T ( n − j )) counts the number of nonempty cells in the left leavesof T ( n ), and hence completes the proof of Theorem 2.1. ESTED RECURSIONS, SIMULTANEOUS PARAMETERS AND TREE SUPERPOSITIONS 9
Lemma 2.2.
Suppose that P is a penultimate node of T ( n ) (and thus a leaf node in T ∗ ( n ) ),and n ≥ j + 2 m + 2 s . Then in T ∗ ( n ) , the number of nonempty cells of P is equal to thenumber of nonempty cells of its left child in T ( n ) .Proof. We begin with the case where P is the first supernode. Note that this is the part ofthe proof where we rely on the assumption that n ≥ j + 2 m + 2 s : the label 4 j + 2 m + 2 s is the last label on the third leaf of T ( n ), which ensures that the left leaf child of P is fullin T ( n ). Also, we have n − j ≥ j + 2 m + 2 s ≥ j + 2 m + s which is the last label onthe right child of P , so all of the cells of the two children of P will have 1 label removedduring the deletion step. Furthermore, note that the second penultimate node (the onejust to the right of P ) has its full complement of j − m labels in T ( n ) since its last labelis 3 j + m + 2 s ≤ j + 2 m + 2 s . Thus, there are at least j − m labels on the tree after thechildren of P , so the j − m labels moved from the end of the tree into P during the initialcorrection step will not come from the children of P . This means that there remain in place2 m labels on the children of P after the deletion step in the pruning process. Therefore inthe pruning process the node P will receive the j − m labels from the end of the tree, plus2 m labels from its children, making P full in the pruned tree T ∗ ( n ) just like its left childis full in T ( n ). This establishes the required result in this special case.We now assume that P is not the first supernode. We require several cases: Case 1:
The label n is on a node before (with respect to preorder) the left child of P in T ( n ), that is, the left child of P has no labels in T ( n ). We want to show that P has nolabels in T ∗ ( n ). In this case, observe that P has at most j − m labels before the pruningoperation, and during the initial correction step, the final j − m labels in T ( n ) are movedinto the first supernode. This means that any labels in P in T ( n ) are removed during thepruning, so P will be empty in T ∗ ( n ). Case 2:
The label n is one of the first j labels on the left child of P in T ( n ). Thus theleft child of P has d total labels in T ( n ), where 0 < d ≤ j (observe that there are no labelsin the right leaf child of P ). This means that between 1 and j of the cells of the left childof P have one label each. We want to show that P will have d total labels in T ∗ ( n ). Sinceall of the labels on the left child of P are larger than n − j , none of them will be deletedduring the deletion step. During the initial correction step, we will move the last j − m labels into the first supernode, and then during the lifting step, the labels remaining onthe left child of P (if any) will be moved up into P . Since P had j − m labels before thepruning operation, and its children had d labels in total, and (the largest) j − m labels wereremoved during the initial correction step, there will be d labels left on P after the liftingstep of the pruning operation, as desired. Case 3:
The last remaining case is when the j th label on the left child of P in T ( n ) issmaller than n . That is, the left child of P in T ( n ) has all j of its cells nonempty and n isnot the first entry in the last cell. We will show that in T ∗ ( n ), the node P also has all ofits cells nonempty. To do so we make use of the result we have just proved for Case 2 with d = j .Let x j be the j th label on the left child of P in T ( n ). By assumption, we have x j < n . ByCase 2, we know that T ∗ ( x j ) has all of the cells of P nonempty. As discussed previously, T ∗ ( x j ) = T ( x j − s − C T ( x j − j )) and T ∗ ( n ) = T ( n − s − C T ( n − j )). If we can prove that x j − s − C T ( x j − j ) ≤ n − s − C T ( n − j ), then we will have shown that T ∗ ( n ) has at least as many labels as T ∗ ( x j ). Thus P will have at least as many labels in T ∗ ( n ) as it does in T ∗ ( x j ), meaning P will have no nonempty cells in T ∗ ( n ).To show that x j − s − C T ( x j − j ) ≤ n − s − C T ( n − j ) we will prove that the function f ( n ) = n − s − C T ( n − j ) is monotone nondecreasing. Note that f ( n + 1) = n + 1 − s − C T ( n + 1 − j ).Since C T counts nonempty leaf cells, either C T ( n + 1 − j ) = C T ( n − j ) (if n + 1 − j is notthe first label of a leaf cell), or C T ( n + 1 − j ) = C T ( n − j ) + 1 (if n + 1 − j is the first labelof a leaf cell). In the former case, we have n + 1 − s − C T ( n + 1 − j ) = n − s − C T ( n − j ) + 1,and in the latter case, we have n + 1 − s − C T ( n + 1 − j ) = n − s − C T ( n − j ). Either way,this establishes that f ( n + 1) ≥ f ( n ), proving the desired inequality. This completes theproof of Case 3 and the lemma, so Theorem 2.1 is established. (cid:3) As we pointed out earlier, the introduction of the simultaneous parameter m in (1 .
4) definesa new recursion family that unifies the results about (1 .
4) and (1 .
5) proved in [10]: when m = 0 (1 .
6) is identical to (1 .
4) while when m = j we have (1 . m plays a key role in the structure of the resulting treeused in solving (1 . m provides thederivation of the solution to (1 .
6) is even easier than the solutions to (1 .
4) and (1 .
5) in[10].We conclude this section by deriving the frequency sequence of the solution that we havejust determined to (1 . Theorem 2.3.
The solution C T to the nested recursion (1 . has frequency sequence φ C T ( v ) = (cid:26) if j (cid:45) v ( j − m ) ν ( v/j ) + m + 1 + s [ vj is a power of 2 ] otherwisewhere ν ( x ) is the 2-adic valuation of x and [ E ] is the indicator function of the set E .Proof. We begin by counting the number of regular nodes between the h th and ( h + 1) st leaves in T . We consider two cases.Suppose h is a power of 2, say h = 2 b . Then observe that the h th leaf comes right beforethe ( b + 1) st supernode. By the tree construction this supernode is the root of a completebinary subtree of height ( b + 1), so it is followed in turn in preorder by b regular nodes (the“leftmost” nodes of the complete binary subtree rooted at the ( b + 1) st supernode), andthen by the ( h + 1) st leaf. Thus, the number of regular nodes between h th and ( h + 1) st leaves is ν ( h ).Assume that h is not a power of 2, say h = a b for some odd integer a >
1. Consider thenode N in T such that the h th leaf is the (2 b +1 − st node in preorder following N . Forexample, in Figure 2.3, for h = 6 the node N contains the labels 22 and 23. Recall that(2 b +1 −
1) is the number of nodes in a complete rooted binary tree of height b . It followsthat the h th leaf is the rightmost node in the left subtree of N and the ( h + 1) st leaf is theleftmost node in the right subtree of N and there are ν ( h ) = b regular nodes in preorderbetween h th and ( h + 1) st leaves. An alternate approach to counting the number of nodes between the h th and ( h + 1) st leaves in T is asfollows (see [5]): in [11] it is shown that the Conolly sequence is the label count on a binary tree with empty ESTED RECURSIONS, SIMULTANEOUS PARAMETERS AND TREE SUPERPOSITIONS 11
Now we proceed with the proof of the theorem. If v is not a multiple of j , then the v th nonempty cell is not the last cell on a leaf. Since cells other than the last cell on a leafhave one label and are followed by another cell, we have φ C T ( v ) = 1 in this case. If v isa multiple of j , then the v th nonempty cell is the last cell on the ( v/j ) th leaf. Thus, thevalue v is assumed on all 1 + m labels in that cell, plus the ν ( v/j ) regular nodes followingit (each with j − m labels), plus another s labels on a supernode if v/j is a power of 2 (andhence the ( v/j ) th leaf comes right before a supernode). This establishes the stated valuesfor the frequency sequence for the solution to recursion (1 . (cid:3) The above argument is a general technique for deriving the frequency sequence for a solutionrelated to labelled trees of this type. This proof technique does not depend on the labellingscheme, but only on the skeleton of the tree, so it can be adapted to other situations suchas the one we will discuss in the following section (see Section 3.4).3.
Simultaneous parameters in higher order nested recursions
In this section we apply our tree-based approach to solve higher order recursions containingsimultaneous parameters. Our starting point is the 2-ary, order p recursion (1.7) that firstappears in [3]. As is discussed there, this recursion can be viewed as an order p extensionof both the order 1 Conolly recursion (1 .
2) (take α = 0 and β = 1) and the H recursion(1 .
3) (take α = 2 and β = 0). Here we show how to construct and solve a natural extensionto this recursion that contains simultaneous parameters s, j and m that each play a roleanalogous to the one they played in (1 . .
6) and (1.7), respectively.In a manner formally similar to the approach taken in [10] for the simultaneous parameters s and j and in the preceding section for the parameter m , we introduce what we will seeare simultaneous parameters s, j and ¯ m into (1.7) in what appears to be a very naturalway, namely: R ( n ) = R ( n − s − p (cid:88) i =1 R ( n − (2 i − j ))+ R ( n − s − ( α + β ) j − ¯ m − p (cid:88) i =1 R ( n − ( α + β ) j − ¯ m − (2 i − j )) , (3.1)with, as in (1.7), α even, β ≥ α + β ≥ p = α/ β ≥ s ≥ j ≥ m that we discuss below.Some modest experimentation with particular values for the parameters in (3 .
1) is sufficientto demonstrate that this parametrization in terms of ( α, β, ¯ m ) is not one to one, that is,different choices of parameters ( α, β, ¯ m ) can generate the same recursion and through itthe same tree and solution sequence. For example, s = 0 , j = 3 , α = 2 , β = 1 , ¯ m = 1 and s = 0 , j = 3 , α = − , β = 3 , ¯ m = 7 generate the same order p = 2 recursion and associatedtree. supernodes. From this it follows that the frequency with which h occurs in the Conolly sequence is preciselythe number of regular nodes between the h th and ( h + 1) st leaves, plus one. Further, it is also shown in [11]that the frequency of the Conolly sequence is ν ( v ) + 1. Therefore, the number of regular nodes betweenthe h th and ( h + 1) st leaves of a binary tree is ν ( h ). This duplication occurs because α, β and ¯ m always appear together in (3 . m = ( α + β − j + ¯ m . Asa result, (3 .
1) becomes R ( n ) = R ( n − s − p (cid:88) i =1 R ( n − (2 i − j ))+ R ( n − s − j − m − p (cid:88) i =1 R ( n − j − m − (2 i − j )) . (3.2)Note that (3.2) reduces to (1.6) when p = 1, so the results of this section generalize thoseof Section 2.Computational evidence to date with (3.2) suggests that whenever this recursion generatesan infinite solution sequence (for some set of initial conditions) then 0 ≤ m ≤ (2 p − j .For this reason we restrict m to this range. Construction of the tree and statement of the main theorem.
The skeletonof the tree T = T s,j,m,p that we use here is the same infinite binary tree as in Section 2, andwe adopt the same terminology and a similar labelling scheme. For n ≥ T ( n ) denote T with the first n labels inserted in preorder according to the following rules: the supernodesof T contain s labels each, every leaf node of T contains j cells with 1 label in each ofthe first j − m labels in the last cell, and all other regular nodes contain x := (2 p − j − m labels each. Continue in this way until we have placed n labels. Figure3.1 shows T (63) for our running example in this section with s = 0 , j = 3 and m = p = 2.Note that in this case x = 7. Figure 3.1.
The labeled tree T (63) for ( s, j, m, p ) = (0 , , , C T (63) = 21.Define the leaf cell counting function C T ( n ) to be the number of nonempty cells in T ( n ). Themain result of this section is that C T ( n ) satisfies (3.2) with appropriate initial conditionsthat follow the tree T . Theorem 3.1.
Suppose that the recursion (3 . has initial conditions R ( n ) = C T ( n ) for n ≤ j + m ) + x + 2 s , that is, the initial conditions follow the tree until the right leaf ofthe second penultimate level node. Then for all n , R ( n ) = C T ( n ) . Before we prove Theorem 3.1 we examine the special endpoint cases m = 0 and m =(2 p − j associated with the range of m . When m = 0 then (3.2) is an order p analogue In fact, this is the range of m for which our tree-based proof below holds. This suggests a heuristicreason for our inability to locate any solutions for the recursion with m outside this range. ESTED RECURSIONS, SIMULTANEOUS PARAMETERS AND TREE SUPERPOSITIONS 13 of (1.4), while if m = (2 p − j then (3.2) is an order p analogue of (1.5). In particular,when s = 0 and j = 1 the first recursion with m = 0 is an order p analogue to the Conollyrecursion (1 .
2) while the second recursion with m = (2 p − j is an order p analogue tothe H recursion (1 .
3) with solution (cid:100) n/ (cid:101) (in both cases, the required initial conditions aregenerated from the associated tree).Even further, as in the case of (1 . m = (2 p − j , s = 0 and j = 1 is a ceiling function, in this case (cid:100) n/ p (cid:101) . Whilea more general version of this result appears in [3], we provide here the first tree-basedderivation of a ceiling function solution for an order p nested recursion. Strategy of the proof : the pruning operation.
We will follow a similar approachto that adopted in Section 2. To prove Theorem 3.1, first we denote by C T,L ( n ) and C T,R ( n )the number of nonempty cells in T ( n ) that are on the left and right leaves, respectively.By definition C T ( n ) = C T,L ( n ) + C T,R ( n ) . Since there are j + m labels in total in a full leaf, there is a natural bijection between thenonempty cells of T ( n ) that are on right leaves and the nonempty cells of T ( n − j − m )that are on left leaves. Thus, C T,R ( n ) = C T,L ( n − j − m ) . Hence, to prove Theorem 3.1 it suffices to show the following result:
Lemma 3.2.
For n > j + m ) + x + 2 s , we have that C T,L ( n ) = C T ( n − s − p (cid:88) i =1 C T ( n − (2 i − j )) . As in Section 2, our proof relies on a pruning technique on T ( n ) that we now describe. Onceagain we use T ∗ ( n ) to denote the pruned tree that results from applying this technique. SeeFigures 3.2, 3.3, 3.4, 3.5 and 3.6 where we illustrate the pruning operation on our runningexample. initial correction step: Remove the s labels from the first supernode of T ( n ) andreplace them with x labels. We do not identify these new labels until we reach therelabelling step below. deletion step: For each i = 1 , . . . , p , consider the tree T ( n − (2 i − j ) as a subtreeof T ( n ). For every nonempty cell in T ( n − (2 i − j ) remove a label from thecorresponding cell in T ( n ). If that cell in T ( n ) has already been emptied by anearlier application of this process then remove a label from the last cell of thecorresponding leaf containing the empty cell (so long as a label is available). Ifboth the cell and the last cell of the leaf containing said cell already have beenemptied by this process, then remove a label from the corresponding parent nodeat the penultimate level. See [3], Theorem 5.2, where a very different methodology is used to characterize all recursions of theform R ( n ) = R ( n − s − (cid:80) pi =1 R ( n − a i )) + R ( n − t − (cid:80) pi =1 R ( n − b i )) with solution (cid:100) n/ p (cid:101) . We believe thatour proof will work for any of the recursions stated in Theorem 5.2 of [3] provided that s = 0 , a i < p, t = 2 p and b i = a i + 2 p . Figure 3.2.
Initial correction step when pruning T (63) where ( s, j, m, p ) = (0 , , , Figure 3.3.
Deletion step when pruning T (63) where ( s, j, m, p ) = (0 , , , Figure 3.4.
Lifting step when pruning T (63) where ( s, j, m, p ) = (0 , , , j cells and there are p subtrees so the maximumnumber of labels that can be attempted to be deleted from a pair of sibling leavesis 2 pj . But the total number of labels within such a pair of sibling leaves and theirparent is 2( j + m ) + x = m + (2 p + 1) j > pj so there are enough labels. ESTED RECURSIONS, SIMULTANEOUS PARAMETERS AND TREE SUPERPOSITIONS 15
Figure 3.5.
End correction when pruning T (63) where ( s, j, m, p ) = (0 , , , Figure 3.6.
Relabelling step when pruning T (63) where ( s, j, m, p ) = (0 , , , lifting step: Lift any remaining labels in a leaf into the corresponding parent at thepenultimate level. Note that any leaf with the first j of its labels all less than n − (2 p − j will be left with m − ( p − j labels to lift so long as m > ( p − j ,and 0 labels otherwise. end correction step: Remove the x largest labels in preorder of T ( n ) that remainafter the lifting step. Note that this removal fully offsets the insertion of x labelsin the first supernode during the initial correction step. relabelling step: Remove the leaves of T ( n ) (which are now empty) and relabel thenew tree in preorder. The former penultimate level nodes are now the leaves of thenew tree. Make j cells for every leaf and assign 1 label in each of the first j − j + m labelsthen fill in its cells as just described, recognizing that some cells may remain emptyor, in the case of the last cell, only partially filled.In pruning T ( n ) we have removed s + (cid:80) pi =1 C T ( n − (2 i − j ) labels. Thus T ∗ ( n ) is atree with the same skeleton structure as T ( n ) and with n − s − (cid:80) pi =1 C T ( n − (2 i − j )labels.The strategy behind our proof of Lemma 3.2 is to show first that the pruning operationon T ( n ) creates a tree T ∗ ( n ) that conforms to the labelling rules that we described above. Together with what we have just observed about the skeleton of the tree T ∗ ( n ), this willimply that T ∗ ( n ) is identical to T ( n − s − (cid:80) pi =1 C T ( n − (2 i − j )), so that the number ofnonempty cells in T ∗ ( n ) is C T ( n − s − (cid:80) pi =1 C T ( n − (2 i − j )). Then we will demonstratea bijection between the nonempty cells of T ∗ ( n ) and the nonempty cells of T ( n ) that areon left leaves. Together these two assertions imply Lemma 3.2. Thus, to prove Lemma 3.2we will establish the following two lemmas: Lemma 3.3.
The pruned tree T ∗ ( n ) is identical to T ( n − s − (cid:80) pi =1 C T ( n − (2 i − j ) . Lemma 3.4.
Let P be a penultimate node of T ( n ) . Then P becomes a leaf node of T ∗ ( n ) and the number of nonempty cells of P in T ∗ ( n ) is equal to the number of nonempty cellsof the left child of P in T ( n ) . Proof of Lemmas 3.3 and 3.4.
We prove both lemmas simultaneously. We beginwith a preliminary discussion of each.As noted above, to prove Lemma 3.3 we need only show that the labelling of T ∗ ( n ) is inaccordance with the rules that we have laid out above. That is, except for the last nonemptynode in T ∗ ( n ), all super nodes of T ∗ ( n ) have s labels, the leaves have j + m labels and theregular nodes have x labels each. Finally, the last nonempty node of T ∗ ( n ) cannot containmore than s, j + m or x labels respectively, depending on its type.As in Section 2, we can think of every node in T ∗ ( n ) as being part of T ( n ). By the designof the pruning operation on T ( n ), all nodes in T ∗ ( n ), except for the leaves of T ∗ ( n ) (whichare the former penultimate nodes of T ( n )) and the last nonempty node (which may or maynot be a leaf of T ∗ ( n )), contain the same number of labels as they do in T ( n ). So to proveLemma 3.3 we need only focus on the leaves of T ∗ ( n ) and its last nonempty node. First, wemake a simple yet important observation that is used several times in the argument. Lemma 3.5.
Let P be a penultimate level node of T ( n ) with left child L and right child R .Suppose that T ( n ) contains at least one completely filled regular node following R . Thenas a leaf of T ∗ ( n ) , the node P contains j + m labels.Proof. The node R has j + m labels on itself in T ( n ). By assumption, the first regular nodein T ( n ) that follows R , say Q , is full with x labels on it (note that it cannot be a leaf).Therefore, there are at least j + m + x = 2 pj labels on or after R in T ( n ). Consequently, R has at least j labels (and thus all j nonempty cells) in each of the subtrees T ( n − j ) , . . . , T ( n − (2 p − j ), and thus so does L . We conclude that the total number of labelsremoved from P , L , and R during the deletion step is the maximum amount, namely 2 pj .Note that since Q has x labels, and comes after R , none of the labels removed for the endcorrection step will come from P , L , or R . Thus, after the lifting step moves all remaininglabels from L and R to P , and pruning is completed, P will have x + 2( j + m ) − pj = j + m labels. (cid:3) Going back to the discussion about Lemma 3.3, consider first the last nonempty node of T ∗ ( n ). Suppose that it is not a leaf of T ∗ ( n ). Then in T ( n ) this node is neither a leaf nor apenultimate node. Therefore the pruning operation on T ( n ) doesn’t add any labels to thisnode (the end correction step of the pruning may remove some labels). After pruning, thislast nonempty node in T ∗ ( n ) has at most the same number of labels that it has in T ( n ),which is what we require. ESTED RECURSIONS, SIMULTANEOUS PARAMETERS AND TREE SUPERPOSITIONS 17
Next we turn to the leaves of T ∗ ( n ). First consider a leaf P of T ∗ ( n ) that is not one ofthe last two nonempty penultimate nodes in T ( n ). Then, in T ( n ), P is a penultimate levelnode with its full complement of x labels, both its children must contain a full complementof j + m labels, and there must be a full penultimate node with two full leaf children thatfollow P . We can thus apply Lemma 3.5 to conclude that the number of labels in P in T ∗ ( n ) is j + m , as required.Finally we consider the two leaves P a and P b of T ∗ ( n ) that are the last two nonemptypenultimate nodes in T ( n ), where P b is to the right of P a . To establish the required resultfor these two nodes we have to show: (1) if P b is not the last nonempty node of T ∗ ( n ) thenboth P a and P b contain the full complement of j + m labels in T ∗ ( n ); (2) if P b is the lastnonempty node in T ∗ ( n ), then P a contains j + m labels and P b contains at most j + m labels in T ∗ ( n ); and (3) if P b is empty in T ∗ ( n ) then P a contains at most j + m labels in T ∗ ( n ).Consider the requirement in case (1). Here we observe that we can apply Lemma 3.5 toboth P a and P b . Indeed, we can apply it to P a since P b will be full in T ( n ). But we can alsoapply it to P b because there is a nonempty node Q in T ∗ ( n ) following P b . Q is not a leafof T ∗ ( n ) by definition of P b . So Q must be located above the penultimate level in T ( n ). Itwill also contain x labels in T ( n ), for otherwise, it would become the last nonempty nodein T ( n ) and all its labels would be deleted during the end correction step. However, thiswould make P b the last nonempty node of T ∗ ( n ). Therefore we have verified Lemma 3.3for P a and P b in case (1).To verify cases (2) and (3) it suffices to show the following. Let P be the last penultimatenode of T ( n ) that has at least one nonempty child in T ( n ). Then P has between 0 to j + m labels in T ∗ ( n ). Indeed, if P b has a nonempty child in T ( n ) (so P = P b ) then we can applyLemma 3.5 to P a in T ( n ) due to P b being full. Hence, P a contains j + m labels as a leafof T ∗ ( n ), and to establish (2) we need to show that P b contains between 0 to j + m labelsin T ∗ ( n ). On the other hand, if both children of P b are empty in T ( n ) (so P = P a ) thenduring the end correction step all labels from P b are removed. This makes P b empty in T ∗ ( n ) and to show (3) we need to worry about the number of labels in P a . Note that inboth cases (2) and (3) node P is the last non leaf node of T ( n ) that is completely filled. Sowe have to show that if T ( n ) is such that it contains a node P which is its last penultimatenode with a nonempty child and also its last filled non leaf node, then P has between 0 to j + m labels in T ∗ ( n ).Before we conclude the proof of Lemma 3.3 we examine in a similar way what must beshown to prove Lemma 3.4. The first part of the lemma has already been covered in theabove discussion, so what remains is to show that the number of nonempty cells of any leaf P in T ∗ ( n ) is equal to the number of nonempty cells of the left child of P in T ( n ).Suppose the node P is the last nonempty penultimate node of T ( n ) and that its left child isempty in T ( n ). Then as a result of the end correction step in the pruning process P will beempty in T ∗ ( n ), as required by Lemma 3.4. If P is not the last nonempty penultimate nodeof T ( n ) then it is followed by a non leaf node Q that is also nonempty in T ( n ). Also, the leftchild of P contains j + m labels in T ( n ). If Q is completely filled in T ( n ) then by Lemma3.5 the node P will contain j + m labels as a leaf of T ∗ ( n ), and hence j cells as required byLemma 3.4. Thus, we are left to consider only one case: P is the last penultimate node of T ( n ) with a nonempty left child and where the first non leaf node Q following P contains less than x labels in T ( n ) ( Q could possibly be empty). We must show that in this case P contains the same number of cells in T ∗ ( n ) as its left child does in T ( n ).Therefore, in order to complete the proofs of both Lemma 3.3 and 3.4 we must consider thefollowing. Suppose the tree T ( n ) contains a penultimate node P with the property that itis the last penultimate node with a nonempty left child in T ( n ) and that it is also the lastcompletely filled non leaf node of T ( n ). Then, we must show that as a leaf of T ∗ ( n ) thenode P contains at most j + m labels, and also that it has same number of nonempty cellsas its left child does in T ( n ). Until the end of this section let P denote such a penultimatenode, and denote by L and R the left and right child leaf of P respectively ( R may be emptyin T ( n )). Our two requirements can be expressed as upper and lower bounds on the netnumber of labels being removed from and lifted into P during the pruning operation.In our proof we will establish these bounds on a case by case basis, where the cases (andsubcases) are determined by the position of the label n in T ( n ). In doing so we will makefrequent use of the following technical lemma: Lemma 3.6.
Let Y be a nonempty leaf in T ( n ) and suppose that there are µ labels in T ( n ) that are situated at or after node Y (so the smallest label in Y is n − µ + 1 ). If µ ≤ j thenthe number of labels removed from Y during the pruning operation is . If µ > j then write µ − j = q (2 j ) + r with ≤ r < j . In this case the number of labels removed from Y andpossibly its parent in T ( n ) is jq + min { r, j } provided that q ≤ p − ; otherwise, the numberof labels removed from Y and possibly its parent in T ( n ) is pj .Proof. If µ ≤ j then Y is empty in all of the subtrees T ( n − j ) , . . . , T ( n − (2 p − j ), so thenumber of labels removed from Y during the pruning operation is 0.If µ > j and µ − j = q (2 j ) + r with q > p − Y are nonempty in theaforementioned p subtrees. In this case we have already seen earlier that the number oflabels removed from Y and possibly its parent is pj .Finally, suppose µ > j and µ − j = q (2 j ) + r with 0 ≤ r < j and q ≤ p −
1. Then the q subtrees T ( n − j ) , . . . , T ( n − (2 q − j ) contain all cells of Y as being nonempty (if q = 0then none do). The subtree T ( n − (2 q + 1) j ) contains only the first min { r, j } cells of Y as being nonempty, and for i > q each of the remaining subtrees T ( n − (2 i + 1) j ) contains Y as an empty leaf. Therefore, the net number of labels deleted from Y and its parent in T ( n ) during the pruning operation is jq + min { r, j } . (cid:3) Now we proceed with the case by case analysis promised above.Case 1. Suppose label n is situated in node L . Subcase 1a:
Label n is the l th label in L with 1 ≤ l ≤ j . Thus L is the last nonemptynode in T ( n ) and it contains l ≤ j labels. The deletion step does not affect any labels in L since L is empty in all the subtrees T ( n − j ) , . . . , T ( n − (2 p − j ). Therefore, in the liftingstep the l labels from L are inserted into P and in the end correction step the x largestlabels are removed from P . This leaves P with l labels in T ∗ ( n ), as required. Subcase 1b:
Label n is the ( j + l ) th label in L with 1 ≤ l ≤ m (if m = 0 then this caseis not needed). By using Lemma 3.6 the number of labels removed from L and its parentis qj + min { r, j } , where l = q (2 j ) + r . Thus, in the lifting step of the pruning operation ESTED RECURSIONS, SIMULTANEOUS PARAMETERS AND TREE SUPERPOSITIONS 19 l + j − qj − min { r, j } labels are lifted into P , while in the end correction step the x labelswith the largest labels are removed from P . We need to verify that j ≤ l + j − qj − min { r, j } ≤ m + j (3.3)to establish Lemmas 3.3 and 3.4. The first inequality follows from l = q (2 j ) + r ≥ qj + min { r, j } and the second follows from m ≥ l .Case 2: Suppose label n is situated in node R . To establish Lemmas 3.3 and 3.4 we mustshow that P contains at least j labels and at most m + j labels after pruning T ( n ). Subcase 2a:
Label n is the l th label in R with l ≤ j . Here no labels are removed from R during the deletion step.If l ≥ (2 p − j − m then L loses pj labels during the deletion step by Lemma 3.6. Inthis case the number of labels in P after pruning T ( n ) is m + l − ( p − j and we need toestablish that j ≤ m + l − ( p − j ≤ m + j . (3.4)Both of these are clear because l + m ≥ (2 p − j ≥ pj and l ≤ j ≤ pj .In the case l < (2 p − j − m , by Lemma 3.6 the number of labels removed from L after thedeletion step is qj + min { r, j } with m + l = q (2 j ) + r . Then we need to verify that j ≤ m + j + l − qj − min { r, j } ≤ m + j . (3.5)The first inequality follows from m + l = q (2 j ) + r ≥ qj + min { r, j } . The second one followsbecause l ≤ qj + min { r, j } if either q > r ≥ j due to l ≤ j . Otherwise, m + l = r andso l ≤ r = min { r, j } , which is the required inequality. Subcase 2b:
Label n is the ( j + l ) th label in R with 1 ≤ l < (2 p − j − m (if the rightmostterm is non positive then this case is not needed). In this case both children of P may loseless than pj labels. To account for the number of labels lost by L and R during the deletionstep, we write m + j + l = q L (2 j ) + r L and l = q R (2 j ) + r R with 0 ≤ r L , r R < j . Then byLemma 3.6, the number of labels removed from nodes L and R are q L j + min { r L , j } and q R j + min { r R , j } , respectively. Note that q R ≤ q L . To prove Lemmas 3.3 and 3.4 we needto show that j ≤ m + j + l + j − ( q L j + min { r L , j } + q R j + min { r R , j } ) ≤ m + j . (3.6)For the first inequality we use the fact that m + j + l = q L (2 j ) + r L to reduce the inequalityto min { r L , j } + min { r R , j } ≤ ( q L − q R ) j + r L . If q L > q R then the latter inequality followseasily. Otherwise, q L = q R and so l = q R (2 j ) + r R implies that m + j + r R = r L . Thus, r L ≥ j and the latter inequality becomes j + min { r R , j } ≤ r L = m + j + r R ; this is clearlytrue.Now we consider the second inequality in (3.6). Here we substitute q L j and q R j withthe equivalent values m + l + j − r L and l − r R respectively. Then after some simplification theinequality becomes j + r R + r L ≤ m + 2 min { r L , j } + 2 min { r R , j } . (3.7)If both the minimums on the right are j then we get r R + r L ≤ m + 3 j , which is true dueto r L < j and r R ≤ m + j . If the minimums are r L and r R (both less than j ) then theinequality holds unless m < j . But when m < j we have q R = 0 and q L ∈ { , } because l + j < j and m + l + 2 j < j . If q L = 0 then r L = m + l + j ≥ j , contradicting that r L < j . Thus, q L = 1 and the second inequality in (3.6) becomes r L ≥ r L or r R and the other is j then the inequality in (3.7)is trivial. Subcase 2c:
Label n is the ( j + l ) th label in R with l ≥ (2 p − j − m . By the choice of l , this case treats the situation where node L loses pj labels. By Lemma 3.6 node R loses qj + min { r, j } labels where l = q (2 j ) + r . After pruning T ( n ) the number of labels in P is m − ( p − j + l + j − qj − min { r, j } . To prove Lemma 3.4 to need to establish that j ≤ m − ( p − j + l + j − qj − min { r, j } . Usingthe assumption (2 p − j − m ≤ l and the fact that l ≤ m , we deduce that m ≥ ( p − j .Also, we have that l = q (2 j ) + r ≥ qj + min { r, j } . Together, these two observationsimply that m + l ≥ ( p − j + qj + min { r, j } . This is the desired inequality above aftersimplification.Now we verify the upper bound m − ( p − j + l + j − qj − min { r, j } ≤ m + j that is requiredfor Lemma 3.3. This is equivalent to l ≤ ( p − j + qj + min { r, j } . Using l = q (2 j ) + r ourupper bound is equivalent to l + r ≤ (2 p − j + 2 min { r, j } . If this inequality fails and min { r, j } = r then we have l > (2 p − j + r . But since l ≤ m ≤ (2 p − j this contradicts l ≡ r ( mod 2 j ). On the other hand if the inequalityfails with min { r, j } = j < r , then since r < j we have that (2 p − j < l ≤ (2 p − j . Thiscontradicts min { r, j } = j < r since then the remainder r is less than or equal to j .Case 3: Suppose label n is located after node R and is the l th label after the final labelin R (thus, we have the understanding that l ≥ P is the last non leaf node in T ( n ) we have l < x . Also, as l < x , the node R does not lose a full set of pj labels. Toestablish Lemmas 3.3 and 3.4 we must show that P contains between j and j + m labelsafter pruning. Subcase 3a: ≤ l < p − j − m ) ≤ x (note that for this to happen we need m < ( p − j ). In this case both children of P may lose less than pj labels. Note that if thiscase does occur then subcase 2c will not because that requires m ≥ ( p − j .By Lemma 3.6, if 2 m + l + j = q L (2 j ) + r L with 0 ≤ r L < j then L loses q L j + min { r L , j } labels. Also, if m + l = q R (2 j ) + r R with 0 ≤ r R < j then R loses q R j + min { r R , j } labels.After the end correction step we remove x − l labels from P . So the number of labels in P after pruning is 2( m + j ) − ( q L j + min { r L , j } + q R j + min { r R , j } ) + l . We need to showthat j ≤ m + j ) − ( q L j + min { r L , j } + q R j + min { r R , j } ) + l ≤ m + j . (3.8)The first inequality is ( q L + q R ) j + min { r L , j } + min { r R , j } ≤ m + j + l = q L (2 j ) + r L .We are done if q L > q R . Otherwise, since q L ≥ q R we must have q L = q R and this implies r L = m + j + r R ≥ j . Then the inequality reduces to j + min { r R , j } ≤ r L = m + j + r R ,which is true. ESTED RECURSIONS, SIMULTANEOUS PARAMETERS AND TREE SUPERPOSITIONS 21
Now we consider the second inequality in (3.8). It reduces to m + j + l ≤ ( q L + q R ) j +min { r L , j } + min { r R , j } , which in turn is the same as( q R + 1) j + r R ≤ q L j + min { r L , j } + min { r R , j } . If q R < q L − q R = q L − { r R , j } = r R , or min { r R , j } = j andmin { r L , j } = j . So we can assume that min { r R , j } = j and min { r L , j } = r L . Then using m + l = ( q L − j ) + r R and 2 m + j + l = q L (2 j ) + r L , we deduce that m = j + r L − r R .However, as m ≥ r R ≤ j + r L as required by the inequality above.If q R = q L then we need to show that j + r R ≤ min { r L , j } + min { r R , j } . Once again wehave r L = m + j + r R , and so min { r L , j } = j . The above then becomes r R ≤ min { r R , j } ,but we do have r R ≤ j for otherwise r L > j . Subcase 3b: p − j − m ) ≤ l < x . In this case L loses all pj labels, and R , by Lemma3.6, loses qj + min { r, j } labels where m + l = q (2 j ) + r . The total number of labels in P after pruning is ( m − ( p − j ) + ( m + j − qj − min { r, j } ) + l . Hence we need to show j ≤ m + j + l − ( p − j − qj − min { r, j } ≤ m + j . (3.9)We consider the first inequality of (3.9). After substituting m + l = q (2 j )+ r and simplifying,the lower bound in (3.9) becomes ( p − − q ) j + min { r, j } ≤ m + r . If q = p − q < p − l ≥ (2 p − j − m , which implies that q (2 j ) + r = m + l ≥ (2 p − j − m .Thus, m + r ≥ p − − q ) j . Since p − − q ≥
1, we conclude that m + r ≥ ( p − − q ) j + j ≥ ( p − − q ) j + min { r, j } as needed.Now consider the second inequality in (3.9). Substituting m + l = q (2 j )+ r we get that qj + r ≤ ( p − j + min { r, j } . As q ≤ p − q < p − r ≤ j . If q = p − r > j , then it follows that all j cells in R are nonempty in T ( n − (2 p − j ); a contradiction since m + l < (2 p − j .With this we have considered all cases and the proofs of Lemma 3.3 and Lemma 3.4 arenow complete. We conclude this section by considering the frequency sequence φ C T of thecell counting function of a fixed tree T = T s,j,m,p .3.4. The frequency sequence.
From [3] we know that the tree-based solution sequence ofthe ( α, β ) Conolly recursion (1.7) has frequency sequence α + βφ C where φ C is the frequencysequence of the Conolly sequence (1.2). This is the linear combination α φ H + βφ C of thefrequency sequences of the H sequence (1.3) and the Conolly sequence. Now the function C T with the choice of simultaneous parameters ( s, j, m, p ) = (0 , j, ( α + β − j, α/ β )gives (3.1), which is (1.7) with the simultaneous parameter j . So it is natural to wonderwhether the frequency sequence of C T with the aforementioned choice of parameters is α φ H j + βφ C j where H j and C j are the tree-based solutions of (1.5) and (1.4) respectivelyfor s = 0.Using the results about the frequency sequences of (1.5) and (1.4) from Theorems 5.1 and5.5 of [10] we can easily compute α φ H j + βφ C j . If ν ( v ) is the 2-adic valuation of v then φ C ( v ) = ν ( v ) + 1 and α φ H j ( v ) + βφ C j ( v ) = (cid:26) α + β if j (cid:45) vβj · ν ( vj ) + α ( j + 1) + β otherwiseOn the other hand, we can derive φ C T using an argument most similar to that of Theorem2.3. The difference is that the non leaf regular nodes now contain x labels instead of j − m .For fixed ( s, j, m, p ) φ C T ( v ) = (cid:26) j (cid:45) v ((2 p − j − m ) · ν ( vj ) + 1 + m + s [ vj is a power of 2] otherwiseFrom this it is easy to see that φ C T (cid:54) = α φ H j + βφ C j when ( s, j, m, p ) = (0 , j, ( α + β − j, α/ β ). In the next section we derive a 2-ary order p recursion whose solution sequence doesindeed have the frequency sequence α φ H j + βφ C j and we give a tree-based proof of thederivation.4. Linear combinations of frequency sequences via tree superpositions
In this section we use the tree-based methodology to derive a nested recursion whose solutionhas a frequency sequence that is a linear combination of the frequency sequences of H ,j ( n )and R ,j ( n ) from (1.5) and (1.4) respectively for s = 0. Our strategy is to construct alabelled infinite binary tree whose cell counting function has the desired property and thenuse the tree along with a pruning operation to derive a nested recursion with the samefrequency function.To motivate our construction we recall the trees whose cell counting functions satisfy recur-sions R ,j ( n ) and H ,j ( n ). The first tree, say T with cell counting function R ,j ( n ), is T ,j, from Section 2, that is, the binary tree corresponding to m = 0. Similarly, the tree withcell counting function H ,j ( n ) is T = T ,j,j from Section 2. To obtain a tree T whose cellcounting function has a frequency sequence α φ H ,j + βφ R ,j we form the “superposition”of the two trees T and T . That is, we place α/ T and β copies of T on top ofeach other. Note that since T and T have the same skeleton, this superposition createsanother infinite binary tree T with the same skeleton (see Figure 2.1). When we superposemultiple copies of T and T we initially treat the labels in each tree as placeholders; asa result, at first the labels in the superposed tree T do not appear in preorder and T hasmultiple occurrences of the same label (see Figure 4.1, where for simplicity we illustratethe superposition of a single copy of each tree). Once we relabel the tree T in preorderit is evident that we obtain a tree whose cell counting function has frequency sequencethat is the desired linear combination α φ H ,j + βφ R ,j of the cell counting functions forthe individual trees. Note that in principle α can be negative; in this case the tree T iswell-defined so long as ( α/ j + 1) + β ≥ T ).4.1. The tree and the pruning operation.
We now give a direct, more general con-struction of a tree T whose cell counting function is, under certain conditions, the desiredlinear combination. This will allow us to solve not only a nested recursion whose solutionhas the desired frequency function, but also a wide spectrum of related recursions. ESTED RECURSIONS, SIMULTANEOUS PARAMETERS AND TREE SUPERPOSITIONS 23
Figure 4.1.
Superposition of T , , and T , , prior to relabelling.Fix simultaneous parameters s, j, m, p . The natural range of these parameters is discussedbelow. The desired tree T = T s,j,m,p has the skeleton of the infinite binary tree from Figure(2.1) with j cells in each leaf. For n ≥ T ( n ) denote T with n labels inserted in preorderas follows: each of the first j − p labels, while the last cell receives p + m labels. All remaining regular nodes in T get pj − m labels each, and the supernodesreceive s labels each. See Figure 4.2, where we use the case s = 0 , j = m = 3 , p = 2 as ourrunning example. Figure 4.2.
The labelled infinite tree T , , , (82); C T (82) = 24.To ensure that each leaf has at least one cell and that cells have a positive number oflabels, we require p, j ≥
1. Likewise, to force regular nodes and supernodes to contain anon-negative number of labels, we need s ≥ ≤ m ≤ pj . Note that for negativevalues of m such that m > − p , the tree T is still well-defined. However, our proof hereonly works for non-negative m , so we restrict the range of this parameter accordingly. SeeSection 6 for further discussion.We let C T ( n ) denote the number of nonempty cells in T ( n ). Since each cell has p labels andeach regular nodes contains pj − m labels, the same argument as in the proof of Theorem2.3 yields that φ C T ( v ) = (cid:26) p if j (cid:45) v ( pj − m ) · ν ( vj ) + p + m + s [ vj is a power of 2] otherwise Observe that when s = 0 and m = bj for some integer b ≥
0, the frequency sequence φ C T = bφ H ,j + ( p − b ) φ R ,j . That is, for m a multiple of j , the resulting tree T is asuperposition of trees T and T as discussed above. Note that the restriction m ≥ φ H ,j and φ R ,j with non-negative coefficients. If m is not a multiple of j then the resulting tree is not asuperposition of trees T and T .Let α, β ≥
0, and set j = 1, p = α/ β and m = α/
2. Then φ C T is exactly α φ H + βφ C ,that is, φ C T is the frequency function of the solution for the recursion (1.7). We nowgeneralize this result by deriving a recursion whose solution has a frequency function thatis a linear combination of the frequency functions for the solutions to R ,j and H ,j .In Section 2 and 3 we have used pruning operations to show that a cell counting functionis the solution to a nested recursion. Here we reverse our approach and use a pruningoperation to derive a recursion whose solution is given by C T ( n ).The two major requirements we place on the pruning operation is that the resulting tree T ∗ ( n ) has the same skeleton as T and that it conforms to the labelling rules describedearlier. In particular, we would like our pruning operation to be defined in such a waythat a “typical” nonempty leaf of T ( n ) loses pj labels in the deletion step. In that case,2 m labels are lifted to its parent to bring the total count of labels in it to pj + m afterpruning (exactly the number of labels in a “typical” nonempty leaf). Using this heuristic,and examples of pruning operations in Section 2 and Section 3, we define the followingpruning operation on T ( n ), n > pj + 3 m + 2 s (the first seven nodes of T are full). SeeFigures 4.3, 4.4, 4.5, 4.6 and 4.7. Figure 4.3.
Initial correction step when pruning T , , , (82). Initial correction:
Remove the s labels from the first supernode and insert pj − m labels in the first supernode (these labels are currently placeholders only; we do notrelabel the tree until the relabelling step). Deletion step:
For each i , 1 ≤ i ≤ p , consider the subtrees T ( n − (2 i − − p ( j − T ( n − (2 i − − p ( j − T ( n ) (to be specific, we remove the largest label in thecell). Note that this is always possible because every nonempty cell has at least p labels. ESTED RECURSIONS, SIMULTANEOUS PARAMETERS AND TREE SUPERPOSITIONS 25
Figure 4.4.
Deletion step when pruning T , , , (82). Figure 4.5.
Lifting step when pruning T , , , (82). Figure 4.6.
End correction when pruning T , , , (82). Lifting step:
Lift all the remaining labels from every nonempty cell of T ( n ) into theparent of the leaf containing that cell. After this step all bottom level leaves of T ( n ) become empty. End correction:
Remove the largest pj − m labels that remain in T ( n ). Figure 4.7.
Relabelling step when pruning T , , , (82). Relabelling step:
Remove all the leaves of T ( n ) and relabel the new tree in preorder.Partition the new leaf labels into j cells in the same manner as was done for theleaves of T and denote this tree by T ∗ ( n ).The number of labels removed from T ( n ) after the pruning operation is s + (cid:80) pi =1 C T ( n − i + 1 − p ( j − T ∗ ( n ) contains n − s − (cid:80) pi =1 C T ( n − i + 1 − p ( j − T , but it is not immediately obvious that T ∗ ( n ) follows thelabelling scheme defined earlier. In other words, we would like to establish the followinglemma: Lemma 4.1.
The pruning of T ( n ) results in a tree T ∗ ( n ) = T ( n − s − (cid:80) pi =1 C T ( n − (2 i − − p ( j − . To derive a recursion satisfied by C T ( n ) we will also need to establish a bijective correspon-dence between the cells of T ∗ ( n ) and the cells of left leaves of T ( n ): Lemma 4.2.
Let P be a nonempty leaf in T ∗ ( n ) (so that P is a penultimate level nodein T ( n ) ). Then the number of nonempty cells of T ∗ ( n ) in P is equal to the number ofnonempty cells of the left child of P in T ( n ) . The main theorem.
Once we establish Lemmas 4.1 and 4.2 we can derive a nestedrecursion as follows. Let C T,L ( n ) and C T,R ( n ) be the number of nonempty cells in T ( n )that are located in the left and right leaves respectively. Thus C T ( n ) = C T,L ( n ) + C T,R ( n ).Since there is a bijection between nonempty cells in the right leaves of T ( n ) and nonemptycells in the left leaves of T ( n − pj − m ), we have that C T,R ( n ) = C T,L ( n − pj − m ). Therefore,by Lemma 4.1 and 4.2 C T ( n ) = C T (cid:32) n − s − p (cid:88) i =1 C T ( n − (2 i − − p ( j − (cid:33) + C T (cid:32) n − s − pj − m − p (cid:88) i =1 C T ( n − (2 i − − m − p (2 j − (cid:33) for n > pj + 3 m + 2 s . That is, Lemma 4.1 and 4.2 together establish the followingresult: ESTED RECURSIONS, SIMULTANEOUS PARAMETERS AND TREE SUPERPOSITIONS 27
Theorem 4.3.
The cell counting function C T ( n ) satisfies the 2-term order p nested recur-sion: R ( n ) = R (cid:32) n − s − p (cid:88) i =1 R ( n − (2 i − − p ( j − (cid:33) (4.1)+ R (cid:32) n − s − pj − m − p (cid:88) i =1 R ( n − (2 i − − m − p (2 j − (cid:33) . In particular, recursion (4 . generates the cell counting function C T ( n ) if it is given pj +3 m + 2 s initial conditions (every node until the right child of the first regular node is full)that agree with the cell counting function. Proof of Theorem 4.3.
We now proceed with the proof of Lemmas 4.1 and 4.2. Thetrees T ∗ ( n ) and T ( n − s − (cid:80) pi =1 R ( n − i + 1 − p ( j − T ∗ ( n ) contain the “correct” number of labels.As in Sections 2 and 3, we can think of every node in T ∗ ( n ) as being part of T ( n ). By thedesign of the pruning operation on T ( n ), all nodes in T ∗ ( n ), except for the leaves of T ∗ ( n )(which are the former penultimate nodes of T ( n )) and the last nonempty node (which mayor may not be a leaf of T ∗ ( n )), contain the same number of labels as they do in T ( n ).So to prove Lemma 4.1 we need only focus on the leaves of T ∗ ( n ) and its last nonemptynode.Consider first the last nonempty node of T ∗ ( n ). Suppose that it is not a leaf of T ∗ ( n ). Thenin T ( n ) this node is neither a leaf nor a penultimate node. Therefore the pruning operationon T ( n ) doesn’t add any labels to this node (the end correction step of the pruning mayremove some labels). After pruning, this last nonempty node in T ∗ ( n ) has at most thesame number of labels that it has in T ( n ), which is what we require.To count the number of labels that remain in leaves of T ∗ ( n ), we have the followinglemma. Lemma 4.4.
Let P be a nonempty penultimate level node in T ( n ) and let l P be the numberof labels in nodes of T ( n ) after P (in preorder). (1) If l P = 0 then P is empty in T ∗ ( n ) . (2) If ≤ l P ≤ pj + m then P contains l P − p (cid:88) i =1 min (cid:18) j, (cid:24) l P − p ( j − − i + 1 p (cid:25) · [ l P − p ( j − − i +1 > (cid:19) − p (cid:88) i =1 min (cid:18) j, (cid:24) l P − p (2 j − − m − i + 1 p (cid:25) · [ l P − p (2 j − − m − i +1 > (cid:19) labels in T ∗ ( n ) . (3) If l P > pj + m then P contains pj + m labels in T ∗ ( n ) . To simplify the notation, let h ( k, x ) = k − p ( j −
1) + 1 − x and d ( l ) = p (cid:88) i =1 min (cid:18) j, (cid:24) h ( l, i ) p (cid:25) · [ h ( l, i ) > (cid:19) Also, when there is no confusion we write l instead of l P . Note that the expression in (2)above reduces to l − d ( l ) − d ( l − pj − m ). Further, the deletion step of the pruning operationcan now be rephrased in terms of the subtrees T ( h ( n, , T ( h ( n, , . . . , T ( h ( n, p )). Proof.
To prove (1), we note that if all the nodes of T ( n ) after P are empty and P is not,then P contains the largest label in T ( n ). During the end correction step of the pruningthe pj − m largest labels are removed from the tree. Since P contains at most pj − m labels,it is emptied by the pruning operation.Next we prove (3). Let a = n − l . Then n − p ( j − − (2 p −
1) = ( l − p ( j − − (2 p − a = h ( l, p ) + a> h (3 pj + m, p ) + a = pj + m + p ( j −
1) + 1 + a since l > pj + m . Thus the label n − p ( j − − (2 p −
1) is no further back in the tree thenthe last label of the right child of P . In other words, every cell of left and right child of P is nonempty in each of T ( h ( n, , T ( h ( n, , . . . , T ( h ( n, p ))Therefore, each cell in the children of P will lose exactly p labels and 2 m labels will belifted to P on the lifting step. Note that none of the labels are removed from P in the endcorrection step of the pruning operation since there are at least pj − m labels in nodes of T ( n ) after the children of P . Thus, P has pj − m + 2 m labels after pruning.Now we prove (2). Let L and R be the left and right child of P respectively. Recall thatwe assume here that 1 ≤ l ≤ pj + m . Note that if h ( l, i ) ≤ i with 1 ≤ i ≤ p ,then L has no nonempty cells in T ( h ( n, i ) and thus no labels in L are pruned when weconsider T ( h ( n, i ). On the other hand, if 0 < h ( l, i ) ≤ pj , then T ( h ( l, i ) will have (cid:100) h ( l, i ) p (cid:101) nonempty cells in L and which is exactly the number of labels removed from L when considering T ( h ( n, i )). Similarly, if h ( l, i ) > pj , then all j cells of L are nonemptyin T ( h ( n, i ) and j labels are removed from L when considering this subtree. Therefore, itfollows that d ( l ) is the number of labels that are removed from L during pruning. Sincethere are l − pj − m labels in nodes of T ( n ) after R , we may repeat this argument to obtainthat d ( l − pj − m ) is the number of labels removed from R during pruning.Let l = l + l where l is the total number of labels in L and R before pruning, and l ≤ pj − m is the number of labels in nodes of T ( n ) after R . Then there will be pj − m + l − d ( l ) − d ( l − pj − m ) labels on P before the end correction step. During the endcorrection step we remove pj − m largest labels from the tree. Namely, l labels will beremoved from the nodes that follow R in preorder and the remaining l − pj + m will betaken from P , leaving exactly l − d ( l ) − d ( l − pj − m ) labels in it. (cid:3) If P is not one of the two last nonempty penultimate nodes in T ( n ), then there are at least3 pj + m labels in the nodes that follow it. Thus, from the lemma, after pruning P will contain pj + m labels. Now, if P and Q are the last and second last nonempty penultimate nodes,respectively, and all children of P are empty, then P will be empty in T ∗ ( n ). If l Q > pj + m ESTED RECURSIONS, SIMULTANEOUS PARAMETERS AND TREE SUPERPOSITIONS 29 then Q contains pj + m labels; otherwise there are only l Q − d ( l Q ) − d ( l Q − pj − m ) labels in Q . If P has a nonempty child then l Q > pj + m and Q will have pj + m labels in T ∗ ( n ). If l P > pj + m then P contains pj + m labels; otherwise there are only l P − d ( l P ) − d ( l P − pj − m )labels in P . Therefore, to prove Lemma 4.1 it remains to verify that if P is a penultimatenode and l P ≤ pj + m , then l P − d ( l P ) − d ( l P − pj − m ) ≤ pj + m .Furthermore, observe that if l P ≤ p ( j −
1) + 1 then d ( l P ) = d ( l P − pj − m ) = 0, i.e. thenumber of labels in P after pruning is l P and in that case both Lemma (4.1) and Lemma(4.2) hold. Therefore, to complete the proof of these lemmas it suffices to check that thefollowing result holds. Lemma 4.5.
For p ( j −
1) + 1 < l ≤ pj + m , p ( j −
1) + 1 ≤ f ( l ) ≤ pj + m where f ( l ) = l − d ( l ) − d ( l − pj − m ) . First we restrict our attention to d ( l ). The following Lemma completely determines be-haviour of d ( l ) for the specified range of l . Lemma 4.6.
For p ( j − ≤ l ≤ pj + m , d ( l ) is a non-decreasing function. In particular,as l changes from p ( j −
1) + 2 to p ( j −
1) + p the function d grows from d ( p ( j −
1) + 2) = 1 to d ( p ( j −
1) + p ) = (cid:98) p (cid:99) . For pj ≥ l > p ( j −
1) + p , if p is odd then d ( l + 1) − d ( l ) = 1 and if p is even then d ( l + 1) − d ( l ) alternates between 0 and 2 if p is even. For l > pj , d ( l ) = pj .Proof. Note that when l increases by 1, each summand in d ( l ) either increases by 1 or staysthe same. It follows, that d ( l ) is a non-decreasing function. To prove the rest of the Lemmawe need to understand how many summands in d ( l ) can increase at the same time. Thatis, we need find how many of the h ( l, , h ( l, , . . . h ( l, p ) can be multiples of p at the sametime.Consider two intervals S = [ h ( l, , h ( l, p )] and S = [ h ( l, p + 1) , h ( l, p )]. The integers in S ∪ S of the form h ( l, i ) correspond to the summands in d ( l ). Also, note that each ofthese intervals contain exactly one multiple of p .If p is odd and h ( l, x ) and h ( l, y ) are multiples of p from the first and second list respectivelythen it follows that y = x + p and hence one of x, y is even and the other one is odd. Wealso note that if l is increased by 1 then the roles of x and y are interchanged, i.e. if x was even and y was odd, then after l is increased x is odd and y is even. Therefore, if p is odd there is always exactly one multiple of p among h ( l, , h ( l, , . . . h ( l, p ), i.e. eachtime l is increased by 1 exactly one of (cid:100) h ( l, p (cid:101) , . . . , (cid:100) h ( l, p ) p (cid:101) increases by 1 as well. We alsonote that the increasing terms alternate between the S i : if an increase in l by 1 leads to anincrease in (cid:100) h ( l, u ) p (cid:101) and h ( l, u ) belongs to S then increasing l again leads to an increasein (cid:100) h ( l, u ) p (cid:101) with some h ( l, u ) in S . The analysis in this paragraph is also valid for p = 1.Similarly, if p is even then x and y have the same parity. Moreover, if x, y are even thenonce l is increased by 1, they both become odd and vice versa. Therefore, if p is even eithernone or exactly two of (cid:100) h ( l, p (cid:101) , . . . , (cid:100) h ( l, p ) p (cid:101) grow by 1 when l increases by 1. Thus, thedifference sequence d ( l + 1) − d ( l ) alternates between 0 and 2.Finally, we are ready to fully describe the behaviour of d ( l ). We observe that when p ( j − < l ≤ p ( j − p the function d grows from d ( p ( j − d ( p ( j − p ) = (cid:98) p (cid:99) . This is because each summand of d ( l ) corresponding to indices in S increases by 1 andeach summand corresponding to indices in S remains zero because the indicator function [ h ( l, i ) > will be zero for those summands. For 2 pj ≥ l > p ( j −
1) + p , d ( l ) either satisfies d ( l + 1) − d ( l ) = 1 if p is odd, or d ( l + 1) − d ( l ) alternates between 0 and 2 if p is even. Aswe have noted earlier, d ( l ) = pj for l > pj . (cid:3) It follows from the Lemma 4.6 that a similar result holds for d ( l − pj − m ). As l increasesfrom pj + m + p ( j −
1) + 2 to pj + m + p ( j −
1) + p , the function d grows from 1 to (cid:98) p (cid:99) .After that d ( l − pj − m ) is either a slowly growing sequence or has successive differencesthat alternate between 0 and 2. Once d reaches pj , it remains constant.Now we are ready to prove Lemma 4.5. Proof.
Recall that we would like to establish that for p ( j − < l ≤ pj + m , p ( j − ≤ f ( l ) ≤ pj + m . It follows from Lemma 4.6 that f ( l ) grows from p ( j − l = p ( j − p ( j − (cid:100) p (cid:101) at l = p ( j − p . For p ( j − p ≤ l min( pj + m + p ( j − , pj ), f ( l ) remainsconstant if p is odd or alternates between p ( j −
1) + (cid:100) p (cid:101) + 1 and p ( j −
1) + (cid:100) p (cid:101) if p is even.Thus, f ( l ) lies within the required bounds for p ( j − < l ≤ min( pj + m + p ( j − , pj ).Now we consider two cases: 2 pj ≤ pj + m + p ( j −
1) + 1 and pj + m + p ( j −
1) + 1 < pj .If 2 pj ≤ pj + m + p ( j −
1) + 1 then f ( l ) is a increasing for l in [2 pj, pj + m + p ( j −
1) + 1]with f ( pj + m + p ( j −
1) + 1) = p ( j −
1) + 1 + m < pj + m since d ( l ) = pj for l ≥ pj . Also,since f ( l ) is increasing in this case, we still have f ( l ) ≥ p ( j −
1) + 1 for l in the given range.If pj + m + p ( j −
1) + 1 < pj then both d ( l ) and d ( l − pj − m ) grow at the same time and f ( l ) can potentially fall below p ( j −
1) + 1. However, note that d ( l − pj − m ) grows onlyup to (cid:98) p (cid:99) between pj + m + p ( j −
1) + 1 and pj + m + p ( j −
1) + p = 2 pj + m by Lemma4.6. Therefore, f ( l ) can only decrease to p ( j −
1) + 1 on this interval. From the previouscase it follows that the upper bound f ( l ) ≤ pj + m still holds in this case as well.Therefore, f ( l ) lies within the required bounds for min( pj + m + p ( j −
1) + 1 , pj ) < l ≤ max( pj + m + p ( j −
1) + 1 , pj ).For max( pj + m + p ( j −
1) + 1 , pj ) ≤ l ≤ p ( j −
1) + p + pj + m , f ( l ) is a non-decreasingfunction with f ( p ( j −
1) + p + pj + m ) = p ( j −
1) + m + (cid:100) p (cid:101) ≤ pj + m . Thus, for l in thisrange f ( l ) also lies within the required bounds.Finally, for l in [ p ( j −
1) + p + pj + m, pj + m ], f ( l ) is either p ( j −
1) + m + (cid:100) p (cid:101) ( p is odd)or alternates between p ( j −
1) + m + (cid:100) p (cid:101) and p ( j −
1) + m + (cid:100) p (cid:101) + 1 ( p is even). Therefore, f ( l ) is within the required range for these values of l as well and the proof is complete. (cid:3) Nested k -ary order p recursions In this section we continue our study of nested recursions via the lens of simultaneousparameters by extending our earlier approach to solve certain k -ary, order p recursionfamilies. We begin our discussion by reviewing previous work on k -ary nested recursions oftype (1.1). ESTED RECURSIONS, SIMULTANEOUS PARAMETERS AND TREE SUPERPOSITIONS 31
Figure 5.1.
The skeleton of the infinite k -ary tree for k = 3.The k -ary Conolly recursion C k ( n ) = k (cid:88) i =1 C k ( n − i + 1 − C k ( n − i )) . (5.1)is studied in [12]. There it is shown that the solution to (5.1), with appropriate initialconditions, counts leaves on the infinite, labelled, k -ary tree which is the natural extensionof the infinite binary tree associated with the solution to the usual Conolly recursion (1.2).Further, it is shown that the frequency sequence of this solution is φ C k ( v ) = ν k ( v ) + 1,where ν k ( v ) is the k -adic valuation of v .As it will be required in what follows, we describe the infinite, labelled k -ary tree used above,which reduces to the binary tree we described earlier when k = 2. There are supernodesalong the leftmost spine and regular nodes. The first supernode has k leaf children; everyother supernode has k − k children. Apart fromthe leaves, all regular nodes also have k children. See Figure 5.1 for the skeleton of thetree for k = 3. The nodes are labelled in preorder, with s labels in each supernode and onelabel in each regular node. For (5.1), s = 0 and C k ( n ) counts the number of leaves up tothe n th label.Recall that the ceiling function (cid:100) n (cid:101) is the solution to the H recursion (1.3) (see, for example,[1, 3]. This result is generalized in [5], where it is shown that (cid:100) nk (cid:101) is the solution to thefollowing k -ary, order k − H k ( n ) = k (cid:88) i =1 H k ( n − ( i − k − k − (cid:88) t =1 H k ( n − ( i − k − t )) . (5.2)Once again a tree-based methodology is used to prove this result. The infinite k -ary treeassociated with (5.2) has the same skeleton as the k -ary tree described in the previousparagraph, but with a different labelling. This tree contains k labels in each leaf and nolabels in any other node. Because the labels are enumerated in preorder, it follows thatthis is just a sequential labelling as one traverses the leaves from left to right. Note thatthe resulting solution sequence (cid:100) nk (cid:99) has the frequency sequence φ H k ( v ) = k .In what follows we use our tree-based methodology to derive and solve a new family of k -ary, order p recursions that includes the above two families. This family of recursionsextends the arity two, ( α, β )-Conolly recursion (3.2) to arity k . Further, we show thatthis family includes certain recursions whose solution sequence has frequency sequence γφ H k + δφ C k = γk + δφ C k . The trees associated with these latter recursions result fromthe superposition of the appropriate number of copies of the trees associated with C k and H k .5.1. The main theorem.
We extend the arity 2, ( α, β )-Conolly recursion (3.2) to arity k by extending the construction of the arity two tree from Section 3. Recall from Section3 that the tree constructed there relied on four parameters s, j, m and p . For ease ofexposition we limit our discussion here to the case s = 0 and j = 1.Fix parameters k ≥ , p ≥ m satisfying p − ≤ m ≤ kpk − −
1; the infinite tree wenow construct will be denoted T = T m,p,k . The tree T has the same skeleton as the infinite k -ary tree described above. The labelling of T is as follows: the leaves of T each contain1 + m labels. All other regular nodes each contain x := pk − ( k − m ) labels. Therange of m ensures that x ≥
0. The labels of the resulting tree are then enumerated inpreorder. Note that for k = 2 we get the construction of Section 3, except that the range of m is more constrained. Refer to Section 5.2, where we talk further about what goes wrongwith our proof for m < p − T ( n ) be the subtree of T with n labels in preorder. If a leaf of T ( n ) is the i th child of its parent at the penultimate level, then we will abbreviate it as a i th leaf. Define C T ( n ) to be the function that counts the number of nonempty leaves of T ( n ). Then ourmain result is that C T ( n ) satisfies a k -ary, order p nested recursion. Theorem 5.1.
With T as defined above let C T ( n ) be the number of nonempty leaves of T ( n ) . For n > k ( p + m ) + p − ( k − m (all labels up to the last child of the secondpenultimate level node must be filled), C T ( n ) satisfies the recursion R ( n ) = k (cid:88) i =1 R (cid:32) n − ( i − m ) − p (cid:88) t =1 R ( n − ( i − m ) − t ) (cid:33) . (5.3) Remark.
For k = 2 the recursion in Theorem 5.1 does not reduce to (3.2). Nonethelesswe will show in Section 5.2 that for any α ≥ m and p such thatthe tree-based solution of (5.3) is the same as that for (3.2) . For α < m ≥ p −
1. See Section 5.2 for additional details.The proof of Theorem 5.1 follows from the definition of C T ( n ) once we establish the followinglemma. Lemma 5.2.
For each ≤ i ≤ k the term C T (cid:32) n − ( i − p + m ) − p (cid:88) t =1 C T ( n − ( i − p + m ) − t ) (cid:33) counts the number of nonempty i th leaves of T ( n ) . To prove Lemma 5.2 we introduce a pruning operation on T ( n ) for n > k ( p + m )+ p − ( k − m (all labels up to the first child of the second penultimate node level must be filled). Thepruned tree will be denoted T ∗ ( n ). ESTED RECURSIONS, SIMULTANEOUS PARAMETERS AND TREE SUPERPOSITIONS 33 initial correction step:
Insert x = pk − ( k − m ) labels in the first supernodeof T ( n ). deletion step: Consider the subtrees T ( n − t ) for 1 ≤ t ≤ p . For each leaf of T ( n ),delete a label from it for every subtree T ( n − t ) in which it is nonempty. lifting step: Lift any remaining labels from the leaves of T ( n ) to their correspondingparent at the penultimate level. After this step all leaves of T ( n ) become empty. end correction step: Delete the largest x = pk − ( k − m ) labels that arepresent in T ( n ) after the last step. This deletion of labels offsets the insertion of x labels into the first supernode during the initial correction step. relabelling step: Delete all the (empty) leaves of T ( n ) and relabel all remaininglabels in preorder. The previous penultimate level nodes of T ( n ) become the newleaves.The total number of labels in the pruned tree T ∗ ( n ) is n − (cid:80) pt =1 C T ( n − t ). The key lemmafollows. Lemma 5.3.
The pruned tree T ∗ ( n ) is identical to the tree T ( n − (cid:80) pt =1 C T ( n − t )) . Fur-thermore, if P is a nonempty leaf of T ∗ ( n ) then the first child of P in T is a nonempty leafof T ( n ) . It follows from Lemma 5.3 that the number of nonempty first leaves of T ( n ) is equal to thenumber of nonempty leaves of T ∗ ( n ). However, the latter number is C T ( n − (cid:80) pt =1 C T ( n − t ))by the first assertion of Lemma 5.3. So the assertion in Lemma 5.2 follows for i = 1. Theassertion for general i follows due to the usual bijection between i -th leaves of T ( n ) andthe first leaves of T ( n − ( i − m )). With Lemma 5.2 established it follows triviallythat C T ( n ) = k (cid:88) i =1 C T (cid:32) n − ( i − m ) − p (cid:88) t =1 C T ( n − ( i − m ) − t ) (cid:33) . We now prove Lemma 5.3. As before, we need only consider the penultimate level node P that is the last penultimate level node of T ( n ) with a nonempty child in T ( n ). We need toshow that P has between 1 to 1 + m labels as a leaf of the pruned tree T ∗ ( n ). All otherpenultimate level nodes of T ( n ) either contain a full set of 1 + m labels or no labels in T ∗ ( n ) according to whether they have a nonempty child in T ( n ) or not. So suppose that P is such a node and condition on the location of label n in T ( n ).For ease of notation in the proof we let µ = m − p + 1. Since p − ≤ m ≤ kk − p −
1, wehave that 0 ≤ µ ≤ p/ ( k − T contains p + µ labels and all other regular nodes contain x = p − ( k − µ labels. Weneed to show that after pruning T ( n ) the node P contains between 1 to p + µ labels in T ∗ ( n ). Case 1:
Label n is located in the i th child of P with 1 ≤ i ≤ k . Suppose that n is the l th label in the i th child. If l ≤ p then the i th child loses l − i − p labels each. If p < l ≤ p + µ (assuming µ ≥ i children of P lose p labels in the deletion step. Therefore, the number of labels in P after the end correction step is ( i − µ + 1 if l ≤ p and ( i − µ + l − p otherwise. Noticethat min i,l (cid:8) [( i − µ + 1] · [ l ≤ p ] + [( i − µ + l − p ] · [ l>p ] (cid:9) ≥ . This implies the second assertion of Lemma 5.3 for this case. Also, using p ≥ ≤ µ ≤ p/ ( k −
1) it follows thatmax i,l (cid:8) [( i − µ + 1] · [ l ≤ p ] + [( i − µ + l − p ] · [ l>p ] (cid:9) ≤ p + µ . Thus, the first assertion of Lemma 5.3 follows as well.
Case 2:
Label n is the l th label following the final label in the last child of P and l ≥ P loses p labels during the deletion step, and during the endcorrection step the number of labels removed from P is p − ( k − µ − min { l, p − ( k − µ } .The total number of labels in P after the relabelling step is kµ + min { l, p − ( k − µ } .However, 1 ≤ kµ + min { l, p − ( k − µ } ≤ p + µ due to l, p ≥ µ ≥
0. This establishesboth assertions of Lemma 5.3 for this case.The two cases above are exhaustive and Lemma 5.3 is thus proved.5.2.
Consequences of Theorem 5.1.
We now relate our k -ary ( α, β )-Conolly general-ization to our tree superposition methodology. Fix k ≥
3. If ( m, p ) = (0 ,
1) then thesolution sequence C T ( n ) is the k -ary Conolly sequence (5.1). On the other hand, with( m, p ) = ( k − , k −
1) the solution sequence C T ( n ) = H k ( n ) because the resulting tree T contains k labels per leaf and 0 labels everywhere else. We can take the trees resulting fromthese two choices of ( m, p ) and superpose them as discussed in Section 4. To do so we fixcoefficients γ, δ ≥ m = kγ + δ − p = ( k − γ + δ . With these choices it is easily verified that p ≥ γ and δ is positive), and that p − ≤ m ≤ kpk − −
1. The resulting tree T m,p,k contains γk + δ labels in each leaf and δ labels in each of the remaining regular nodes. It isthe superposition of γ copies of T k − ,k − ,k and δ copies of T , ,k . Therefore, the frequencysequence φ C T of the solution sequence C T ( n ) is γφ H k ( n ) + δφ C k = γk + δφ C k .Finally, because we require m ≥ p − m and p that we implicitly restrict γ to positive values. Observe thatin the above k -ary generalization, γ corresponds to α/ δ corresponds to β from the2-ary case. In the latter, negative values of α are permitted under appropriate constraints.This leads naturally to the question whether frequency sequences with negative values of γ can also be obtained. We comment further on this open question in the next section.6. Future directions
In the course of our investigation of nested recursions in this paper we have identifiedseveral areas where there are questions about the possibility of extending our results. Inthis concluding section we collect and further discuss these open problems, which providepossible directions for future research in this area.
ESTED RECURSIONS, SIMULTANEOUS PARAMETERS AND TREE SUPERPOSITIONS 35
Tree Superposition.
Recall that in Section 4, T s,j,m,p is defined to be the tree withthe skeleton of the infinite binary tree from Figure (2.1) with j cells in each leaf and labellingscheme as follows: each of the first j − p labels, while the lastcell receives p + m labels. All remaining regular nodes in T s,j,m,p get pj − m labels each,and the supernodes receive s labels each. To ensure that each leaf has at least one celland that cells have a positive number of labels, we require p, j ≥
1. Likewise, to forceregular nodes and supernodes to contain a non-negative number of labels, we need s ≥ ≤ m ≤ pj .For such s, j, m, p , Theorem 4.3 states that the cell counting sequence for T s,j,m,p solvesnested recursion (4.1) with sufficient number of initial conditions that follow the tree. InSection 4, we observed that T s,j,m,p remains well-defined for negative m as long as m > − p .However, for − p < m <
0, the pruning operation on T s,j,m,p ( n ) defined in Section 4 does notproduce a tree T ∗ s,j,m,p ( n ) that conforms to the labelling rules described earlier, so this prun-ing operation can not be used to derive a nested recursion whose solution sequence is thecell counting sequence for this tree (for a specific example consider pruning T , , − , (168)).However, this does not exclude the possibility that there exists an alternative pruning op-eration that does lead to suitable nested recursions in case − p < m <
0. This leads to ourfirst question:
Open problem . Fix s, j, m, p such that p, j ≥ , s ≥ , − p < m < . Does there exista 2-ary order p recursion of the form (1.1) that has a solution sequence given by the cellcounting sequence of T s,j,m,p ? For s = 0 and m = bj for some b , T s,j,m,p is a superposition of T ,j,j and T ,j, from Section2 and its frequency function is a linear combination bφ H ,j + ( p − b ) φ R ,j . Therefore, theanswer to the above question would allow to determine whether there exists a 2-ary, order p recursion of the form (1.1) whose solution sequence is a linear combination of φ H ,j and φ R ,j with negative coefficients.6.2. Linear Combinations of Frequency Sequences with Negative Coefficients.
In order to consider this possibility, we begin by investigating the possibility of k -ary, order p recursions R ( n ) with slow solutions that have frequency sequences of the form λ + δφ C k ,where λ and δ are constants. In this case, some necessary conditions must be met by λ and δ .As R ( n ) is slow we require that λ + δφ C k ( v ) ≥ n . Since φ C k is unbounded itmust be the case that δ ≥
0, and since φ C k ( v ) = 1 for many values of v , it must be that λ + δ ≥ h v be the last occurrence of v in the sequence R ( n ). Since R ( n ) is slow we havethat h v = (cid:80) vi =1 ( λ + δφ C k ( i )). It can be easily verified that lim v →∞ v (cid:80) vn =1 φ C k ( i ) = k/ ( k −
1) because C k ( n ) /n converges to ( k − /k (see [12]). Therefore, lim v →∞ h v /v = λ + δ kk − .But note that R ( h v ) = v , and so we have that lim v →∞ R ( h v ) /h v = k − k − λ + kδ .It is not hard to see that R ( n ) /n must have a limit (since it is slow with frequency sequence λ + δφ C ). On the other hand, as R ( n ) is a solution to a k -ary order p recursion of the form(1.1), its recursive structure implies that any limit of R ( n ) /n must be either zero or k − kp (see,for example, [3] Theorem 2.1). In our case, the limit can not be zero since the subsequential limit k − k − λ + kδ is nonzero for k ≥
2. Thus, the limit of R ( n ) /n must be k − kp , and equatingthis to the subsequential limit k − k − λ + kδ implies that ( k − λ = k ( p − δ ). So k divides λ since k is relatively prime to k −
1. Therefore, λ = γk with γk + δ ≥ δ ≥
0. Thisexplains why in the previous section we limited our consideration to frequency sequencesof the form γφ H k + δφ C k = γk + δφ C k , as well as the close connection of this material totree superpositions of the trees associated with H k and C k .The conditions γk + δ ≥ δ ≥ γ . In fact, if thefrequency sequence of R ( n ) is γk + δφ C with γ < R ( n ) is the leaf counting function C T ( n ) of the tree T m,p,k with p = ( k − γ + δ ≥ m = p − γ < p −
1. Our proofof Theorem 5.1 does not work for m < p − T m,p,k is welldefined. This leads to the following open problem of finding a k -ary, order p recursion thatis satisfied by C T ( n ) for such T . Open problem . Fix γ < and δ ≥ such that γk + δ ≥ . Does there exits a k -ary order p recursion of the form (1.1) that has a slow solution with frequency sequence γk + δφ C k ? In fact, for fixed choices of γ and δ , there can be a multitude of recursions whose solutionsequences have the common frequency sequence γk + δφ C k . Classifying all such recursionsis nontrivial. Initial empirical evidence suggests that the following recursion may be a goodcandidate with p = ( k − γ + δ : R ( n ) = k (cid:88) i =1 R ( n − ( i − p + γ ) − R ( n − − | γ | (cid:88) t =1 R ( n − − tk ) − p −| γ |− (cid:88) t =1 R ( n − − | γ | k − t )) . Ceiling Function Solutions to k -ary Order p Recursions.
In [3] it is shown that (cid:100) n/ p (cid:101) is the solution of (3.1). In the k -ary recursion (5.3), if we set p = ( k − q and m = kq − (cid:100) n/kq (cid:101) (this follows because the resulting tree associatedwith this recursion contains kq labels in each leaf, and no labels elsewhere).Observe that if the ceiling function (cid:100) anb (cid:101) is a solution of any k -ary order p recursions ofthe form (1.1) with integers a and b , then it must be that ab = k − kp . This follows from thefact that R ( n ) /n converges to ab while the recursive structure of (1.1) implies that such alimit must be either 0 or k − kp (see [3]). This leads to the following open question that firstappears in [6]: Open problem . For k ≥ and p not dividing k − , does the ceiling function (cid:100) ( k − nkp (cid:101) satisfy a k -ary order p recursion of the form (1.1)? It is conjectured in [6] that no ceiling function solutions can occur unless p = ( k − q .6.4. More Simultaneous Parameters.
In [5] the simultaneous parameter q is introducedin the following 2-ary recursion: R s,j, − q ( n ) = R s,j, − q ( n − s − R s,j, − q ( n − j )) + R s,j, − q ( n − s − j − R s,j, − q ( n − j + q )) (6.1)where s is a nonnegative integer, j is a positive integer, and q is an integer with 0 ≤ q ≤ j .This recursion is then solved using a tree-based methodology. ESTED RECURSIONS, SIMULTANEOUS PARAMETERS AND TREE SUPERPOSITIONS 37
It is shown that the parameter q is analogous to the parameter m , and as with m , thesolution counts the number of nonempty cells in the leaves of an appropriately constructedinfinite binary tree. Note the use of the negative subscript on q in (6 . .
1) as the “negative” end of recursion (1 . m = q = 0,the two coincide and are the same as recursion (1 . . .
1) to q < q > j appears to never lead to well-defined solution sequences.Since there are strong analogies between the parameters m and q for certain arity 2 recur-sions, and since we showed in Section 5 that it is possible to introduce the simultaneousparameter m into arity k recursions that can be solved by our tree-based methods, it isnatural to ask if we can do something similar for q . As a possible candidate for the arity k recursion family we now introduce the simultaneous parameter j into the k -ary Conollyrecursion 5.1 as follows: C s,j,k ( n ) = k (cid:88) i =1 C s,j,k ( n − s − ( i − j − C s,j,k ( n − ij )) (6.2)We pose the following open question: Open problem . For k ≥ is it possible to find a k -ary family of recursions based on (6.2)that can be solved by tree-based methods and that contain an analogue to the simultaneousparameter q ? References [1] B. Balamohan, Z. Li, and S. Tanny, A combinatorial interpretation for certain relatives of the Conollysequence,
J. Integer Seq. (2008), Article 08.2.1.[2] B.W. Conolly, Fibonacci and meta-Fibonacci sequences, in: S. Vajda. ed., Fibonacci & Lucas Numbersand the Golden Section: Theory and Applications , E. Horwood Ltd., Chichester, 1989, 127–139.[3] A. Erickson, A. Isgur, B.W. Jackson, F. Ruskey and S. Tanny, Nested Recurrence Relations withConolly-like Solutions,
Siam J. Discrete Math , (1) (2012), 206–238[4] D. R. Hofstadter, G¨odel, Escher, Bach: An Eternal Golden Braid , Random House, 1979.[5] A. Isgur, Solving nested recursions with trees,
Ph. D. thesis , 2012, University of Toronto.[6] A. Isgur, V. Kuznetsov, and S. Tanny, Nested recursions with ceiling function solutions,
J.Difference Equations and Applications , J. of Difference Equations and Applications , 2011, 1–10(DOI:10.1080/10236198.2012.662967).[7] A. Isgur, V. Kuznetsov, and S. Tanny, A combinatorial approach for solving certain nestedrecursions with non-slow solutions,
J. Difference Equations and Applications , 2012, 1–10 DOI:10.1080/10236198.2012.662967.[8] A. Isgur, M. Rahman, On variants of Conway and Conolly’s Meta-Fibonacci recursions,
Electron. J.Combin. (1) (2011), P96.[9] A. Isgur, M. Rahman, and S. Tanny, Solving non-homogeneous nested recursions using trees, Annalsof Combinatorics , to appear; arXiv:1105.2351v2[10] A. Isgur, D. Reiss, and S. Tanny, Trees and meta-Fibonacci sequences,
Electron. J. Combin. (2009),R129.[11] B. Jackson and F. Ruskey, Meta-Fibonacci sequences, binary trees and extremal compact codes, Elec-tron. J. Combin. (2006), R26.[12] F. Ruskey and C. Deugau, The combinatorics of certain k -ary meta-Fibonacci sequences, J. IntegerSeq. (2009), Article 09.4.3.(Abraham Isgur, Mustazee Rahman, and Stephen Tanny) Department of Mathematics, University ofToronto, 40 St. George Street, Toronto, ON M5S 2E4, Canada (Vitaly Kuznetsov)
Courant Institute of Mathematical Sciences, New York University, 251Mercer Street, New York, NY 10012-1185, USA
E-mail address , Abraham Isgur: [email protected]
E-mail address , Vitaly Kuznetsov: [email protected]
E-mail address , Mustazee Rahman: [email protected]
E-mail address , Stephen Tanny:, Stephen Tanny: