Q-analogues of the Fibo-Stirling numbers Quang T. Bach Roshil Paudyal Department of Mathematics Department of Mathematics University of California, San Diego 7 La Jolla, CA 92093-0112. USA Howard University 1 qtbach@ucsd.edu roshil.paudyal@bison.howard.edu 0 2 Jeffrey B. Remmel n Department of Mathematics a University of California, San Diego J La Jolla, CA 92093-0112. USA 5 remmel@math.ucsd.edu 2 Submitted: Date 1; Accepted: Date 2; Published: Date 3. ] O MR Subject Classifications: 05A15, 05E05 keywords: Fibonacci numbers, Stirling numbers, Lah numbers C . h t a Abstract m [ Let Fn denote the nth Fibonacci number relative to the initial conditions F0 = 0 and F1 =1. In[2],weintroducedFibonaccianaloguesoftheStirlingnumberscalledFibo-Stirling 1 numbers of the first and second kind. These numbers serve as the connection coefficients v between the Fibo-falling factorial basis {(x) : n ≥ 0} and the Fibo-rising factorial basis 5 ↓F,n 1 {(x)↑F,n : n ≥ 0} which are defined by (x)↓F,0 = (x)↑F,0 = 1 and for k ≥ 1, (x)↓F,k = 5 x(x−F1)···(x−Fk−1) and (x)↑F,k = x(x+F1)···(x+Fk−1). We gave a general rook 7 theory model which allowed us to give combinatorial interpretations of the Fibo-Stirling 0 numbers of the first and second kind. . 1 There are two natural q-analogues of the falling and rising Fibo-factorial basis. That is, 0 let [x] = qx−1. Then we let [x] =[x] =[x] = [x] =1 and, for k > 0, we 7 q q−1 ↓q,F,0 ↓q,F,0 ↑q,F,0 ↑q,F,0 let 1 v: [x]↓q,F,k =[x]q[x−F1]q···[x−Fk−1]q, [x]↓q,F,k =[x]q([x]q −[F1]q)···([x]q −[Fk−1]q), Xi [x]↑q,F,k =[x]q[x+F1]q···[x+Fk−1]q, and [x]↑q,F,k =[x]q([x]q +[F1]q)···([x]q +[Fk−1]q). In this paper, we show we can modify the rook theory model of [2] to give combinatorial r interpretations for the two different types q-analogues of the Fibo-Stirling numbers which a arise as the connection coefficients between the two different q-analogues of the Fibonacci falling and rising factorial bases. 1 Introduction Let Q denote the rational numbers and Q[x] denote the ring of polynomials over Q. Many classical combinatorial sequences can be defined as connection coefficients between various basis of the polynomial ring Q[x]. There are three very natural bases for Q[x]. The usual power basis {xn : n ≥ 0}, the falling factorial basis {(x) : n ≥ 0}, and the rising factorial basis ↓n {(x) : n ≥ 0}. Here we let (x) = (x) = 1 and for k ≥ 1, (x) = x(x−1)···(x−k+1) ↑n ↓0 ↑0 ↓k and (x) = x(x+1)···(x+k−1). Then the Stirling numbers of the first kind s , the Stirling ↑ n,k k 1 numbers of the second kind S and the Lah numbers L are defined by specifying that for n,k n,k all n ≥ 0, n n n (x) = s xk, xn = S (x) , and (x) = L (x) . ↓n n,k n,k ↓k ↑n n,k ↓k k=1 k=1 k=1 X X X The signless Stirling numbers of the first kind are defined by setting c = (−1)n−ks . Then n,k n,k it is well known that c , S , and L can also be defined by the recursions that c = S = n,k n,k n,k 0,0 0,0 L = 1, c = S = L = 0 if either n < k or k < 0, and 0,0 n,k n,k n,k c = c +nc , S = S +kS , and L = L +(n+k)L n+1,k n,k−1 n,k n+1,k n,k−1 n,k n+1,k n,k−1 n,k for all n,k ≥ 0. There are well known combinatorial interpretations of these connection coefficients. That is, S is the number of set partitions of [n] = {1,...,n} into k parts, c is the n,k n,k number of permutations in the symmetric group S with k cycles, and L is the number of n n,k ways to place n labeled balls into k unlabeled tubes with at least one ball in each tube. In [2], we introduced Fibonacci analogues of the number s , S , and L . We started n,k n,k n,k with the tiling model of the F of [11]. That is, let FT denote the set of tilings a column n n of height n with tiles of height 1 or 2 such that bottom most tile is of height 1. For example, possible tiling configurations for FT for i≤ 4 are shown in i F1 = 1 F2 = 1 F3 = 2 F4 = 3 Figure 1: The tilings counted by F for 1≤ i ≤ 4. i For each tiling T ∈ FT , we let one(T) is the number of tiles of height 1 in T and two(T) n is the number of tiles of height 2 in T and define F (p,q) = qone(T)ptwo(T). n T∈XFTn ItiseasytoseethatF (p,q) = q,F (p,q)= q2 andF (p,q) = qF (p,q)+pF (p,q)forn ≥ 2 1 2 n n−1 n−2 so that F (1,1) = F . We then defined the p,q-Fibo-falling factorial basis {(x) : n ≥ 0} n n ↓F,p,q,n and the p,q-Fibo-rising factorial basis {(x) : n ≥ 0} by setting (x) = (x) = 1 ↑F,p,q,n ↓F,p,q,0 ↑F,p,q,0 and setting (x) = x(x−F (p,q))···(x−F (p,q)) and ↓F,p,q,k 1 k−1 (x) = x(x+F (p,q))···(x+F (p,q)) ↑F,p,q,k 1 k−1 for k ≥ 1. Ourideatodefinep,q-FibonaccianaloguesoftheStirlingnumbersofthefirstkind,sf (p,q), n,k the Stirling numbers of the second kind, Sf (p,q), and the Lah numbers, Lf (p,q), is to de- n,k n,k fine them to be the connection coefficients between the usual power basis {xn : n ≥ 0} and 2 the p,q-Fibo-rising factorial and p,q-Fibo-falling factorial bases. That is, we define sf (p,q), n,k Sf (p,q), and Lf (p,q) by the equations n,k n,k n (x) = sf (p,q) xk, (1) ↓F,p,q,n n,k k=1 X n xn = Sf (p,q) (x) , and (2) n,k ↓ F,p,q,k k=1 X n (x) = Lf (p,q) (x) (3) ↑F,p,q,n n,k ↓F,p,q,k k=1 X for all n ≥ 0. It is easy to see that these equations imply simple recursions for the connection coefficients sf (p,q)s, Sf (p,q)s, and Lf (p,q)s. That is, sf (p,q)s, Sf (p,q)s, and Lf (p,q)s can n,k n,k n,k n,k n,k n,k be defined by the following recursions sf (p,q) = sf (p,q)−F (p,q)sf (p,q), n+1,k n,k−1 n n,k Sf (p,q) = Sf (p,q)+F (p,q)Sf (p,q), and n+1,k n,k−1 k n,k Lf (p,q) = Lf (p,q)+(F (p,q)+F (p,q))Lf (p,q) n+1,k n,k−1 k n n,k plus the boundary conditions sf (p,q) = Sf (p,q) = Lf (p,q) =1 0,0 0,0 0,0 and sf (p,q) = Sf (p,q) = Lf (p,q) = 0 n,k n,k n,k if k > n or k < 0. If we define cf (p,q) := (−1)n−ksf (p,q), then cf (p,q)s can be defined n,k n,k n,k by the recursions cf (p,q) = cf (p,q)+F (p,q)cf (p,q) (4) n+1,k n,k−1 n n,k plus the boundary conditions cf (p,q) = 1 and cf (p,q) = 0 if k > n or k < 0. It also follows 0,0 n,k that n (x) = cf (p,q) xk. (5) ↑F,p,q,n n,k k=1 X In [2], we developed a new rook theory model to give a combinatorial interpretation of the cf (p,q)s and the Sf (p,q)s and to give combinatorial proofs of their basic properties. This n,k n,k new rook theory model is a modification of the rook theory model for S and c except that n,k n,k we replace rooks by Fibonacci tilings. Themain goal of this paperis to show how that modelcan bemodifiedto give combinatorial interpretations to two new q-analogues of the cf (1,1)s and the Sf (1,1)s. Let [0] = 1 and n,k n,k q [x] = 1−qx. When n is a positive integer, then [n] = 1+q+···+qn−1 is the usual q-analogue q 1−q q of n. Then there are two natural analogues of the falling and rising Fibo-factorial basis. First we let [x] = [x] = [x] = [x] = 1. For k > 0, we let k > 0, ↓q,F,0 ↓q,F,0 ↑q,F,0 ↑q,F,0 [x] = [x] [x−F ] ···[x−F ] , ↓q,F,k q 1 q k−1 q [x] = [x] ([x] −[F ] )···([x] −[F ] ), ↓ q q 1 q q k−1 q q,F,k [x] = [x] [x+F ] ···[x+F ] , and ↑q,F,k q 1 q k−1 q [x] = [x] ([x] +[F ] )···([x] +[F ] ). ↑ q q 1 q q k−1 q q,F,k 3 Then we define cF (q) and cF (q) by the equations n,k n,k n [x] = cF (q)[x]k (6) ↑q,F,n n,k q k=1 X and n [x] = cF (q)[x]k. (7) ↑q,F,n n,k q k=1 X Similarly, we define SF (q) and SF (q) by the equations n,k n,k n [x]n = SF (q)[x] (8) q n,k ↓q,F,k k=1 X and n [x]n = SF (q)[x] (9) q n,k ↓ q,F,k k=1 X One can easily find recursions for these polynomials. For example, n+1 n [x]n+1 = SF (q)[x] = SF (q)[x] [x] q n+1,k ↓q,F,k n,k ↓q,F,k q k=1 k=1 X X n = SF (q)[x] ([F ] +qFk[x−F ] ) n,k ↓ k q k q q,F,k k=1 X n n = [F ] SF (q)[x] + qFkSF (q)[x] . k q n,k ↓ n,k ↓ q,F,k q,F,k+1 k=1 k=1 X X Taking the coefficient of [x] [x] on both sides shows that ↓ q q,F,k SFn+1,k(q) = qFk−1SFn,k−1(q)+[Fk]qSFn,k(q) (10) for 0 ≤ k ≤ n+1. It is then easy to check that the SF (q)s can be defined by the recursions n,k (10) with the initial conditions that SF (q) = 1 and SF (q) = 0 if k < 0 or n < k. A similar 0,0 n,k argument will show that SF (q) can be defined by the initial conditions that SF (q) =1 and n,k 0,0 SF (q) = 0 if k < 0 or n < k and the recursion n,k SF (q) = SF (q)+[F ] SF (q). (11) n+1,k n,k−1 k q n,k for 0≤ k ≤ n+1. Similarly, cF (q) can be defined by the initial conditions that cF (q) = 1 n,k 0,0 and cF (q) = 0 if k < 0 or n < k and the recursion n,k cF (q) = qFn−1cF (q)+[F ] cF (q), (12) n+1,k n,k−1 n q n,k for 0 ≤ k ≤ n + 1, and cF (q) can be defined by the initial conditions that cF (q) and n,k 0,0 cF (q) = 0 if k < 0 or n c +1. Indeed, Zeckendorf’s theorem says that the greedy algorithm give us the proper i+1 i P representation. That is, given n, find k such that F ≤ n < F , then the representation for n k k+1 is gotten by taking the representation for n−F and adding F . For example, suppose that we k k want to find T such that rank (T) = 100. Then 13 6 1. F = 89 ≤ 100 < F = 144 so that we need to find the Fibonacci representation of 11 12 100−89 = 11. 2. F = 8 ≤ 11 < F = 13 so that we need to find the Fibonacci representation of 11−8= 3. 6 7 3. F = 3 ≤ 3 < F = 5. 4 5 Thus we can represent 100 = F +F +F = 3+8+89 so that 4 6 11 seq(T)= (1,1,1,0,2,0,2,1,1,1,0,2,1). 3 The rook theory model for the SF (q)s and the cF (q)s. n,k n,k In this section, we shall give a rook theory model which will allow us to give combinatorial interpretations for the SF (q)s and the cF (q)s. This rook theory model is based on the n,k n,k one which Bach, Paudyal, and Remmel used in [2] to give combinatorial interpretations to the Sf (p,q)s and the cf (p,q)s. Thus, we shall briefly review the rook theory model in [2]. n,k n,k A Ferrers board B = F(b ,...,b ) is a board whose column heights are b ,...,b , reading 1 n 1 n from left to right, such that 0 ≤ b ≤ b ≤ ··· ≤ b . We shall let B denote the Ferrers board 1 2 n n F(0,1,...,n −1). For example, the Ferrers board B = F(2,2,3,5) is pictured on the left of Figure 4 and the Ferrers board B is pictured on the right of Figure 4 4 F(2,2,3,5) = B 4 = F(0,1,2,3) = Figure 4: Ferrers boards. Classically, there are two type of rook placements that we consider on a Ferrers board B. First we let N (B) be the set of all placements of k rooks in B such that no two rooks lie in k the same row or column. We shall call an element of N (B) a placement of k non-attacking k rooks in B or just a rook placement for short. We let F (B) be the set of all placements of k k rooks in B such that no two rooks lie in the same column. We shall call an element of F (B) k a file placement of k rooks in B. Thus file placements differ from rook placements in that file placements allow two rooks to be in the same row. For example, we exhibit a placement of 3 non-attacking rooks in F(2,2,3,5) on the left in Figure 5 and a file placement of 3 rooks on the right in Figure 5. X X X X X X Figure 5: Examples of rook and file placements. 7 Given a Ferrers board B = F(b ,...,b ), we define the k-th rook number of B to be 1 n r (B) = |N (B)| and the k-th file number of B to be f (B) = |F (B)|. Then the rook theory k k k k interpretation of the classical Stirling numbers is S = r (B ) for all 1 ≤ k ≤ n and n,k n−k n c = f (B ) for all 1 ≤ k ≤ n. n,k n−k n Theideaof[2]istomodifythesetsN (B)andF (B)toreplacerookswithFibonaccitilings. k k The analogue of file placements is very straightforward. That is, if B = F(b ,...,b ), then we 1 n let FT (B) denote the set of all configurations such that there are k columns (i ,...,i ) of B k 1 k where 1 ≤ i < ··· < i ≤ n such that in each column i , we have placed one of the tilings T 1 k j ij for the Fibonacci number F . We shall call such a configuration a Fibonacci file placement and bij denote it by P = ((i ,T ),...,(i ,T )). 1 i1 k ik Letone(P)denotethenumberoftilesofheight1thatappearinP andtwo(P)denotethenumber of tiles of height 2 that appear in P. Then in [2], we defined the weight of P, WF(P,p,q), to be qone(P)ptwo(P). For example, we have pictured an element P of FT (F(2,3,4,4,5)) in Figure 6 3 whose weight is q7p2. Then we defined the k-th p,q-Fibonacci file polynomial of B, fT (B,p,q), k by setting fT (B,p,q) = WF(P,p,q). k P∈FXTk(B) If k = 0, then the only element of FT (B) is the empty placement whose weight by definition k is 1. Figure 6: A Fibonacci file placement. Then in [2], we proved the following theorem concerning Fibonacci file placements in Ferrers boards. Theorem 2. Let B = F(b ,...,b ) be a Ferrers board where 0 ≤ b ≤ ··· ≤ b and b > 0. Let 1 n 1 n n B− = F(b ,...,b ). Then for all 1 ≤ k ≤ n, 1 n−1 fT (B,p,q)= fT (B−,p,q)+F (p,q)fT (B−,p,q). (16) k k bn k−1 To obtain the q-analogues that we desire for this paper, we define a new weight functions for elements of FT (B) where B = F(b ,...,b ) is Ferrers board. That is given a Fibonacci k 1 n file placement P = ((i ,T ),...,(i ,T )) in FT (B), let (j ,...,j ) be the sequence of 1 i1 n−k in−k n−k 1 k columns in B which have no tilings, reading from left to right. Then we define wB,q(P) = qPns=−1krankbis(Tis)+Pkt=1Fbjt and wB,q(P) = qPns=−1krankbis(Tis) 8 Note that the only difference between these two weight functions is that if b is column that i does not contain a tiling in P, then it contributes a factor of qFbi to wB,q(P) and a factor of 1 to w (P). We then define FT (B,q) and FT (B,q), by setting B,q k k FT (B,q) = w (P) and k B,q P∈FXTk(B) FT (B,q) = w (P). k B,q P∈FXTk(B) If k = 0, then the only element of FTk(B)is the empty placement ∅ so that wB,q(∅) = qPni=1Fbi and w (∅) = 1. B,q Then we have the following analogue of Theorem 2. Theorem 3. Let B = F(b ,...,b ) be a Ferrers board where 0 ≤ b ≤ ··· ≤ b and b > 0. Let 1 n 1 n n B− = F(b ,...,b ). Then for all 1 ≤ k ≤ n, 1 n−1 FTk(B,q) = qFbnFTk(B−,q)+[Fbn]qFTk−1(B−,p,q) (17) and FT (B,q) =FT (B−,q)+[F ] FT (B−,p,q). (18) k k bn q k−1 Proof. We claim (17) results by classifying the Fibonacci file placements in FT (B) accord- k ing to whether there is a tiling in the last column. If there is no tiling in the last column of P, then removing the last column of P produces an element of FT (B−) . Thus such place- k ments contribute qFbnFTk(B−,q) to FTk(B,q) since the fact that the last column has no tiling means that it contributes a factor of qFbn to wB,q(P). If there is a tiling in the last column, then the Fibonacci file placement that results by removing the last column is an element of FT (B−) and the sum of the weights of the possible Fibonacci tilings of height b for the last k−1 n column is T∈Fbn qrankbn(T) = [Fbn]q. Hence such placements contribute [Fbn]qFTk−1(B−,q) to FT (B,q). Thus k P FTk(B,q)= qFbnFTk(B−,q)+[Fbn]qFTk−1(B−,p,q). A similar argument will prove (18). If B = F(b ,...,b ) is a Ferrers board, then we let B denote the board that results by 1 n x adding x rows of length n below B. We label these rows from top to bottom with the numbers 1,2,...,x. We shall call the line that separates B from these x rows the bar. A mixed file placement P on the board B consists of picking for each column b either (i) a Fibonacci tiling x i T of height b above the bar or (ii) picking a row j below the bar to place a rook in the cell i i in row j and column i. Let M (B ) denote set of all mixed rook placements on B. For any n x P ∈ M (B ), we let one(P) denote the number of tiles of height 1 that appear in P and two(P) n x denote the set tiles of height 2 that appear in P. Then in [2], we defined the weight of P, WF(P,p,q), to be qone(P)ptwo(P). For example, Figure 7 pictures a mixed placement P in B x where B = F(2,3,4,4,5,5) and x is 9 such that WF(P,p,q)= q7p2. Also in [2], we proved the following theorem by counting WF(P,p,q) in two P∈Mn(Bx) different ways. P 9 bar 1 2 3 4 5 X X 6 7 X 8 9 Figure 7: A mixed file placement. Theorem 4. Let B = F(b ,...,b ) be a Ferrers board where 0≤ b ≤ ··· ≤ b and b > 0. 1 n 1 n n n (x+F (p,q))(x+F (p,q))···(x+F (p,q)) = fT (B,p,q)xn−k. (19) b1 b2 bn k k=0 X To obtain the desired q-analogues for this paper, we must define new weight functions for mixed placements P ∈ M (B ). That is, suppose that P ∩B is the Fibonacci tile placement n x Q = ((i ,T ),...,(i ,T )), and suppose that, for the rooks below the bar in columns 1 ≤ 1 i1 k in−k j < ...j ≤ n, the rook in column j is in row d for s = 1,...,k. Then we define 1 k s js wBx,q(P) = wB,q(P)qPkt=1djt−1 = qPns=−1krankbis(Tis)+Pkt=1Fbjt+djt−1 and wBx,q(P) = wB,q(P)qPkt=1djt−1 = qPns=−1krankbis(Tis)+Pkt=1djt−1. Thatis,foreachcolumnithechoiceofaFibonaccitilingT ofheightb abovethebarcontributes i i a factor of qrankbi(Ti) to wBx,q(P) and the choice of picking a row j below the bar to place a rook in the cell in row j and column i contributes a factor of qFbi+j−1 to wBx,q(P). Similarly, for each column b the choice of a Fibonacci tiling T of height b above the bar contributes a i i i factor of qrankbi(Ti) to wBx,q(P) and the choice of picking a row j below the bar to place a rook in the cell in row j and column i contributes a factor of qj−1 to w (P). Bx,q Then we have the following analogue of Theorem 4. Theorem 5. Let B = F(b ,...,b ) be a Ferrers board where 0 ≤ b ≤ ··· ≤ b and b > 0. 1 n 1 n n Then for all positive integers x, n [x+F ] [x+F ] ···[x+F ] = FT (B,q)[x]n−k (20) b1 q b2 q bn q k q k=0 X and n ([x] +[F ] )([x] +[F ] )···([x] +[F ] ) = FT (B,q)[x]n−k (21) q b1 q q b2 q q bn q k q k=0 X 10

Q-analogues of the Fibo-Stirling numbers

Similar Q-analogues of the Fibo-Stirling numbers