Q-analogues of the Fibo-Stirling numbers

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by  Quang T. Bach

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 [email protected] [email protected]
0
2 Jeffrey B. Remmel

n Department of Mathematics
a University of California, San Diego
J La Jolla, CA 92093-0112. USA

5 [email protected]
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 coeffi-
cients. 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 <k and the recursion
n,k
cF (q) = cF (q)+[F ] cF (q) (13)
n+1,k n,k−1 n q n,k
4
for 0 ≤ k ≤ n+1.
The main goal of this paper is to give a rook theory model for the polynomials cF (q),
n,k
cF (q), SF (q), and SF (q). Our rook theory model will allow us to give combinatorial
n,k n,k n,k
proofs of the defining equations (6), (7), (8), and (9) as well as combinatorial proofs of the
recursions(10), (11), (12), and(13). WeshallseethatourrooktheorymodelcF (q), cF (q),
n,k n,k
SF (q),andSF (q)isessentiallythesameasthetherooktheorymodelusedin[2]tointerpret
n,k n,k
the Sf (p,q)s and Sf (p,q)s but with a different weighting scheme.
n,k n,k
The outline of the paper is as follows. In Section 2, we describe a ranking and unranking
theory for the set of Fibonacci tilings which will a crucial element in our weighting scheme for
our rook theory model that we shall useto give combinatorial interpretations of the polynomials
cF (q),cF (q),SF (q),andSF (q). Insection3,weshallreviewtherooktheorymodelin
n,k n,k n,k n,k
[2]andshowhowitcanbemodifiedforourpurposes. InSection4,weshallprovegeneralproduct
formulasforFerrersboardsinournewmodelwhichwillspecialize(6),(7),(8),and(9)inthecase
where the Ferrers board is the staircase board whose column heights are 0,1,...,n−1 reading
from left to right. In Section 5, we shall prove various special properties of the polynomials
cF (q), cF (q), SF (q), and SF (q).
n,k n,k n,k n,k
2 Ranking and Unranking Fibonacci Tilings.
Thereisawelldeveloped theoryforrankingandunrankingcombinatorial objects. Seeforexam-
ple, Williamson’s book [14]. That is, give a collection of combinatorial objects O of cardinality
n, one wants to define bijections rank : O → {0,...,n−1} and unrank : {0,...,n−1} → O
which are inverses of each other. In our case, we let F denote the set of Fibonnaci tilings of
n
height n. Then we construct a tree which we call the Fibonacci tree for F . That is, we start
n
from the top of a Fibonacci tiling and branch left if we see a tile of height 1 and branch right if
we see a tiling of height 2. For example, the Fibonacci tree for F is pictured in Figure 2.
5
5
4
3
2
1
0
Figure 2: The tree for F
5
Then for any tiling T ∈ F , we define the rank of T for F , rank (T), to be the number of
n n n
paths to the left of the path for T in the Fibonacci tree for F . Clearly
n
{rank (T) :T ∈ F } = {0,1,2,...,F −1}
n n n
sothat qrankn(T) = 1+q+···+qFn−1 = [F ] . Itis, infact, quiteeasy toseecomputethe
T∈Fn n q
functions rank andunrank inthis situation. Thatis, supposethatwe representthetiling T as
n n
P
a sequence seq(T) = (t ,...,t ) where reading the tiles starting at the bottom, t = 1 if there is
1 n i
5
a tiling t of height 1 that ends at level i in T, t = 2 if there is t of height 2 that ends at level i
i i i
in T, and t = 0 if there is no tile t that ends at level i in T. For example, the tiling of T height
i i
9 pictured in Figure 3 would be represented by the sequence seq(T) = (1,0,2,1,1,1,0,2,1).
Figure 3: A tiling in F .
9
For any statement A, we let χ(A) = 1 is A is true and χ(A) = 0 if A is false. Then we have
the following lemma.
Lemma 1. Suppose that T ∈ F is a Fibonacci tiling such that seq(T) = (t ,...,t ). Then
n 1 n
rank (T)= n F χ(t = 2).
n i=1 i−1 i
Proof. The tPheorem is easy to prove by induction. It is clearly true for n = 1 and n = 2. Now
suppose n ≥ 3. Then it is easy to see from the Fibonacci tree for F that if t = 2 so that
n n
t = 0, then the tree that starts at level n−1 which represents taking the path to the left
n−1
starting at level n is justthe Fibonacci tree for F andhence this tree will contain F leaves
n−1 n−1
which will all be to the left of path for the tiling T. Then the tree that starting at level n−2
which represents taking the path to the right starting at level n is just the Fibonacci tree for
F and the number of paths in this tree which lie to the left of the path for T is just that the
n−2
number of paths to the left of the tiling T′ such that seq(T′) = (t ,...,t ) in the Fibonacci
1 n−2
tree for F . Thus in this case
n−2
rank (T) = F +rank (T′) =F +rank (t ,...,t ). (14)
n n−1 n−2 n−1 n−2 1 n−2
On the other hand if t = 1, then we branch left at level n so that the number of paths to the
n
left of the path for T in the Fibonacci tree for F will just be the number of paths to the left
n
of the tiling T′′ such that seq(T′′) = (t ,...,t ) in the Fibonacci tree for F . Thus in this
1 n−1 n−1
case
rank (T)= rank (T′′) = rank (t ,...,t ). (15)
n n−1 n−1 1 n−1
For example, for the tiling T in Figure 3, rank (T)= F +F = 1+21 = 22.
9 2 8
For the unrank function, we must rely on Zeckendorf’s theorem [15] which states that ev-
ery positive integer n is uniquely represented as sum n = k F where each c ≥ 2 and
i=0 ci i
c > 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