aa r X i v : . [ m a t h . C O ] F e b ANTI-PALINDROMIC COMPOSITIONS
GEORGE E. ANDREWS, MATTHEW JUST, AND GREG SIMAY
Abstract.
A palindromic composition of n is a composition of n which canbe read the same way forwards and backwards. In this paper we define ananti-palindromic composition of n to be a composition of n which has nomirror symmetry amongst its parts. We then give a surprising connectionbetween the number of anti-palindromic compositions of n and the so-calledtribonacci sequence, a generalization of the Fibonacci sequence. We concludeby defining a new q -analogue of the Fibonacci sequence, which is related tocertain equivalence classes of anti-palindromic compositions. Introduction
Let σ = ( σ , σ , . . . , σ s ) be a sequence of positive integers such that P σ i = n .The sequence σ is called a composition of n of length s . The numbers σ i are calledthe parts of the composition. The number of compositions of n equals 2 n − , whilethe number of compositions of n into s parts equals (cid:0) n − s − (cid:1) . The empty compositionis often considered the only composition of 0, having length equal to 0.1.1. Palindromic and anti-palindromic compositions. If σ i = σ s − i +1 for all i , then σ is called a palindromic composition. It is well known [11] that if pc ( n )is the number of palindromic compositions, then pc ( n ) = 2 ⌊ n ⌋ . For instance, the pc (5) = 4 palindromic compositions of 5 are(5) , (1 , , , (2 , , , and (1 , , , , . Recent work of the authors [3, 12] generalize this result to compositions that arepalindromic modulo m , where the condition σ i = σ s − i +1 is replaced with the weakercondition σ i ≡ σ s − i +1 (mod m ).If σ i = σ s − i +1 for all i = s +12 , then we say σ is an anti-palindromic composition.Let ac ( n ) be the number of anti-palindromic compositions of n . Then the ac (4) = 5anti-palindromic compositions of 4 are(4) , (1 , , (3 , , (1 , , , and (2 , , . Furthermore, let ac ( n, s ) be the number of anti-palindromic compositions of n oflength s , ac ( n ) be the number of anti-palindromic compositions of n of even length,and ac ( n ) be the number of anti-palindromic compositions of odd length (thus ac ( n ) = ac ( n ) + ac ( n )).Notice that for each anti-palindromic composition of n of length s , we can form2 ⌊ s ⌋ flip-equivalent anti-palindromic compositions of n of length s by switching anynumber of the pairs σ i and σ s − i +1 ( i = s +12 ). For instance, the anti-palindromiccompositions(1 , , , , , (1 , , , , , (4 , , , , , and (4 , , , , n ac ( n ) ac ( n ) ac ( n ) rac ( n ) rac ( n ) rac ( n )0 1 0 1 1 0 11 0 1 1 0 1 12 0 1 1 0 1 13 2 1 3 1 1 24 2 3 5 1 2 35 4 5 9 2 3 56 8 9 17 3 5 87 14 17 31 5 8 138 26 31 57 8 13 219 48 57 105 13 21 3410 88 105 193 21 34 55 Table 1.
Values of ac ( n ), ac ( n ), ac ( n ), rac ( n ), rac ( n ), and rac ( n ) for n ≤ n form a partition of the set of all anti-palindromiccompositions of n , and we refer to each equivalence class as a reduced anti-palindromiccomposition of n of length s . Let rac ( n ) equal the total number of reduced anti-palindromic compositions of n , and rac ( n, s ) equal the number of reduced anti-palindromic compositions of n of length s . Furthermore, let rac ( n ) and rac ( n )equal the total number of even and odd reduced anti-palindromic compositions of n , respectively. Clearly we have rac ( n ) = rac ( n )+ rac ( n ). Since each equivalenceclass contains 2 ⌊ s ⌋ anti-palindromic compositions, it follows that rac ( n, s ) = ac ( n, s )2 ⌊ s ⌋ . Our primary results regarding the formulae for these functions come from obser-vations made in Table 1 and Table 2.1.2.
The k -binocci numbers. Recall the n th Fibonacci numer is given by f ( n ) =0 for n < f (1) = 1, and f ( n ) = f ( n −
1) + f ( n −
2) for all n ≥ . The n th tribonacci number is given by f ( n ) = 0 for n < f (1) = 1, and f ( n ) = f ( n −
1) + f ( n −
2) + f ( n −
3) for n ≥
2. The sequence begins0 , , , , , , , , . . . see OEIS A000073 [15]. It has been suggested tribonacci numbers appear in Dar-win’s Origins of Species in a similar relation to elephant population growth asFibonacci numbers bear to rabbit populations [13]. In general, we can define the n th k -binocci number by f k ( n ) = 0 for n < f k (1) = 1, and f k ( n ) = k X i =1 f k ( n − i )for n >
1. Connections between k -bonacci numbers for various k have been studiedby Bravo and Luca [5]. In a paper of Benjamin, Chinn, Scott, and Simay [4], In some applications the offset f (0) = f (1) = 1 is used. NTI-PALINDROMIC COMPOSITIONS 3 n ac ( n, ac ( n, ac ( n, ac ( n, ac ( n, ac ( n, n rac ( n, rac ( n, rac ( n, rac ( n, rac ( n, rac ( n, Table 2.
Values of ac ( n, s ) (top) and rac ( n, s ) (bottom) for n ≤ k -binocci are developed. For instance, we have f ( n + 1) = ⌊ n/ ⌋ X j =0 ( − j (cid:18) n − jj (cid:19) n − jn − j n − j − . The k -binocci numbers also play a role in computing the probability of flippingexactly k consecutive heads in n flips of a fair coin.1.3. Formulae for anti-palindromic compositions.
Our first result gives a sur-prising connection between the tribonacci numbers and anti-palindromic composi-tions of even length.
Theorem 1.
For all n ≥ we have ac ( n ) = 2 · f ( n − . This theorem can be deduced by a careful inspection of the identity (cid:16) q − q (cid:17) − q − q − (cid:20)(cid:16) q − q (cid:17) − q − q (cid:21) = 2 q − ( q + q + q ) . Indeed, the left hand side is X n ≥ ac ( n ) q n GEORGE E. ANDREWS, MATTHEW JUST, AND GREG SIMAY since every even-length anti-palindromic composition is a sequence of pairs of dis-tinct positive integers, and the right hand side is X n ≥ · f ( n ) q n +2 . We will give an algebraic (Section 2.1) and combinatorial (Section 2.2) proof of thisresult. Note that ac (0) = ac (0) = 1, as the empty composition is vacuously anti-palindromic. Our next result gives the number of anti-palindromic compositions of n . Theorem 2.
For all n ≥ , ac ( n ) = f ( n ) + f ( n − . We will prove Theorem 2 in Section 2.3, and also observe that (for n ≥ ac ( n ) = f ( n −
1) + f ( n − . In Section 2.4, we prove the following results which give the formulae for ac ( n, s ). Theorem 3.
Let s ≥ be a fixed integer, and G ( q, s ) = X n ≥ ac ( n, s ) q n . Then for | q | < G ( q, s ) = 2 ⌊ s/ ⌋ q ⌊ s/ ⌋ (1 − q ) s (1 + q ) ⌊ s/ ⌋ . For instance, G ( q,
0) = 1, G ( q,
1) = q − q = q + q + q + . . . , and G ( q,
2) = 2 q − q − q + q = 2 q + 2 q + 4 q + . . . , which give the (verifiable) formulae ac (0 ,
0) = 1, ac ( n,
0) = 0 for n > ac ( n,
1) = 1for n >
0, and ac ( n,
2) = 2 · ⌊ n − ⌋ for n >
1. For s ≥
2, we have the followingcorollary.
Corollary 1.
Let a be a positive integer. If s = 2 a , then ac ( n, s ) = X r +2 t = n − a a (cid:18) a + r − r (cid:19)(cid:18) a + t − t (cid:19) , and if s = 2 a + 1 , then ac ( n, s ) = X r +2 t = n − a a (cid:18) a + rr (cid:19)(cid:18) a + t − t (cid:19) , By the observation in Section 1.1 regarding rac ( n, s ) and ac ( n, s ), we also havea formula for rac ( n, s ) by dividing by the appropriate power of 2. NTI-PALINDROMIC COMPOSITIONS 5
An observation regarding the Fibonacci numbers.
Recall the followingtwo q -analogues of the Fibonacci numbers, F n ( q ) = n = 0 , n = 1 ,F n − ( q ) + q n − F n − ( q ) n > , and ˆ F n ( q ) = n = 0 , n = 1 , ˆ F n − ( q ) + q n − ˆ F n − ( q ) n > . These are referred to as q -analogues due to the property that F n ( q ) → f ( n ) andˆ F n ( q ) → f ( n ) as q → − . Properties of these two sequences of polynomials havebeen studied extensively, see for instance [1, 6, 7, 14].We define a new q -analogue of the Fibonacci numbers, which will have a con-nection to the anti-palindromic compositions. Define φ n ( q ) = q n = 1 ,q n = 2 ,q + q n = 3 ,φ n − ( q ) + φ n − ( q ) + ( q − φ n − ( q ) n > . Clearly φ n ( q ) → f ( n ) as q → − for all n ≥
1, and our final result gives acombinatorial description of the coefficients of these polynomials. For convenience,we set φ ( q ) = 1. Theorem 4.
The coefficient of q s in the polynomial φ n ( q ) equals rac ( n, s ) . We will give a proof of Theorem 4 in Section 2.5. The first few polynomials φ n ( q ) are given below, where the coefficients can be compared with Table 2. φ ( q ) = 1 φ ( q ) = q + 2 q + 2 q φ ( q ) = q φ ( q ) = q + 2 q + 4 q + q φ ( q ) = q φ ( q ) = q + 3 q + 6 q + 2 q + q φ ( q ) = q + q φ ( q ) = q + 3 q + 9 q + 5 q + 3 q φ ( q ) = q + q + q φ ( q ) = q + 4 q + 12 q + 8 q + 8 q + q . Also in Section 2.5, we deduce the following corollary.
Corollary 2.
For n ≥ we have rac ( n ) = f ( n − , rac ( n ) = f ( n − , and rac ( n ) = f ( n ) . We summarize our results regarding ac ( n ) and rac ( n ) for sufficiently large n below, illustrating the elegance of the formulae. ac ( n ) = 2 · f ( n − rac ( n ) = f ( n − ac ( n ) = f ( n −
1) + f ( n − rac ( n ) = f ( n − ac ( n ) = f ( n ) + f ( n − rac ( n ) = f ( n ) . GEORGE E. ANDREWS, MATTHEW JUST, AND GREG SIMAY Proofs of theorems
Algebraic proof of Theorem 1.
In light of the fact that ac (1) = ac (2) =0, ac (3) = ac (4) = 2, and ac (5) = 4, we see that the theorem is true for n < n ≥
6. Clearly we can construct an anti-palindromic composition of n from one of two fewer parts by inserting j at the beginning and k at the end(making sure j = k ), where if the inner composition is a composition of m , then j + k must equal m − n . Hence ac ( n ) = n − X m =0 ( n − m − − χ ( n − m )) ac ( m ) , where χ ( j ) = 1 if j is even and 0 if j is odd. The term ( n − m − − χ ( n − m ))accounts for the number of j and k . Hence ac ( n ) − ac ( n −
1) = n − X m =0 ( n − m − − χ ( n − m )) ac ( m ) − n − X m =0 ( n − − m − − χ ( n − − m )) ac ( m )= (2 − χ (3)) ac ( n −
3) + n − X m =0 ( n − m − − χ ( n − m )) ac ( m ) − n − X m =0 ( n − − m − − χ ( n − − m )) ac ( m )=2 ac ( n −
3) + 2 n − X m =0 χ ( n − m − ac ( m ) . Thus ac ( n ) − ac ( n − − ac ( n −
3) = 2 n − X m =0 χ ( n − m − ac ( m ) . Let r ( n ) = ac ( n ) − ac ( n − − ac ( n − r ( n ) + r ( n −
1) =2 n − X m =0 χ ( n − m − ac ( m )+ 2 n − X m =0 χ ( n − m − ac ( m )=2 n − X m =0 ac ( m )since χ ( n ) + χ ( n −
1) = 1 and χ (3) = 0. Therefore, r ( n ) + r ( n − − ( r ( n −
1) + r ( n − ac ( n − , and simplifying we obtain ac ( n ) − m ( n − − ac ( n − − ac ( n −
3) = 0 . NTI-PALINDROMIC COMPOSITIONS 7
This is the defining recurrence for f ( n ), and since 2 · f ( n −
2) = ac ( n ) for n > ac ( n ) = 2 · f ( n −
2) for all n ≥ Combinatorial proof of Theorem 1.
We begin with a lemma regardingthe tribonacci numbers.
Lemma 1.
For n ≥ , the tribonacci number f ( n ) equals number of compositionsof n − with parts equal to 1, 2, or 3.Proof. First note that f (2) = 1, f (3) = 2, and f (4) = 4. Since the compositionsof 1, 2, and 3 only consist of parts equal to 1, 2, or 3, and the number of compositionsof n is equal to 2 n − , the lemma holds for n ≤
4. Now for n >
4, each compositionof n − n − n − n − n − f ( n −
3) + f ( n −
2) + f ( n −
1) = f ( n ). (cid:3) We will now show that for n ≥ n − n . Since ac (1) = 0 = 2 · f ( −
1) and ac (2) = 0 = 2 · f (0) this will establish the theorem. Proof of Theorem 1.
For n = 2 we see that ac (2) = 2 · f (0) = 0, so for any n ≥ σ of n − σ to construct a sequence of pairs of distinct positive integers with sum equalto n .Now recall a partition of n is a composition of n where the parts are written innon-increasing order. Let σ + τ denote sequence concatenation, as in (1 , ,
5) =(1 , , , σ , we can find partitions λ , λ , . . . , λ r with partsequal to 1 or 2 (or the empty partition, ∅ ) such that σ = λ + σ + λ . . . + σ r + λ r , where each σ j is either equal to the composition (3) or equal to the composition(1 , σ = (2 , , , , , , , , , , , σ as λ = (2) σ = (3) λ = (1) σ = (1 , λ = (2 , , σ = (1 , λ = (1) σ = (3) λ = ∅ . It is not difficult to see that this decomposition is unique; the only way a segmentin the composition that is a partition with parts equal to 1 or 2 terminates is withthe segment (3) or the segment (1 , GEORGE E. ANDREWS, MATTHEW JUST, AND GREG SIMAY
Now given the decomposition σ = λ + σ + λ + . . . + σ r + λ r , form a sequenceof pairs ( s , λ ), ( s , λ ), . . . , ( s r , λ r ) where s = +3, s j = +3 if σ j = (3), and s j = − σ j = (1 , , (2)) , (+3 , (1)) , ( − , (2 , , , ( − , (1)) , (+3 , ∅ ) . For each pair ( s j , λ j ) we now form a new pair ( b j , c j ) in the following way. Startwith b j = 2 and c j = 1. For each 2 in the partition λ j increase both b j and c j byone. For each 1 in the partition λ j increase b j by one. We now have pairs ( b j , c j )of positive integers such that b j > c j . Now if s j = +3 we are done. If s j = −
3, weswitch the numerical values of b j and c j so that b j < c j , and then we are done.Finally form the anti-palindromic composition τ = ( τ , τ , . . . , τ r ) by setting τ j = b j and τ r − j +1 = c j . Notice that though we started with a composition of n − n ; the addition of 3 came from inserting s = +3. Inour toy example we have τ = (3 , , , , , , , , , . We have now embedded the compositions of n − n . We still need to embed a second,disjoint copy. To do this we return to the pairs ( s j , λ j ) and make a new collectionof pairs ( s ′ j , λ j ) by setting s ′ j = − s j . Now following the same procedure as beforewe construct an anti-palindromic word τ ′ that is, in fact, the reverse of τ . Againlooking at our example from before we have τ ′ = (2 , , , , , , , , , . To show that these two embedded sets are disjoint, notice that for a composition τ formed by using s = +3 we have τ > τ r , and that for a word τ ′ formed byusing s = − τ < τ r .Showing this process reverses and that we can send the pairs { τ, τ ′ } of an anti-palindromic composition of n and its reverse back to a composition of n − (cid:3) Proof of Theorem 2.
In this section we develop the formula for ac ( n ). Westart by proving some initial observations regarding ac ( n ), ac ( n ), ac ( n ), and ac ( n, s ). Proposition 1.
For all n ≥ , we have ac ( n ) = f ( n −
1) + f ( n − . Proof.
This is just two applications of the defining recurrence for f ( n ), recallingthat f ( n ) = 0 for n < f ( n −
1) + f ( n −
5) = f ( n −
2) + f ( n −
3) + f ( n −
4) + f ( n − f ( n −
2) + f ( n − · f ( n − ac ( n ) (cid:3) Proposition 2.
We have ac (0 ,
0) = 1 , ac (0 ,
1) = 0 , and for all n ≥ and s ≥ ac ( n, s ) + ac ( n, s + 1) = ac ( n + 1 , s + 1) . NTI-PALINDROMIC COMPOSITIONS 9
Proof.
When n = 0, there is only one composition (the empty composition) whichhas length 0.Now any anti-palindromic composition σ of n + 1 of length 2 s + 1 has a centralpart σ s +1 . If σ s +1 = 1, this composition can be formed from an anti-palindromiccomposition of n of length 2 s by adding a central part equal of 1. If σ s +1 >
1, thiscomposition can be formed from an anti-palindromic composition of n of length2 s + 1 by adding 1 to the central part. Therefore, ac ( n, s ) + ac ( n, s + 1) = ac ( n + 1 , s + 1). (cid:3) Proposition 3.
For n ≥ and s ≥ ac ( n, s + 1) = n − X j =0 ac ( j, s ) , where in the case n = 0 we take the empty sum to be 0.Proof. Let n >
0. Then by applying Proposition 2 n times, we have ac ( n, s + 1) = ac ( n − , s + 1) + ac ( n − , s )= ac ( n − , s + 1) + ac ( n − , s ) + ac ( n − , s )...= ac (0 , s + 1) + n − X j =0 ac ( j, s ) . Since ac (0 , s + 1) = 0 for all s ≥
0, the result follows. (cid:3)
Proposition 4.
For all n ≥ ac ( n ) = ac ( n + 1) . Proof. If n = 0, we see that ac (0) = ac (1) = 1. If n >
0, by definition we have ac ( n ) = X s ≥ ac ( n, s )= X j ≥ ( ac ( n, j ) + ac ( n, j + 1))= X j ≥ ac ( n + 1 , j + 1)by Proposition 2. But this last expression is equal to ac ( n + 1). (cid:3) Proposition 5.
For n ≥ ac ( n ) = n − X j =0 ac ( j ) , where in the case n = 0 we take the empty sum to be 0.Proof. For n >
0, we have by Proposition 4 ac ( n ) = ac ( n − ac ( n −
1) + ac ( n − . Now if n = 1, we are done since ac (0) = 0. If n >
1, we can again apply Proposition4 to get ac ( n ) = ac ( n −
1) + ac ( n − . Repeating the same argument n − (cid:3) Proposition 6.
For all n ≥ , we have n X j =0 f ( j ) = f ( n ) + f ( n + 2) − . Proof.
We give a proof by mathematical induction. For n = 0, f (0) = 0 = f (0) + f (2) − . Now for n >
0, suppose the proposition holds for all k < n . Then n X j =0 f ( j ) = n − X j =0 f ( j ) + f ( n )= f ( n −
1) + f ( n + 1) −
12 + f ( n )= f ( n + 2) − f ( n ) −
12 + f ( n )= f ( n ) + f ( n + 2) − . (cid:3) Proposition 7.
For all n ≥ , ac ( n ) = f ( n −
3) + f ( n − . Proof.
By Proposition 5, Theorem 1, and Proposition 6 we have ac ( n ) = n − X j =0 ac ( j )=2 n − X j =0 f ( j −
2) + ac (0)= 2 n − X j =0 f ( j ) + ac (0)= f ( n −
3) + f ( n − − ac (0) . Since ac (0) = 1, the result follows. (cid:3) Proof of Theorem 2.
Theorem 2 now immediately follows from Proposition 7, since ac (1) = 1 = f (1) + f ( − n ≥ ac ( n ) = ac ( n ) + ac ( n )= 2 · f ( n −
2) + f ( n −
3) + f ( n − f ( n ) + f ( n − . (cid:3) NTI-PALINDROMIC COMPOSITIONS 11
Proof of Theorem 3 and Corollary 1.
In this section we develop theformulae for ac ( n, s ) by deriving the ordinary generating function G ( q, s ) for afixed s ≥
0. It is easier to split into the cases when s is even and odd.Suppose s = 2 a , where a ≥
0. An anti-palindromic composition of n of length2 a consists of a sequence of a ordered pairs of distinct positive integers. If d ( n ) isthe number of distinct pairs of positive integers that sum to n , then D ( q ) := X n ≥ d ( n ) q n = (cid:18) q − q (cid:19) − q − q = 2 q (1 − q )(1 − q ) . To see why this is the case, notice that (cid:18) q − q (cid:19) = (cid:0) q (cid:1) + (cid:0) q + q (cid:1) + (cid:0) q + q + q (cid:1) + . . . and q − q = q + q + q + . . . , so we are taking all pairs of positive integers and subtracting the repeated pairs.To form a sequence of a such pairs, we multiple D ( q ) by itself a times, showingthat G ( q, a ) = [ D ( q )] a = 2 a q a (1 − q ) a (1 − q ) a . To prove the first half of Corollary 1, recall that for a > − q ) a = X n ≥ (cid:18) a − nn (cid:19) q n and 1(1 − q ) a = X n ≥ (cid:18) a − nn (cid:19) q n . Multiplying these two series and reindexing gives the result.Now suppose s = 2 a + 1, where a ≥
0. An anti-palindromic composition of n oflength 2 a + 1 still consists of a ordered pairs of distinct positive integers, with anadditional central part. Therefore, G ( q, a + 1) = G ( q, a ) · q − q = 2 a q a (1 − q ) a (1 − q ) a . The second half of Corollary 1 follows the same way as the first half once we observe1(1 − q ) a +1 = X n ≥ (cid:18) a + nn (cid:19) q n . Proof of Theorem 4 and Corollary 2.
We begin with a lemma.
Lemma 2.
For n ≥ and s ≥ we have ac ( n, s ) = ac ( n − , s ) + ac ( n − , s ) + 2 · ac ( n − , s − − ac ( n − , s ) . Proof. If σ is an anti-palindromic composition of n ≥ s ≥
2, let m σ := σ + σ s . Observe that m σ ≥ δ ( m σ ) ≤ | σ − σ s | ≤ m σ − , where δ ( m σ ) = 1 if m σ is odd and δ ( m σ ) = 2 if m σ is even. Let us first count the number of anti-palindromic compositions of n of length s with m σ = 3. Each one of these compositions can be formed by taking an anti-palindromic composition of n − s − n of length s with m σ = 3 equals2 · ac ( n − , s − n of length s with m σ >
3. Now for any anti-palindromic composition τ of n − s ,we can form an anti-palindromic composition of n of length s by adding 1 to τ if τ > τ s , or adding 1 to τ s if τ s > τ . Now in this way, we have constructed all theanti-palindromic compositions of n of length s with m σ > | σ − σ s | > δ ( m σ ).For any composition γ of n − s , form an anti-palindromic compositionof n of length s by adding 1 to γ and 1 to γ s . In this way, we have constructed all theanti-palindromic compositions of n of length s with m σ > | σ − σ s | ≤ m σ − n of length s with m σ > δ ( m σ ) ≤ | σ − σ s | ≤ m σ − apc ( n − , s ) + apc ( n − , s )minus the anti-palindromic compositions of n of length s with m σ > δ ( m σ ) < | σ − σ s | ≤ m σ −
4, as we have counted these compositions exactly twice. To provethe lemma, we now must show that the number of compositions that we countedtwice equals apc ( n − , s ).Let ρ be an anti-palindromic composition of n − s . Form an anti-palindromic composition of n of length s by adding 2 to ρ and 1 to ρ s if ρ > ρ s ,or 1 to ρ and 2 to ρ s if ρ s > ρ . In this way, we have constructed all of the anti-palindromic compositions of n of length s with δ ( m σ ) < | σ − σ s | ≤ m σ − (cid:3) Proof of Theorem 4.
The theorem can be verified for all n and s with n + s < φ ( q ) = rac (0 , · q + rac (0 , · q + rac (0 , · q = 1 · q + 0 · q + 0 · q φ ( q ) = rac (1 , · q + rac (1 , · q + rac (1 , · q = 0 · q + 1 · q + 0 · q φ ( q ) = rac (2 , · q + rac (2 , · q + rac (2 , · q = 0 · q + 1 · q + 0 · q φ ( q ) = rac (3 , · q + rac (3 , · q + rac (3 , · q = 0 · q + 1 · q + 1 · q . Let [ q s ] φ n ( s ) be the coefficient of q s in the polynomial φ n ( s ). Now for n ≥ s ≥
2, using the defining recurrence for φ n ( q ) we have[ q s ] φ n ( q ) = [ q s ] φ n − ( q ) + [ q s ] φ n − ( q ) + [ q s − ] φ n − ( q ) − [ q s ] φ n − ( q )= rac ( n − , s ) + rac ( n − , s ) + rac ( n − , s − − rac ( n − , s )by induction. Using the relationship between rac ( n, s ) and ac ( n, s ),[ q s ] φ n ( q ) = ac ( n − , s )2 ⌊ s ⌋ + ac ( n − , s )2 ⌊ s ⌋ + ac ( n − , s − ⌊ s − ⌋ − ac ( n − , s )2 ⌊ s ⌋ = ac ( n − , s )2 ⌊ s ⌋ + ac ( n − , s )2 ⌊ s ⌋ + 2 · ac ( n − , s − ⌊ s ⌋ − ac ( n − , s )2 ⌊ s ⌋ = ac ( n, s )2 ⌊ s ⌋ by Lemma 2. Therefore, [ q s ] φ n ( s ) = rac ( n, s ). (cid:3) NTI-PALINDROMIC COMPOSITIONS 13
Proof of Corollary 2.
Notice that by Theorem 4 we have rac ( n ) = X s ≥ rac ( n, s ) = φ n (1) = f ( n ) . As for rac ( n ), we have rac (1) = rac (2) = 0, rac (3) = 1, and for n ≥ rac ( n ) = X s ≥ rac ( n, s ) = φ n (1) + φ n ( − , again using Theorem 4. By the definition of φ n ( q ), this equals φ n − (1) + φ n − ( − φ n − (1) + φ n − ( − rac ( n −
1) + rac ( n − . This is the defining recurrence relation for the Fibonacci numbers, thus we concludethat rac ( n ) = f ( n − rac ( n ), we have rac (0) = 1, rac (2) = rac (3) = 1, and for n ≥ rac ( n ) = X s ≥ rac ( n, s + 1) = φ n (1) − φ n ( − φ n ( q ), this equals φ n − (1) − φ n − ( − φ n − (1) − φ n − ( − rac ( n −
1) + rac ( n − . This is the defining recurrence relation for the Fibonacci numbers, thus we concludethat rac ( n ) = f ( n − (cid:3) Acknowledgements
The second author (M.J.) was partially supported by the Research and TrainingGroup grant DMS-1344994 funded by the National Science Foundation. We thankRobert Schneider and Drew Sills for helpful comments.
References [1] G. E. Andrews, Fibonacci numbers and the Rogers–Ramanujan identities,
Fibonacci Quart. (2004), 3-19.[2] K. Alladi and V. E. Hoggatt, Jr., Compositions with ones and twos, Fibonacci Quart. (1975), 233-239.[3] G. Andrews and G. Simay, Parity Palindrome Compositions, submitted[4] A. T. Benjamin, P. Chinn, J. N. Scott, and G. Simay, Combinatorics of two-toned tilings, Fibonacci Quart. (2011).[5] J. J. Bravo and F. Luca, Coincidences in generalized Fibonacci sequences, J. Number Theory (2013), 2121-2137.[6] L. Carlitz, Fibonacci notes. III: q -Fibonacci numbers, Fibonacci Quart. (1974), 317–322.[7] J. Cigler, q -Fibonacci polynomials, Fibonacci Quart. (2003), 31-40.[8] S. Eger, Restricted weighted integer compositions and extended binomial coefficients, Journalof Integer Sequences , (2013).[9] S. Heubach and T. Mansour, Compositions of n with parts in a set, Congressus Numerantium. (2004), 33–51.[10] S. Heubach and T. Mansour,
Combinatorics of Compositions and Words , CRC Press, 2010.[11] V. E. Hoggatt, Jr. and M. Bicknell, Palindromic compositions,
Fibonacci Quart. , (1975), 350-356.[12] M. Just, Compositions that are palindromic modulo m , submitted.[13] ´A. Kun, J. Podani and J. Szil´agyi, How fast does Darwin’s elephant population grow?, J HistBiol (2018), 259-281. [14] H. Pan, Arithmetic properties of q -Fibonacci numbers and q -pell numbers, Discrete Mathe-matics306