A generalization of Eulerian numbers via rook placements
Esther Banaian, Steve Butler, Christopher Cox, Jeffrey Davis, Jacob Landgraf, Scarlitte Ponce
aa r X i v : . [ m a t h . C O ] N ov A generalization of Eulerian numbersvia rook placements
Esther Banaian ∗ Steve Butler † Christopher Cox ‡ Jeffrey Davis § Jacob Landgraf ¶ Scarlitte Ponce k Abstract
We consider a generalization of Eulerian numbers which count the numberof placements of cn “rooks” on an n × n board where there are exactly c rooksin each row and each column, and exactly k rooks below the main diagonal.The standard Eulerian numbers correspond to the case c = 1. We show thatfor any c the resulting numbers are symmetric and give generating functions ofthese numbers for small values of k . Rook placements on boards have a wonderful and rich history in combinatorics (see,e.g., Butler, Can, Haglund and Remmel [1]). Traditionally the rooks are placed ina non-attacking fashion (i.e., at most one rook in each row and column) and thecombinatorial aspects come from considering variations on the board shapes.Instead of varying the board, we could also change the restrictions on how manyrooks are allowed in each row and each column. If we have a square board and thenumber of rooks in each column and row is fixed, then this corresponds to count-ing non-negative matrices with fixed row and column sums (c.f.
A000681 , A001500 , A257493 , etc., in the OEIS [6]).In this paper, we will look at this latter case of placing multiple rooks in eachrow and column more closely. We begin in Section 2 by exploring the connectionsbetween these rook placements and juggling patterns. In Section 3 we look at Euleriannumbers (which correspond to the number of non-attacking rook placements on an ∗ College of St. Benedict, Collegeville, MN 56321, USA [email protected] † Iowa State University, Ames, IA 50011, USA [email protected] ‡ Carnegie Mellon University, Pittsburgh, PA 15213, USA [email protected] § University of South Carolina, Columbia, SC 29208, USA [email protected] ¶ Michigan State University, East Lansing, MI 48824, USA [email protected] k California State University, Monterey Bay, Seaside, CA 93955, USA [email protected] × n board with a fixed number of rooks below the main diagonal) and in Section 4generalize to the case in which c rooks are placed in each row and each column.In Section 5, we provide generating functions for special cases of these generalizedEulerian numbers. We end with concluding remarks and open problems in Section 6. Juggling patterns can be described by a siteswap sequence listing the throws that thepattern requires, i.e., t t . . . t n where at time s ≡ i (mod n ) we throw the ball so thatit will land t i beats in the future. A sequence of throws can be juggled if and only ifthere are no collisions, i.e., two balls landing at the same time, which is equivalent to1 + t , t , . . . , n + t n being distinct modulo n . One well known property of siteswapsequences is that the average of the throws is the number of balls needed to jugglethe pattern (see [4, 5]).A minimal juggling pattern is a valid juggling pattern t t . . . t n with 0 ≤ t i ≤ n − n arise by starting from some minimal juggling pattern and adding multiplesof n to the various throws (such additions do not affect modular conditions). Moreabout this approach is found in Buhler, Eisenbud, Graham and Wright [4].This naturally leads to the problem of enumerating minimal juggling patterns.This is done by relating such patterns to rook placements on a square board. Inparticular we will consider the n × n board B n , with labels on each cell ( i, j ) givenby the following rule: (cid:26) j − i if j ≥ i,n + j − i if j < i. We can interpret the rows of B n as the throwing times (modulo n ) and the columnsof B n as the landing times (modulo n ). The label of the cell ( i, j ) is then the smallestpossible throw required to throw at time i and land at time j .Given a minimal juggling pattern t t . . . t n we form a rook placement by placinga rook in row i on the cell labeled t i for 1 ≤ i ≤ n (note that this forces the rookto be placed in the column corresponding to the landing time modulo n ). Sincelanding times are unique modulo n no two rooks will be in the same column, sothis forms a non-attacking rook placement with n rooks. Conversely, given a non-attacking rook placement with n rooks we can form a minimal juggling pattern byreading off the cell labels of the covered square starting at the first row and readingdown. This establishes the bijective relationship between minimal juggling patternsand non-attacking rook patterns on B n . An example of this is shown in Figure 1 forthe minimal juggling pattern 24234.We can extract information about the minimal juggling pattern by properties ofthe rook placements, including, for example, the number of balls.2123440123340122340112340Figure 1: A non-attacking rook placement on B corresponding to the minimal jug-gling pattern 24234. Proposition 1.
The number of rooks below the main diagonal in a non-attackingrook placement on B n is the same as the number of balls necessary to juggle thecorresponding minimal juggling pattern.Proof. Suppose there are k rooks below the main diagonal in a placement of n non-attacking rooks on B n . Then when we sum the labels of all the cells covered by arook, i.e. we sum the throw heights for the juggling sequence, we have n X ℓ =1 t ℓ = kn + n X j =1 j − n X i =1 i = kn. Since the average of the throws is the number of balls needed for the sequence, theclaim follows.Note that in Figure 1 there are three rooks below the main diagonal and that thejuggling pattern 24234 requires three balls to juggle. c -rook placements A natural variation in juggling is to allow multiple balls to be caught and thrown ata time. This is known as multiplex juggling , and we will see that many of the basicideas generalize well to this setting.We will let c denote a hand capacity, i.e. at each beat we make c throws (allowingsome of the throws to be 0, which happens when the number of actual balls thrownis less than c ). Siteswap sequences of period n now correspond to a sequence of n sets, T T . . . T n , where each T i is a (multi-)set of the form { t i, , t i, , . . . , t i,c } , denotedin shorthand notation as [ t i, t i, . . . t i,c ]. A multiplex juggling sequence is valid if andonly if the juggling modular condition is satisfied. Namely, every 1 ≤ ℓ ≤ n appearsexactly c times in the multiset { t i,j + i (mod n ) } ≤ i ≤ n ≤ j ≤ c . In other words, no more than c balls land at each time. As in standard jugglingpatterns, the number of balls b needed to juggle the pattern relates to an average. In A 0 throw indicates a ball is not landing. n n X i =1 c X j =1 t i,j = b. We say a multiplex juggling sequence is a minimal multiplex juggling sequence if andonly if 0 ≤ t i,j ≤ n − t i,j .There is a relationship between period n , hand capacity c multiplex juggling se-quences and placements of “rooks” on B n . This is done by generalizing from non-attacking rook placements to c -rook placements , placements of cn rooks with exactly c rooks in every row and column, where multiple rooks are allowed in cells.There is a bijection between minimal multiplex juggling patterns of period n withhand capacity c and c -rook placements on B n . In particular, for each i we place c rooksin the i -th row corresponding to t i, , . . . , t i,c . Conversely, given a c -rook placement wecan form a minimal multiplex juggling sequence by letting T i denote the cells coveredby the rooks in row i (with appropriate multiplicity). An example of this is shown inFigure 2 for the minimal multiplex juggling pattern [24][02][14][22][03].0123440123340122340112340Figure 2: A 2-rook placement on B corresponding to the minimal multiplex jugglingpattern [24][02][14][22][03]By the same argument used for Proposition 1 we have the following. Proposition 2.
The number of rooks below the main diagonal in a c -rook placementon B n is the same as the number of balls necessary to juggle the corresponding minimalmultiplex juggling pattern. For example, the multiplex juggling pattern in Figure 2 requires four balls tojuggle.
The Eulerian numbers, denoted (cid:10) nk (cid:11) , are usually defined as the number of permutationsof [ n ], π = π π . . . π n , with k ascents ( π i < π i +1 ), or equivalently the number ofpermutations with k descents ( π i > π i +1 ). There is a bijection between permutationsof [ n ] with k descents and permutations with k drops ( i > π i ), so that (cid:10) nk (cid:11) also countspermutations of [ n ] with k drops (see [4]). Given an n × n board, with rows and4olumns labeled 1 , , . . . , n , we can use our permutation to form a non-attacking rookplacement by placing rooks at positions ( i, π i ). A drop in the permutation correspondsto a rook below the main diagonal, so we will call any rook below the main diagonala drop .By Proposition 1, the number of drops in a non-attacking rook placement equalsthe number of balls necessary for the corresponding juggling pattern. Therefore, (cid:10) nk (cid:11) also counts the number minimal juggling patterns of period n using k balls. (cid:10) nk (cid:11) k =0 k =1 k =2 k =3 k =4 k =5 k =6 n =1 1 n =2 1 1 n =3 1 4 1 n =4 1 11 11 1 n =5 1 26 66 26 1 n =6 1 57 302 302 57 1 n =7 1 120 1191 2416 1191 120 1Table 1: The Eulerian numbers for 1 ≤ n ≤ (cid:10) nk (cid:11) = (cid:10) nn − k − (cid:11) . This can be shown bynoting if we start with a permutation with k ascents and reverse the permutation, wenow have n − − k ascents (i.e., ascents go to descents and vice-versa; and there are n − Proposition 3.
The Eulerian numbers satisfy (cid:10) nk (cid:11) = ( n − k ) (cid:10) n − k − (cid:11) + ( k + 1) (cid:10) n − k (cid:11) . This recurrence is again proven using permutations and ascents. Here, we providean alternate proof using rook placements and drops.
Proof.
Start by considering a non-attacking rook placement on an ( n − × ( n − k − n -th row and n -th column, and place a rook inposition ( n, n ). The newly added rook is not below the diagonal and so we havenot created any new drops. We can now create one additional drop by taking anyrook (other than the one just added) which is on or above the main diagonal, say inposition ( i, j ), move that rook to position ( n, j ) and move the rook in position ( n, n )to position ( i, n ). This moves the rook in the j -th column below the main diagonalcreating a new drop. Since no other rook moves we now have precisely k drops and anon-attacking rook placement. Note that there are ( n − − ( k −
1) = n − k ways wecould have chosen which rook to move, so that in total this gives ( n − k ) (cid:10) n − k − (cid:11) boardsof size n × n with k drops. 5ow, consider a non-attacking rook placement on an ( n − × ( n −
1) board with k drops. Add an n -th row and n -th column, and place a rook in position ( n, n ). Asbefore we switch, but now only switch with a rook which is below the main diagonal(i.e., a drop). This will not change the number of drops, so the result is a non-attacking rook placement on an n × n board with k drops. There are k rooks we canchoose to switch with, or alternatively, we can leave the n -th rook in position ( n, n );thus, there are k + 1 ways to build the desired rook placement, so that in total thisgives ( k + 1) (cid:10) n − k (cid:11) boards of size n × n with k drops.Finally, we note that each n × n board with k drops is formed uniquely from oneof these operations. This can be seen by taking such a board and then noting thelocation of the rook(s) in the last row and in the last column. Suppose these are inpositions ( i, n ) and ( n, j ), respectively. We then move these rooks to positions ( i, j )and ( n, n ). This can at most decrease the number of drops by one (i.e, moving therook in the last column does not affect the number of drops). Now removing thelast row and column gives an ( n − × ( n −
1) board having a non-attacking rookplacement with either k or k − The generalized Eulerian numbers , denoted (cid:10) nk (cid:11) c , are the number of c -rook placementson the n × n board with k drops. Just as the Eulerian numbers count the number ofminimal juggling patterns of period n with k balls, the generalized Eulerian numberscount the number of minimal multiplex juggling patterns of period n with k balls andhand capacity c . Notice that the generalized Eulerian numbers reduce to the Euleriannumbers when c = 1. In Table 2 we give some of the generalized Eulerian numbersfor c = 2 and 3.These numbers appear to satisfy a symmetry property similar to Eulerian num-bers. We will give two proofs of this symmetry, one in terms of rook placements andthe other using minimal multiplex juggling patterns. Theorem 4.
Let n , k and c be non-negative integers. Then (cid:10) nk (cid:11) c = (cid:10) nc ( n − − k (cid:11) c .Proof. We construct a bijection between the rook placements with k rooks below themain diagonal and those with c ( n − − k rooks below the diagonal. Consider a rookplacement with c rooks in every row and column, and k rooks below the diagonal.Now, shift every rook one space to the right cyclically. Let us consider the numberof rooks which are strictly above the main diagonal. • All c rooks in the last column were shifted to the first column. So, none of theserooks are above the main diagonal. • All of the k rooks that were initially below the main diagonal are now either onor still below the main diagonal. 6 nk (cid:11) k =0 k =1 k =2 k =3 k =4 k =5 k =6 k =7 k =8 n =1 1 n =2 1 1 1 n =3 1 4 11 4 1 n =4 1 11 72 114 72 11 1 n =5 1 26 367 1492 2438 1492 367 26 1 (cid:10) nk (cid:11) k =0 k =1 k =2 k =3 k =4 k =5 k =6 k =7 k =8 k =9 n =1 1 n =2 1 1 1 1 n =3 1 4 11 23 11 4 1 n =4 1 11 72 325 595 595 325 72 11 1Table 2: Small values of the generalized Eulerian numbers for c = 2 and 3. • All other rooks will be above the diagonal.Since there are cn rooks on the board total, there are cn − c − k = c ( n − − k rooksabove the diagonal after this shift. Finally, we switch the rows and columns of theboard. This flips the rook placement across the main diagonal. After this transforma-tion, there are now c ( n − − k rooks below the main diagonal. This composition oftransformations is invertible by switching rows and columns then shifting every rookleft one space. Thus, the transformation gives a bijection, completing the proof.Before we can give the second proof, we must first establish some basic propertiesof (multiplex) juggling sequences. Lemma 5. If T T . . . T n satisfies the juggling modular conditions with hand capacity c , and α ∈ Z n with gcd( α, n ) = 1 , then ( αT α − )( αT α − ) . . . ( αT nα − ) , where αT i := { αt i, , . . . , αt i,c } , and the subscripts are taken modulo n , also satisfies the juggling modular conditions.Proof. We have A = { αt iα − ,j + i } ≤ i ≤ n ≤ j ≤ c = { α (cid:0) t iα − ,j + iα − (cid:1) } ≤ i ≤ n ≤ j ≤ c = { α (cid:0) t i ′ ,j + i ′ (cid:1) } ≤ i ′≤ n ≤ j ≤ c , where we use that gcd( α, n ) = 1 so that α is invertible modulo n and as i rangesbetween 1 and n , then so does i ′ := iα − . Since { t i,j + i } ≤ i ≤ n ≤ j ≤ c has c occurrences eachof 1 through n then scaling by α and taking terms modulo n we also have that A willhave c occurrences each of 1 through n . 7 emma 6. If T T . . . T n satisfies the juggling modular conditions of hand capacity c ,and β ∈ Z , then ( T + β )( T + β ) . . . ( T n + β ) , where T i + β := [ t i, + β, t i, + β, . . . , t i,c + β ] , still satisfies the juggling modular conditions.Proof. The multiset A = { ( t i,j + β ) + i } ≤ i ≤ n ≤ j ≤ c is found by taking { t i,j + i } ≤ i ≤ n ≤ j ≤ c andshifting each element by β . Since T T . . . T n satisfy the juggling modular conditionsthen so also must A . Juggling proof of Theorem 4.
We show there is a bijection between the minimal mul-tiplex juggling sequences using k balls and those using c ( n − − k balls for a fixedlength n and hand capacity c . So let T T . . . T n be a valid minimal multiplex jug-gling sequence with k balls and hand capacity c . By Lemma 5 and Lemma 6, if wescale each T i by − n − n − − T n )( n − − T n − ) . . . ( n − − T ) . We also note the resulting throws all lie between 0 and n − n n X i =1 c X j =1 (cid:0) ( n − − t i,j (cid:1) = 1 n (cid:18) cn ( n − − n X i =1 c X j =1 t i,j (cid:19) = c ( n − − k. Finally, we note that this operation is its own inverse, and thus gives the desiredbijection. k We now look at determining the values of the generalized Eulerian numbers (cid:10) nk (cid:11) c forsmall k . This depends of course on both n and c . However, for a fixed k there areonly finitely many c that need to be considered. This is a consequence of the followinglemma. Lemma 7.
For c ≥ k we have (cid:10) nk (cid:11) c = (cid:10) nk (cid:11) k .Proof. It will suffice to establish the following claim.
Claim.
Every c -rook placement with k drops has at least c − k rooks in every entryon the main diagonal. 8e proceed to establish this by using induction on k + c . For k + c = 1, the onlypossible case is k = 0 and c = 1 for which there is only one placement, namely onerook in each cell on the main diagonal.Now assume that we have established the claim for all k, c with k + c < ℓ , andlet k + c = ℓ . Let S be a c -rook placement with k drops. We can interpret therook placement as an incidence relationship of a regular bipartite graph. By Hall’sMarriage Theorem, we know we can find a perfect matching in this bipartite graphwhich corresponds to T , a 1-rook placement contained in S . Suppose there are i drops in T . Then, S − T is a ( c − k − i drops. Since( c −
1) + ( k − i ) < c + k = ℓ , by our induction hypothesis, there are at least c − k + i − S − T , and hence also in S . If i ≥
1, weare done. If i = 0, then T is again the unique 1-rook placement where every rook ison the main diagonal, so S still has at least c − k rooks on each entry on the maindiagonal.This can also be established in terms of minimal multiplex juggling patterns. Juggling proof of Lemma 7.
If there are k balls, then at each step we can throw atmost k balls, i.e., each T i has at least c − k entries of 0. It follows that in thecorresponding c -rook placement each row has at least c − k rooks on the diagonal.We will be looking at the generalized Eulerian numbers (cid:10) nk (cid:11) c for k = 1 , ,
3. ByLemma 7 this reduces down to only six cases to consider, namely, (cid:10) n (cid:11) , (cid:10) n (cid:11) , (cid:10) n (cid:11) , (cid:10) n (cid:11) , (cid:10) n (cid:11) and (cid:10) n (cid:11) . Since (cid:10) nk (cid:11) = (cid:10) nk (cid:11) , then (cid:10) n (cid:11) , (cid:10) n (cid:11) and (cid:10) n (cid:11) have been previouslydetermined (see A000295 , A000460 and
A000498 , respectively, in the OEIS [6]). Sothat leaves (cid:10) n (cid:11) , (cid:10) n (cid:11) and (cid:10) n (cid:11) and in Table 3 we give the generating function for thesethree sequences. In the remainder of this section we will demonstrate the techniquesused to determine the generating functions by working through the case for (cid:10) n (cid:11) . We break the problem of counting c -rook placements into several sub-problems accord-ing to the way the rooks below the main diagonal are placed relative to one another(i.e., relative placements instead of absolute placements). Given some generic place-ment of the k rooks below the main diagonal we can determine the number of waysto place the remaining rooks on or above the main diagonal. We then combine theresults over all possible generic placements.We will carefully work through the rook placement shown in Figure 3 which con-sists of two rooks below the main diagonal and where both rooks are in the samecolumn and different rows. Here a , b , c and d are the number of rows between thevarious transition points (a transition point to passing a rook, or rooks, in a row ora column as we move along the main diagonal).9 n ≥ (cid:10) n (cid:11) x n = x + 11 x + 72 x + 367 x + 1630 x + 6680 x + 26082 x + · · · = x − x − x − x + 5 x (1 − x ) (1 − x ) (1 − x + 5 x ) X n ≥ (cid:10) n (cid:11) x n = 4 x + 114 x + 1492 x + 13992 x + 109538 x + 769632 x + · · · = 4 x + 2 x − x + 1748 x − x + 7058 x − x +4397 x − x + 625 x (1 − x ) (1 − x ) (1 − x + 5 x ) (1 − x + 13 x ) X n ≥ (cid:10) n (cid:11) x n = x + 23 x + 325 x + 3368 x + 28819 x + 218788 x + · · · = x − x + 39 x − x + 1844 x − x + 9697 x − x + 7532 x − x + 2435 x − x (1 − x ) (1 − x ) (1 − x + 5 x ) (1 − x + 27 x − x )Table 3: Generating functions for some of the generalized Eulerian numbers.We place the remaining rooks one row at a time starting from the bottom andgoing to the top. For each new row, the way we place rooks will depend on all of thechoices we have made previously. However, it suffices to know only what is happeninglocally. In particular, we only need to know how many columns can have rooks placedinto them, as well as the respective numbers that can go into those columns. We canrepresent these by a partition of what we will call the excess (the total number ofrooks that can still be placed in the columns after the row has had its rooks placed).As we move one row up the board we will gain a new column (from the diagonal) andthe excess will change in one of several ways. • There are no rooks below or to the left of the new diagonal cell. Initially wenow have a new column that can take up to c rooks, and we place c rooks inthe row. The excess remains unchanged. • There are τ rooks below the new diagonal cell. Initially we have the new column,but that can only take up to c − τ rooks (i.e., τ rooks have already gone intothe column), and we still have to place c rooks in the row. The excess decreasesby τ . • There are σ rooks to the left of the new diagonal cell. Initially we have the newcolumn that can take up to c rooks, and we place c − σ rooks in the row (i.e.,10 × abcd Figure 3: A 2-rook placement with two rooks below the main diagonal where bothrooks are in the same column and different rows. σ rooks have already gone into the row). The excess increases by σ .We note that it is possible for the last two situations to occur simultaneously.In going from row to row we will transition from partitions of the old excess topartitions of the new excess. We illustrate this with an example in which case theexcesses are both 2. We indicate a column which can still have r rooks placed intoit by r , then underneath look at all possible ways we can place 2 rooks into thosecolumns, and finally note the resulting set of columns contributing to the new excess.2 22 0 →
20 2 →
21 1 → → → → → columns of the transitionmatrix correspond to the excess of the original row and the rows of the transitionmatrix correspond to the partitions of the excess of the new row. ! . Repeating this for all possible situations that might arise for transitioning betweenexcesses 0, 1, or 2, we get the transition matrices in the following table.11ransition from T r a n s i t i o n t o ∅ ∅ (1) (1) (cid:0) (cid:1) (cid:0) (cid:1)
21 1 (cid:18) (cid:19) (cid:18) (cid:19) (cid:18) (cid:19) We now start below the bottom row (in 1 possible way) and we move up from rowto row and multiply on the left by the transition that we perform between the tworows. At any point we stop, the resulting vector will denote the number of ways tofill up the board to that row such with a particular excess. In particular, if we carrythis procedure all the way to the top we will get a 1 × a , b , c and d rows as well as three othertransitions to make, the resulting product that gives our count is as follows(1) d (cid:0) (cid:1) (cid:18) (cid:19) c (cid:18) (cid:19) (2) b (1)(1) a . Finally, for this generic rook placement we sum over all possible choices of a , b , c and d that gives an n × n board, i.e., X a + b + c + d = n − (1) d (cid:0) (cid:1) (cid:18) (cid:19) c (cid:18) (cid:19) (2) b (1)(1) a . In order to help evaluate this sum, we will add in an extra parameter x that keepstrack of how many of each transition we made, or viewed another way the power of x corresponds to the number of rows we have. Therefore when counting the numberof placements on an n × n board, we are interested in the coefficient of x n of theexpression X a,b,c,d ≥ ( x ) d x (cid:0) (cid:1) (cid:18) x xx x (cid:19) c x (cid:18) (cid:19) (2 x ) b x ( x ) a . This sum can be decomposed as a combination of geometric sums giving X a,b,c,d ≥ ( x ) d x (cid:0) (cid:1) (cid:18) x xx x (cid:19) c x (cid:18) (cid:19) (2 x ) b x ( x ) a = x (cid:18) X d ≥ x d (cid:19) (cid:0) (cid:1) (cid:18) X c ≥ (cid:18) x xx x (cid:19) c (cid:19) (cid:18) (cid:19) (cid:18) X b ≥ (2 x ) b (cid:19)(cid:18) X a ≥ x a (cid:19) = x · − x · (cid:0) (cid:1) (cid:18) I − (cid:18) x xx x (cid:19) (cid:19) − (cid:18) (cid:19) · − x · − x x (1 − x ) (1 − x ) · (cid:0) (cid:1) (cid:18) − x + 5 x (cid:18) − x xx − x (cid:19) (cid:19) (cid:18) (cid:19) = x (2 − x )(1 − x ) (1 − x )(1 − x + 5 x ) . This is the generating function for one of the generic ways to place rooks. Wecan now repeat this procedure for every way in which we can place rooks below themain diagonal and add the individual generating functions together. All the sevengeneric cases, with their corresponding generating functions, are shown in Figure 4.Adding the individual generating functions together then gives us the overall gener-ating function that was given in Table 3.This same process works for determining the generating function of (cid:10) nk (cid:11) c for anyfixed k and c . The main challenge lies in that the number of generic cases that haveto be considered grows drastically as we increase c and k . This is demonstrated inTable 4. It is possible to automate this process, which was used for determining thegenerating functions for k = 3 given in Table 3. c =1 c =2 c =3 c =4 c =5 c =6 c =7 k =1 1 k =2 4 7 k =3 26 68 75 k =4 236 940 1090 1105 k =5 2752 16645 20360 20790 20821 k =6 39208 360081 464111 477242 478376 478439 k =7 660032 9202170 12492277 12933423 12974826 12977688 12977815Table 4: The number of generic c -rook placements with k rooks below the maindiagonal. The generalized Eulerian numbers are a natural extension of the Eulerian numbers,at least in regards to the interpretation coming from rook placements. We have alsoseen that these numbers exhibit a symmetry similar to that of the Eulerian numbers.It would be interesting to know which other properties and relationships involvingEulerian numbers generalize. Some natural candidates to try and generalize includethe following. • Is there a generalization of the recurrence in Proposition 3 for Eulerian numbersto generalized Eulerian numbers? Related to this, is there a simple generatingfunction for the generalized Eulerian numbers?13 × ×× × × x (1 − x ) (1 − x ) x (1 − x )(1 − x ) (1 − x + 5 x ) 2 x (1 − x ) (1 − x ) ×× × × x (5 − x )(1 − x ) (1 − x ) (1 − x + 5 x ) x (5 − x )(1 − x ) (1 − x ) (1 − x + 5 x ) ×× ×× x (2 − x )(1 − x ) (1 − x )(1 − x + 5 x ) x (2 − x )(1 − x ) (1 − x )(1 − x + 5 x )Figure 4: All generic 2-rook placements and corresponding generating functions. • Is there a generalization of Worpitzky’s identity, x n = P k (cid:10) nk (cid:11)(cid:0) x + kn (cid:1) , to gener-alized Eulerian numbers? Worpitzky’s identity is used in counting the numberof juggling patterns (see [4]), so a generalization might be useful in countingmultiplex juggling patterns. 14 Is there a generalization of the identity of Chung, Graham and Knuth [2], X k (cid:0) a + bk (cid:1)(cid:10) ka − (cid:11) = X k (cid:0) a + bk (cid:1)(cid:10) kb − (cid:11) (This uses the convention (cid:10) (cid:11) = 0.)More information about the Eulerian numbers and various identities and relationshipsthat could be considered are given in Graham, Knuth, Patashnik [3, Section 6.2].We also note the original motivation for investigating these numbers was lookinginto the mathematics of multiplex juggling. There is a close connection between themathematics of juggling and the mathematics of rook placements. We hope to seethis relationship strengthened in future work. Acknowledgments.
The research was conducted at the 2015 REU program heldat Iowa State University which was supported by NSF DMS 1457443.
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Rook Theory Notes ,manuscript. Available online at [2] Fan Chung, Ron Graham, and Donald Knuth,
A symmetrical Eulerian identity ,Journal of Combinatorics (2010), 29–38.[3] Ron Graham, Donald Knuth, and Oren Patashnik, Concrete Mathematics: AFoundation for Computer Science (2nd ed.) , Addison-Wesley Longman Publish-ing Co., Inc., Boston, 1994.[4] Joe Buhler, David Eisenbud, Ron Graham, and Colin Wright,
Juggling Dropsand Descents , American Math Monthly (1994) 507–519.[5] Burkhard Polster,
The Mathematics of Juggling , Springer, New York, 2000.[6] Neil Sloane,
The On-Line Encyclopedia of Integer Sequences . Available online at oeis.orgoeis.org