Q-analogues of the Fibo-Stirling numbers
Quang T. Bach Roshil Paudyal Jeffrey B. Remmel


Non-colocated Time-Reversal MUSIC:
 High-SNR Distribution of Null Spectrum 
    D. Ciuonzo, Senior Member, IEEE and P. Salvo Rossi, Senior Member, IEEEManuscript received 2nd December 2016; accepted 25th January 2017.

D. Ciuonzo is with DIETI, University of Naples “Federico II”,
Naples, Italy. 

P. Salvo Rossi is with the Dept. of Electronics and Telecommunications,
NTNU, Trondheim, Norway. 

E-mail: {domenico.ciuonzo, salvorossi}@ieee.org.


    Received: date / Accepted: date
===================================================================================================================================================================================================================================================================================================================================================================================









 

Let F_n denote the n^th Fibonacci number relative to the initial 
conditions F_0=0 and F_1=1. 
In <cit.>, we introduced Fibonacci analogues of the Stirling 
numbers called Fibo-Stirling numbers 
of the first and second kind. These numbers serve as the connection coefficients 
between the Fibo-falling factorial basis 
{(x)_↓_F,n:n ≥ 0} and the Fibo-rising factorial 
basis {(x)_↑_F,n:n ≥ 0} which are defined by  
(x)_↓_F,0 = (x)_↑_F,0 = 1 and for 
k ≥ 1, (x)_↓_F,k = x(x-F_1) ⋯ (x-F_k-1) and 
(x)_↑_F,k = x(x+F_1) ⋯ (x+F_k-1). 
We gave a general rook theory model which allowed us to give combinatorial 
interpretations of the Fibo-Stirling numbers of the first and second kind.  

There are two natural 
q-analogues of the falling and rising Fibo-factorial basis. That is, let 
[x]_q = q^x-1/q-1. Then we let 
[x]_↓_q,F,0 = [x]_↓_q,F,0 = 
[x]_↑_q,F,0 = [x]_↑_q,F,0=1 and, for 
k > 0, we let 

[x]_↓_q,F,k = [x]_q [x-F_1]_q ⋯ [x-F_k-1]_q, 
[x]_↓_q,F,k= [x]_q ([x]_q-[F_1]_q) ⋯ ([x]_q-[F_k-1]_q), 

[x]_↑_q,F,k=  [x]_q [x+F_1]_q ⋯ [x+F_k-1]_q, and   
[x]_↑_q,F,k= [x]_q ([x]_q+[F_1]_q) ⋯ 
([x]_q+[F_k-1]_q). 

In this paper, we show we can modify the  
rook theory model of <cit.> to give combinatorial interpretations 
for the two different types q-analogues of the Fibo-Stirling numbers which 
arise as the connection coefficients between the two different q-analogues 
of the Fibonacci falling and rising factorial bases. 






§ 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 {x^n: n≥ 0}, 
the falling factorial basis  {(x)_↓_n: n≥ 0}, 
and the rising factorial basis  {(x)_↑_n: n≥ 0}. 
Here we let (x)_↓_0 = (x)_↑_0 = 1 and for 
k ≥ 1, (x)_↓_k = x(x-1) ⋯ (x-k+1) and 
(x)_↑_k = x(x+1) ⋯ (x+k-1).
Then the Stirling numbers of the first kind s_n,k, 
the Stirling numbers of the second kind S_n,k and 
the Lah numbers L_n,k are defined by specifying 
that for all n ≥ 0, 

    (x)_↓_n = ∑_k=1^n s_n,k x^k,   
    x^n = ∑_k=1^n S_n,k (x)_↓_k,    
    (x)_↑_n = ∑_k=1^n L_n,k (x)_↓_k.

The signless Stirling numbers of the first kind 
are defined by setting c_n,k = (-1)^n-k s_n,k. 
Then it is well known that c_n,k, S_n,k, and L_n,k 
can also be defined by the recursions that 
c_0,0 = S_0,0 = L_0,0 = 1, 
c_n,k = S_n,k = L_n,k = 0 if either n < k or 
k < 0, and 

    c_n+1,k = c_n,k-1+ n c_n,k,   
    S_n+1,k =  S_n,k-1+kS_n,k,     
    L_n+1,k = L_n,k-1 +(n+k) L_n,k

for all n,k ≥ 0. 
There are well known combinatorial interpretations of 
these connection coefficients. That is, 
S_n,k is the number of set partitions of [n] = {1, …, n} 
into k parts, c_n,k is the number of permutations 
in the symmetric group S_n with k cycles, and 
L_n,k is the number of ways to place n labeled balls into k unlabeled 
tubes with at least one ball in each tube. 


In <cit.>, we introduced Fibonacci analogues of 
the number s_n,k, S_n,k, and L_n,k.  
We started with the tiling model of the F_n of <cit.>. 
That is, let ℱ𝒯_n denote the set of tilings  
a column 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_i for i ≤ 4 are shown in 

TilingsThe tilings counted by F_i for 1 ≤ i ≤ 4.


For each tiling T ∈ℱ𝒯_n, we let 
one(T) is the number of tiles of 
height 1 in T and two(T) is the number of tiles of 
height 2 in T and define 

    F_n(p,q) = ∑_T ∈ℱ𝒯_n q^one(T)p^two(T).
 
It is easy to see that 
 F_1(p,q) =q, F_2(p,q) =q^2 and F_n(p,q) =q F_n-1(p,q)+ 
pF_n-2(p,q) for n ≥ 2 so that F_n(1,1) =F_n.
We then 
defined the p,q-Fibo-falling 
factorial basis 
{(x)_↓_F,p,q,n:n ≥ 0} and the p,q-Fibo-rising factorial 
basis {(x)_↑_F,p,q,n:n ≥ 0} by setting 
(x)_↓_F,p,q,0 = (x)_↑_F,p,q,0 = 1 and setting

    (x)_↓_F,p,q,k   =    x(x-F_1(p,q)) ⋯ (x-F_k-1(p,q))  
     
    (x)_↑_F,p,q,k   =    x(x+F_1(p,q)) ⋯ (x+F_k-1(p,q))

for k ≥ 1. 

Our idea to define p,q-Fibonacci analogues of 
the Stirling numbers of the first kind, 𝐬𝐟_n,k(p,q), 
the Stirling numbers of the second kind, 𝐒𝐟_n,k(p,q), 
and the Lah numbers, 𝐋𝐟_n,k(p,q), is to 
define them   
to be the connection coefficients between the usual power basis 
{x^n: n ≥ 0} and the p,q-Fibo-rising factorial and 
p,q-Fibo-falling factorial bases. 
That is, we define 𝐬𝐟_n,k(p,q), 𝐒𝐟_n,k(p,q), 
and 𝐋𝐟_n,k(p,q) by the equations  

    (x)_↓_F,p,q,n = ∑_k=1^n 𝐬𝐟_n,k(p,q)  x^k,


    x^n = ∑_k=1^n 𝐒𝐟_n,k(p,q)  (x)_↓_F,p,q,k,


    (x)_↑_F,p,q,n = ∑_k=1^n 𝐋𝐟_n,k(p,q)  (x)_↓_F,p,q,k

for all n ≥ 0. 

It is easy to see that these equations imply simple recursions for 
the connection coefficients 𝐬𝐟_n,k(p,q)s, 𝐒𝐟_n,k(p,q)s, and 𝐋𝐟_n,k(p,q)s. 
That is, 𝐬𝐟_n,k(p,q)s, 𝐒𝐟_n,k(p,q)s, and 𝐋𝐟_n,k(p,q)s can be defined by the following recursions 

    𝐬𝐟_n+1,k(p,q)    =   𝐬𝐟_n,k-1(p,q)- F_n(p,q) 𝐬𝐟_n,k(p,q), 
    𝐒𝐟_n+1,k(p,q)    =   𝐒𝐟_n,k-1(p,q)+ F_k(p,q) 𝐒𝐟_n,k(p,q),  
    𝐋𝐟_n+1,k(p,q)    =   𝐋𝐟_n,k-1(p,q)+ (F_k(p,q) +F_n(p,q))𝐋𝐟_n,k(p,q)

plus the boundary 
conditions 

    𝐬𝐟_0,0(p,q)=𝐒𝐟_0,0(p,q)=𝐋𝐟_0,0(p,q)=1
 
and 

    𝐬𝐟_n,k(p,q) =𝐒𝐟_n,k(p,q) =𝐋𝐟_n,k(p,q) =0

 if k > n or k < 0. 
If we define 𝐜𝐟_n,k(p,q):= (-1)^n-k𝐬𝐟_n,k(p,q), then 
𝐜𝐟_n,k(p,q)s can be defined by the recursions   

    𝐜𝐟_n+1,k(p,q) = 𝐜𝐟_n,k-1(p,q)+F_n(p,q) 𝐜𝐟_n,k(p,q)
  plus the boundary 
conditions 𝐜𝐟_0,0(p,q)=1 and 𝐜𝐟_n,k(p,q) =0 if k > n or k < 0. It also follows that 

    (x)_↑_F,p,q,n = ∑_k=1^n 𝐜𝐟_n,k(p,q)  x^k.

In <cit.>, we developed a new rook theory model to give a 
combinatorial interpretation of  
the 𝐜𝐟_n,k(p,q)s and the 𝐒𝐟_n,k(p,q)s and to give 
combinatorial  proofs of their basic properties. 
This new rook theory model is 
a modification of the rook theory model for S_n,k 
and c_n,k except that we replace rooks by Fibonacci 
tilings. 


 The main goal of this paper 
is to show how that model can be modified to 
give combinatorial interpretations to two new  
q-analogues of the 𝐜𝐟_n,k(1,1)s and the 
𝐒𝐟_n,k(1,1)s.
Let [0]_q =1 and [x]_q = 1-q^x/1-q.  When 
n is a positive integer, then [n]_q = 1+ q+ ⋯ +q^n-1 is 
the usual q-analogue of n.  Then there are two natural 
analogues of the falling and rising Fibo-factorial basis. First we let  
[x]_↓_q,F,0 = [x]_↓_q,F,0 = 
[x]_↑_q,F,0 = [x]_↑_q,F,0=1.  For k > 0, 
we let 
k > 0, 

    [x]_↓_q,F,k   =    [x]_q [x-F_1]_q ⋯ [x-F_k-1]_q, 
     [x]_↓_q,F,k   =    [x]_q ([x]_q-[F_1]_q) ⋯ ([x]_q-[F_k-1]_q), 
     [x]_↑_q,F,k   =    [x]_q [x+F_1]_q ⋯ [x+F_k-1]_q,  
     [x]_↑_q,F,k   =    [x]_q ([x]_q+[F_1]_q) ⋯ 
    ([x]_q+[F_k-1]_q).

Then we define 𝐜𝐅_n,k(q) and 𝐜𝐅_n,k(q) 
by the equations 

    [x]_↑_q,F,n= ∑_k=1^n 𝐜𝐅_n,k(q) [x]_q^k

and 

    [x]_↑_q,F,n= ∑_k=1^n 
    𝐜𝐅_n,k(q)[x]_q^k.

Similarly, we define 𝐒𝐅_n,k(q) and 
𝐒𝐅_n,k(q) 
by the equations 

    [x]_q^n= ∑_k=1^n 𝐒𝐅_n,k(q) [x]_↓_q,F,k

and 

    [x]_q^n= ∑_k=1^n 
    𝐒𝐅_n,k(q) [x]_↓_q,F,k



One can easily find recursions for these polynomials. 
For example, 

    [x]_q^n+1   =   ∑_k=1^n+1𝐒𝐅_n+1,k(q) 
    [x]_↓_q,F,k = ∑_k=1^n 𝐒𝐅_n,k(q) [x]_↓_q,F,k[x]_q 
       =   ∑_k=1^n 𝐒𝐅_n,k(q) [x]_↓_q,F,k([F_k]_q 
    + q^F_k[x-F_k]_q)
       =   ∑_k=1^n [F_k]_q 𝐒𝐅_n,k(q) [x]_↓_q,F,k
    +∑_k=1^n q^F_k𝐒𝐅_n,k(q) [x]_↓_q,F,k+1.

Taking the coefficient of [x]_↓_q,F,k[x]_q on both sides 
shows that 

    𝐒𝐅_n+1,k(q)=  q^F_k-1𝐒𝐅_n,k-1(q)+[F_k]_q 
    𝐒𝐅_n,k(q)

for 0 ≤ k ≤ n+1.
It is then easy to check that the  𝐒𝐅_n,k(q)s can be defined  
by the recursions (<ref>) with the initial conditions 
that 𝐒𝐅_0,0(q)=1 and  𝐒𝐅_n,k(q)=0 if k < 0 or 
n < k. 
A similar argument  will show that 
𝐒𝐅_n,k(q) can be defined 
by the initial conditions that 
𝐒𝐅_0,0(q)=1 and  𝐒𝐅_n,k(q)=0 if k < 0 or 
n <  k and the recursion 

    𝐒𝐅_n+1,k(q)=  𝐒𝐅_n,k-1(q)+[F_k]_q 
    𝐒𝐅_n,k(q).

for 0 ≤ k ≤ n+1. 
Similarly, 𝐜𝐅_n,k(q) can be defined 
by the initial conditions that 
𝐜𝐅_0,0(q)=1 and  𝐜𝐅_n,k(q)=0 if k < 0 or 
n <  k and the recursion

    𝐜𝐅_n+1,k(q)=  q^F_n-1𝐜𝐅_n,k-1(q)+[F_n]_q 
    𝐜𝐅_n,k(q),

for 0 ≤ k ≤ n+1, and  
𝐜𝐅_n,k(q) can be defined 
by the initial conditions that 
𝐜𝐅_0,0(q) and  𝐜𝐅_n,k(q)=0 if k < 0 or 
n < k and the recursion 

    𝐜𝐅_n+1,k(q)=  𝐜𝐅_n,k-1(q)+
    [F_n]_q 𝐜𝐅_n,k(q)

for 0 ≤ k ≤ n+1. 


The main goal of this paper is to give a rook theory model 
for the polynomials 𝐜𝐅_n,k(q), 
𝐜𝐅_n,k(q), 𝐒𝐅_n,k(q), and 
𝐒𝐅_n,k(q).   Our rook theory 
model will allow us to give combinatorial proofs of 
the defining equations (<ref>), (<ref>), (<ref>), 
and (<ref>) as well as combinatorial proofs of 
the recursions (<ref>), (<ref>), (<ref>), and (<ref>).
We shall see that our rook theory model 
𝐜𝐅_n,k(q), 
𝐜𝐅_n,k(q), 𝐒𝐅_n,k(q), and 
𝐒𝐅_n,k(q) is essentially the same 
as the the rook theory model used in <cit.> to interpret 
the 𝐒𝐟_n,k(p,q)s and 𝐒𝐟_n,k(p,q)s 
but with a different weighting scheme. 


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 use to give 
combinatorial interpretations of   
the polynomials 𝐜𝐅_n,k(q), 
𝐜𝐅_n,k(q), 𝐒𝐅_n,k(q), and 
𝐒𝐅_n,k(q).  In section 3, we shall 
review the rook theory model in <cit.> and show how it 
can be modified for our purposes.  In Section 4, 
we shall prove general product formulas for Ferrers boards 
in our new model which will specialize 
(<ref>), (<ref>), (<ref>), 
and (<ref>) in the case 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_n,k(q), 
𝐜𝐅_n,k(q), 𝐒𝐅_n,k(q), and 
𝐒𝐅_n,k(q).




§ RANKING AND UNRANKING FIBONACCI TILINGS.



There is a well developed theory for ranking and unranking 
combinatorial objects. See for example, Williamson's book 
<cit.>. 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_n denote the set of Fibonnaci tilings of height n. Then 
we construct a tree which we call the Fibonacci tree for F_n. 
That is, we start 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_5 is pictured 
in Figure <ref>. 

FtreeThe tree for F_5


Then for any tiling T ∈ℱ_n, we define 
the rank of T for F_n, _n(T), to be the number of paths to the left of the path for T in the Fibonacci tree for F_n. Clearly 

    {_n(T): T ∈ℱ_n}= {0,1,2, … , F_n-1}
 
so that ∑_T ∈ℱ_n q^_n(T) = 1+q+ ⋯ +q^F_n-1 = 
[F_n]_q. It is, in fact, quite easy to see compute 
the functions _n and unrank_n in this situation. 
That is, suppose that we represent the tiling T as a sequence 
seq(T) = (t_1, … ,t_n) where reading the tiles starting at the bottom, 
t_i = 1 if there is a tiling t_i of height 1 that ends at level i in 
T, t_i =2 if there is t_i of height 2 that ends at level i in T, 
and t_i =0 if there is no tile t_i that ends at level i in T. 
For example, the tiling of T height 9  pictured in Figure <ref>
would be represented by the sequence seq(T)= (1,0,2,1,1,1,0,2,1). 

F9A 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. 

 Suppose that T ∈ℱ_n is a Fibonacci tiling 
such that seq(T) = (t_1, …, t_n). Then 
_n(T) = ∑_i=1^n F_i-1χ(t_i=2). 


The theorem 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_n that if t_n=2 so that t_n-1 =0, 
then the tree 
that starts at level n-1 which represents taking the path to the 
left starting at level n is just the Fibonacci tree for F_n-1 and 
hence this tree will contain F_n-1 leaves 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_n-2 and 
the number of paths in this tree which lie to the left of 
the path for T is just that the number of paths to the left of 
the tiling T' such that seq(T') = (t_1,…,t_n-2) in 
the Fibonacci tree for F_n-2.  
Thus in this case 

    _n(T) = F_n-1+_n-2(T')= 
    F_n-1+_n-2(t_1, …, t_n-2).

On the other hand if t_n =1, then we branch left at level n so 
that the number of paths to the left of the path for T in the 
Fibonacci tree for F_n will just be the number of paths to 
the left of the tiling T” such that seq(T”) = (t_1, …, t_n-1) 
in the 
Fibonacci tree for F_n-1. 
Thus in this case 

    _n(T) = _n-1(T”)= _n-1(t_1, …, t_n-1).



For example, for the tiling T in  Figure <ref>, 
_9(T) = F_2+F_8 = 1+21 =22. 


For the unrank function, we must rely on Zeckendorf's theorem 
<cit.> which states that every positive integer n is uniquely represented 
as sum n = ∑_i=0^k F_c_i where each c_i ≥ 2 and 
c_i+1 > c_i +1. Indeed, Zeckendorf's theorem 
says that the greedy algorithm give us the proper representation. 
That is, given n, find k such that F_k ≤ n < F_k+1, then 
the representation for n is gotten by taking the representation 
for n-F_k and adding F_k. For example, suppose that we want to find 
T such that _13(T) =100. Then
 

  * F_11 =89 ≤ 100< F_12=144 so that we need to find 
the Fibonacci representation of 100-89 =11.

  * F_6 = 8 ≤  11 < F_7 =13 so that we need to find the Fibonacci 
representation of 11-8 =3. 

  * F_4 = 3 ≤ 3 < F_5 =5. 

Thus we can represent 
100 =  F_4+ F_6 + F_11 = 3+ 8 + 89 so that 

    seq(T) =(1,1,1,0,2,0,2,1,1,1,0,2,1).
















§ THE ROOK THEORY MODEL FOR THE 
𝐒𝐅_N,K(Q)S AND THE 𝐜𝐅_N,K(Q)S.



In this section, we shall give a rook theory model which 
will allow us to give combinatorial interpretations for 
the 𝐒𝐅_n,k(q)s and the 𝐜𝐅_n,k(q)s. 
This rook theory 
model is based on the one 
which Bach, Paudyal, and Remmel used in <cit.> to give combinatorial interpretations to 
the 𝐒𝐟_n,k(p,q)s and the 𝐜𝐟_n,k(p,q)s.  
Thus, we shall briefly review the rook theory model in <cit.>.


A Ferrers board B=F(b_1, …, b_n) is 
a board whose column heights are b_1, …, b_n, reading 
from left to right, such that 0≤ b_1 ≤ b_2 ≤⋯≤ b_n. 
We shall let B_n denote the Ferrers board F(0,1, …, n-1). 
For example, the Ferrers board B = F(2,2,3,5) is 
pictured on the left of Figure <ref> 
and the Ferrers board B_4 is pictured on the right of 
Figure <ref> 

FerrersFerrers boards.


Classically, there are two type of rook placements that we 
consider on a Ferrers board B.  First we let 
𝒩_k(B) be the set of all placements of 
k rooks in B such that no two rooks lie in the same 
row or column.  We shall call an element  of 
𝒩_k(B) a placement of k non-attacking rooks 
in B or just a rook placement for short.  We let 
ℱ_k(B) be the set of all placements of 
k rooks in B such that no two rooks lie in the same 
column.  We shall call an element  of 
ℱ_k(B) 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 
<ref> and a file placement of 3 rooks on 
the right in  Figure <ref>.


placementsExamples of rook and file placements.




Given a Ferrers board B = F(b_1, …, b_n), we 
define the k-th rook number of B to be 
r_k(B) = |𝒩_k(B)| and the k-th file number 
of B to be f_k(B) = |ℱ_k(B)|. Then the rook 
theory interpretation of the classical Stirling numbers is 

    S_n,k   =    r_n-k(B_n)   1 ≤ k ≤ n  
    
    c_n,k   =    f_n-k(B_n)   1 ≤ k ≤ n.


 
The idea of <cit.> is to modify the sets 𝒩_k(B) and 
ℱ_k(B) to replace rooks with Fibonacci tilings. 
The analogue of file placements is very straightforward. 
That is, if B=F(b_1, …, b_n), then we let 
ℱ𝒯_k(B) denote the set of all configurations such that 
there are k columns (i_1, …, i_k) of B where 
1 ≤ i_1 < ⋯ < i_k ≤ n such that in each 
column i_j, we have placed one of the tilings T_i_j for the Fibonacci 
number F_b_i_j.  We shall call such a configuration 
a Fibonacci file placement and denote it by 

    P = ((i_1,T_i_1),  …, (i_k,T_i_k)).

 Let 
one(P) denote the number of tiles of height 1 that appear 
in P and two(P) denote the number of tiles of height 2 that appear 
in P.  Then in <cit.>, we defined the weight of P, WF(P,p,q), to be 
q^one(P)p^two(P).  For example, we have 
pictured an element P of ℱ𝒯_3(F(2,3,4,4,5)) in 
Figure <ref> whose weight is q^7 p^2. Then 
we defined the k-th p,q-Fibonacci file polynomial of B, 𝐟𝐓_k(B,p,q), 
by setting 

    𝐟𝐓_k(B,p,q) = ∑_P ∈ℱ𝒯_k(B) WF(P,p,q).

If k =0, then the only element of ℱ𝒯_k(B) is the empty placement 
whose weight by definition is 1. 







FibfileA Fibonacci file placement.




Then in <cit.>, we proved the following theorem concerning 
Fibonacci file placements in Ferrers boards. 

 
Let B =F(b_1, …, b_n) be a Ferrers 
board where 0 ≤ b_1 ≤⋯≤ b_n and b_n > 0. 
Let B^- = F(b_1, …, b_n-1). Then for all 
1 ≤ k ≤ n, 

    𝐟𝐓_k(B,p,q) = 𝐟𝐓_k(B^-,p,q)+ F_b_n(p,q) 𝐟𝐓_k-1(B^-,p,q).




To obtain the q-analogues that we desire for this paper, we 
define a new weight functions for elements of ℱ𝒯_k(B) where 
B=F(b_1, …, b_n) is Ferrers board. That is given 
a Fibonacci file placement 
P = ((i_1,T_i_1),  …, (i_n-k,T_i_n-k)) in 
ℱ𝒯_n-k(B), let (j_1, …, j_k) be 
the sequence of columns in B which have no tilings, reading 
from left to right. Then we define 

    𝐰_𝐁,𝐪(P)    =    q^∑_s=1^n-k_b_i_s(T_i_s) + 
    ∑_t=1^k F_b_j_t 
    𝐰_𝐁,𝐪(P)    =    
    q^∑_s=1^n-k_b_i_s(T_i_s)

Note that the only difference between these two weight functions 
is that if b_i is column that does not contain a tiling in 
P, then it contributes a factor of 
q^F_b_i to 𝐰_𝐁,𝐪(P) and a factor of 1 to 
𝐰_𝐁,𝐪(P).
We then define 𝐅𝐓_k(B,q) and 𝐅𝐓_k(B,q), 
by setting 

    𝐅𝐓_k(B,q)    =   ∑_P ∈ℱ𝒯_k(B)𝐰_𝐁,𝐪(P) 
     
    𝐅𝐓_k(B,q)    =   ∑_P ∈ℱ𝒯_k(B)𝐰_𝐁,𝐪(P).

If k =0, then the only element of ℱ𝒯_k(B) is the empty 
placement ∅ so that  
𝐰_𝐁,𝐪(∅) =q^∑_i=1^n F_b_i and 
𝐰_𝐁,𝐪(∅) =1.  
 


Then we have the following analogue of Theorem 
<ref>.

 
Let B =F(b_1, …, b_n) be a Ferrers 
board where 0 ≤ b_1 ≤⋯≤ b_n and b_n > 0. 
Let B^- = F(b_1, …, b_n-1). Then for all 
1 ≤ k ≤ n, 

    𝐅𝐓_k(B,q) = q^F_b_n𝐅𝐓_k(B^-,q)+ [F_b_n]_q 𝐅𝐓_k-1(B^-,p,q)

and 

    𝐅𝐓_k(B,q) = 𝐅𝐓_k(B^-,q)+ [F_b_n]_q 𝐅𝐓_k-1(B^-,p,q).


 
We claim (<ref>) 
results by classifying the Fibonacci file placements 
in ℱ𝒯_k(B) according 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_k(B^-) . Thus such placements 
contribute q^F_b_n𝐅𝐓_k(B^-,q) to 𝐅𝐓_k(B,q) 
since the fact that the last column has no tiling means that 
it contributes 
a factor of q^F_b_n to 𝐰_𝐁,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_k-1(B^-) and the sum of 
the weights of the possible 
Fibonacci tilings of height b_n for the last column 
is ∑_T ∈ℱ_b_nq^_b_n(T) = 
[F_b_n]_q. Hence such placements 
contribute [F_b_n]_q 𝐅𝐓_k-1(B^-,q) to 
𝐅𝐓_k(B,q). Thus 

    𝐅𝐓_k(B,q) = q^F_b_n𝐅𝐓_k(B^-,q)+ [F_b_n]_q 𝐅𝐓_k-1(B^-,p,q).


A similar argument will prove (<ref>). 



If B=F(b_1, …, b_n) is a Ferrers board, 
then we let B_x denote the board that results by 
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_x consists 
of picking for each column b_i either (i) a Fibonacci tiling 
T_i of height b_i above the bar or (ii) picking 
a row j below the bar to place a rook in the cell in row j 
and column i.  Let ℳ_n(B_x) denote set of all 
mixed rook placements on B. For any P ∈ℳ_n(B_x), 
we let one(P) denote the number of tiles of height 1 that appear 
in P and two(P) denote the set tiles of height 2 that appear 
in P.  Then in <cit.>, we defined the weight of P, WF(P,p,q), to be 
q^one(P)p^two(P).
For example, 
Figure <ref> 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) =  q^7p^2. 

mixedA mixed file placement.


Also in <cit.>, we proved the following theorem by counting  
∑_P ∈ℳ_n(B_x) WF(P,p,q) in two different ways. 


Let B =F(b_1, …, b_n) be a Ferrers 
board where 0 ≤ b_1 ≤⋯≤ b_n and b_n > 0. 

    (x+F_b_1(p,q))(x+F_b_2(p,q)) ⋯ (x+F_b_n(p,q)) =
    ∑_k=0^n 𝐟𝐓_k(B,p,q) x^n-k.



 
To obtain the desired q-analogues for this paper, we 
must define new weight functions 
for mixed placements P ∈ℳ_n(B_x). 
That is, suppose that P ∩ B is the Fibonacci tile placement 
Q = ((i_1,T_i_1),  …, (i_k,T_i_n-k)), and suppose that, for the rooks below the bar in columns 
1 ≤j_1<  …j_k≤ n, the rook in column 
j_s is in row d_j_s for s =1, …, k.  
Then we define

    𝐰_𝐁_𝐱,𝐪(P)    =   𝐰_𝐁,𝐪(P)q^∑_t=1^k d_j_t -1 = 
    q^∑_s=1^n-k_b_i_s(T_i_s) + ∑_t=1^k F_b_j_t+d_j_t-1 
    𝐰_𝐁_𝐱,𝐪(P)    =   𝐰_𝐁,𝐪(P)q^∑_t=1^k d_j_t -1 = 
    q^∑_s=1^n-k_b_i_s(T_i_s) + ∑_t=1^k d_j_t-1.

That is, for each column i the choice of a Fibonacci tiling 
T_i of height b_i above the bar contributes a factor 
of q^_b_i(T_i) to 𝐰_𝐁_𝐱,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 
q^F_b_i+j-1 to 𝐰_𝐁_𝐱,𝐪(P). Similarly, 
for each column b_i the choice of a Fibonacci tiling 
T_i of height b_i above the bar contributes a factor 
of q^_b_i(T_i) to 𝐰_𝐁_𝐱,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 
q^j-1 to 𝐰_𝐁_𝐱,𝐪(P). 



Then we have the following analogue of Theorem <ref>.




Let B =F(b_1, …, b_n) be a Ferrers 
board where 0 ≤ b_1 ≤⋯≤ b_n and b_n > 0. 
Then for all positive integers x, 

    [x+F_b_1]_q [x+F_b_2]_q ⋯ [x+F_b_n]_q =
    ∑_k=0^n 𝐅𝐓_k(B,q) [x]_q^n-k

and 

    ([x]_q+[F_b_1]_q) ([x]_q+[F_b_2]_q) ⋯ ([x]_q+[F_b_n]_q) =
    ∑_k=0^n 𝐅𝐓_k(B,q) [x]_q^n-k



To prove (<ref>),   
fix x to be a positive integer and consider 
the sums 

    S   =   ∑_P ∈ℳ_n(B_x)𝐰_𝐁_𝐱,𝐪(P)  
    S   =   ∑_P ∈ℳ_n(B_x)𝐰_𝐁_𝐱,𝐪(P).



For S, in  a given column i, our choice of the Fibonacci tiling 
of height b_i will contribute a factor 
of ∑_T ∈ℱ_nq^_b_i(T) =[F_b_i]_q 
to S.  Our choice of placing a rook below the bar in 
column i contribute a factor of 

    ∑_j=1^x q^F_b_i+j-1 = q^F_b_i(1+q+q^2 + ⋯ q^x-1) =
    q^F_b_i[x]_q
 
to S.  As [F_b_i]_q + q^F_b_i[x]_q =[x+F_b_i]_q,   
each column of b_i of B contributes 
a factor of [x+ F_b_i]_q to S so that 

    S = ∏_i=1^n [x +  F_b_i]_q.


For S, in  a given column i, our choice of the Fibonacci tiling 
of height b_i will contribute a factor 
of ∑_T ∈ℱ_nq^_b_i(T) =[F_b_i]_q 
to S.  Our choice of placing a rook below the bar in 
column i contribute a factor of 

    ∑_j=1^x q^j-1 = [x]_q
 
to S. Thus each column b_i contributes 
a factor of [x]_q+ [F_b_i]_q to S so that 

    S = ∏_i=1^n ([x]_q + [F_b_i]_q).




On the other 
hand, suppose that we fix a Fibonacci file placement  
P ∈ℱ𝒯_k(B). 
Then we want to compute S_P = ∑_Q  ∈ℳ_n(B), 
Q ∩ B = P𝐰_𝐁_𝐱,𝐪(Q) which is the sum of 
𝐰_𝐁_𝐱,𝐪(Q) over all 
mixed placements Q such that Q intersect B equals P. 
It it easy to see that such a Q arises by choosing 
a rook to be placed below the bar for each column 
that does not contain a tiling.  Each such column contributes 
a factor of 1+q+ ⋯ +q^x-1 =[x]_q in addition 
to the weight 𝐰_𝐁,𝐪(P). Thus it follows that 
S_P =𝐰_𝐁,𝐪(P) [x]_q^n-k. 
Hence it follows that 

    S    =   ∑_k=0^n ∑_P ∈ℱ𝒯_k(B) S_P 
       =   ∑_k=0^n [x]_q^n-k∑_P ∈ℱ𝒯_k(B)𝐰_𝐁,𝐪(P) 
       =   ∑_k=0^n 𝐅𝐓_k(B,q)  [x]_q^n-k.


The same argument will show that 

    S = ∑_k=0^n 𝐅𝐓_k(B,q)  [x]_q^n-k.











Now consider the special case of the previous two theorems 
when B_n = F(0,1,2, …, n-1). Then (<ref>) implies 
that 

    𝐅𝐓_n+1-k(B_n+1,q) = q^F_n𝐅𝐓_n+1-k(B_n,p,q) + 
    [F_n]_q 
    𝐅𝐓_n-k(B_n,q).

It then easily follows that for all 0 ≤ k ≤ n, 

    𝐜𝐅_n,k(q) = 𝐅𝐓_n-k(B_n,q).

Note that 𝐜𝐅_n,0(q) = 0 for all n ≥ 1 since 
there are no Fibonacci file placements in 
ℱ𝒯_n(B_n) since there are only n-1 non-zero columns. 
Moreover such a situation, we see that (<ref>) 
implies that 

    [x]_q [x+F_1]_q [x+F_2]_q ⋯ [x+F_n-1]_q = 
    ∑_k=1^n 𝐜𝐅_n,k(q)  [x]_q^k.

Thus we have given a combinatorial proof of 
(<ref>). 


Similarly (<ref>) implies 
that 

    𝐅𝐓_n+1-k(B_n+1,q) = 𝐅𝐓_n+1-k(B_n,p,q) + 
    [F_n]_q 
    𝐅𝐓_n-k(B_n,q).

It then easily follows that for all 0 ≤ k ≤ n, 

    𝐜𝐅_n,k(q) = 𝐅𝐓_n-k(B_n,q).

Moreover such a situation, we see that (<ref>) 
implies that 

    [x]_q ([x]_q+[F_1]_q) ([x]_q+[F_2]_q) ⋯ ([x]_q+[F_n-1]_q) = 
    ∑_k=1^n 𝐜𝐅_n,k(q)  [x]_q^k.

Thus we have given a combinatorial proof of 
(<ref>). 


The Fibonacci analogue of rook  placements defined in 
<cit.> is a slight 
variation of Fibonacci file placements. The main 
difference is that each 
tiling will cancel some of the top most cells in each column 
to its right that has not been canceled by a tiling 
which is further to the left.  Our goal is to ensure 
that if we start with a Ferrers board B =F(b_1, …, b_n), 
our cancellation scheme will ensure that the number of 
uncanceled cells in the empty columns are b_1, …, b_n-k, 
reading from left to right. 
That is, if B=F(b_1, …, b_n), then we let 
𝒩𝒯_k(B) denote the set of all configurations such that 
that there are k columns (i_1, …, i_k) of B where 
1 ≤ i_1 < ⋯ < i_k ≤ n such that the 
following conditions hold. 

  1. In column i_1, we place a Fibonacci tiling 
T_i,1 of height b_i_1 and for each j > i_1, 
this tiling cancels the top b_j-b_j-1 cells at the top 
of column j. This cancellation has the effect of 
ensuring that the number of uncanceled cells in the columns 
without tilings at this point is 
b_1, …, b_n-1, reading from left to right. 

  2. In column i_2, our cancellation due to the tiling 
in column i_1 ensures that there are b_i_2-1 uncanceled 
cells in column i_2. Then we place a Fibonacci tiling 
T_i,2 of height b_i_2-1 and for each j > i_2, 
we cancel the top b_j-1-b_j-2 cells in column j 
that has not been canceled by the tiling in column i_1. 
This cancellation has the effect of 
ensuring that the number of uncanceled cells in columns 
without tilings at this point 
is b_1, …, b_n-2, reading from left to right.

  3. In general, when we reach column i_s, we assume 
that the cancellation due to the tilings in columns  
i_1, …, i_j-1 ensure that the number of uncanceled 
cells in the columns without tilings is b_1, …, b_n-(s-1), 
reading from left to right. Thus there will be 
b_i_s -(s-1) uncanceled cells in column i_s at this point. 
Then we place a Fibonacci tiling 
T_i,s of height b_i_s-(s-1) and for each j > i_s, 
this tiling will 
cancel the top b_j-(s-1)-b_j-s cells in column 
j that has not been canceled by the tilings in 
columns i_1, …, i_s-1. 
This cancellation has the effect of 
ensuring that the number of uncanceled cells in columns 
without tilings at this point 
is b_1, …, b_n-s, reading from left to right.



We shall call such a configuration 
a Fibonacci rook placement and denote it by 

    P = ((i_1,T_i_1),  …, (i_k,T_i_k)).
 
Let one(P) denote the number of tiles of height 1 that appear 
in P and two(P) denote the number of tiles of height 2 that appear 
in P.  Then in <cit.>, we  defined the weight of P, WF(P,p,q), to be 
q^one(P)p^two(P).  For example, on the left in 
Figure <ref>, we have 
pictured an element P of 𝒩𝒯_3(F(2,3,4,4,6,6)) 
whose weight is q^5 p^2. In 
Figure <ref>, we have 
indicated the canceled cells by the tiling in column 
i by placing an i in the cell.  
We note in the special case where B = F(0,k,2k, …, (n-1)k), then 
our cancellation scheme is quite simple. That is, each tiling 
just cancels the top k cells in each column to its right which 
has not been canceled by tilings to its left. 
For example, on the right in 
Figure <ref>, we have 
pictured an element P of 𝒩𝒯_3(F(0,1,2,3,4,5)) 
whose weight is q^6 p. Again, we have 
indicated the canceled cells by the tiling in column 
i by placing an i in the cell.



FibrookA Fibonacci rook placement.

We define the k-th p,q-Fibonacci rook polynomial of B, 𝐫𝐓_k(B,p,q), 
by setting 

    𝐫𝐓_k(B,p,q) = ∑_P ∈𝒩𝒯_k(B) WF(P,p,q).

If k =0, then the only element of ℱ𝒯_k(B) is the empty placement 
whose weight by definition is 1. 


Then in <cit.>, we proved the following two theorems concerning 
Fibonacci rook placements in Ferrers boards. 

 
Let B =F(b_1, …, b_n) be a Ferrers 
board where 0 ≤ b_1 ≤⋯≤ b_n and b_n > 0. 
Let B^- = F(b_1, …, b_n-1). Then for all 
1 ≤ k ≤ n, 

    𝐫𝐓_k(B,p,q) = 𝐫𝐓_k(B^-,p,q)+ F_b_n-(k-1)(p,q) 𝐫𝐓_k-1(B^-,p,q).



 
Let B =F(b_1, …, b_n) be a Ferrers 
board where 0 ≤ b_1 ≤⋯≤ b_n and b_n > 0. 

    x^n =
    ∑_k=0^n 𝐫𝐓_n-k(B,p,q) (x-F_b_1(p,q))(x-F_b_2(p,q))⋯ 
    (x-F_b_k(p,q)).



To obtain the q-analogues that we want for this paper, we need to define two 
new weight functions on Fibonacci rook tilings. 
That is, suppose that B =F(b_1, …, b_n) is a Ferrers board 
and P = ((i_1,T_i_1),  …, (i_k,T_i_k)) is an Fibonacci 
rook tiling in 𝒩𝒯_k(B). Then we know 
that the number of uncanceled cells in the n-k columns 
which do not have tilings are b_1, … ,b_n-k reading from 
left to right. Suppose that the number of uncanceled cells in 
the columns with tilings are e_1, …, e_k reading from left 
to right so that tiling T_i_j is of height e_j for j =1, 
…, k.  The we define 

    𝐖_𝐁,𝐪(P)    =    q^∑_s=1^k _e_s(T_i_s) + 
    ∑_t=1^n-k F_b_t 
    𝐖_𝐁,𝐪(P)    =    q^∑_s=1^k _e_s(T_i_s).
 
For example, if B=(2,3,4,4,5,5) and P=((1,T_1),(3,T_3),(5,T_5)) is the rook tiling pictured 
in Figure <ref>, then e_1 =2, e_2 = 3 and e_3 =4 and 
one can check that _2(T_1) =0, _3(T_3) = F_2 =1, and _4(T_5) = F_3 = 2. Thus 
𝐖_𝐁,𝐪(P)=q^0+1+2+F_2+F_3+F_4 = q^9 and 
𝐖_𝐁,𝐪(P)=q^0+1+2 = q^3.
If k =0, then the only element of ℱ𝒯_k(B) is the empty placement 
∅ which means that 𝐖_𝐁,𝐪(∅) = 
q^∑_i=1^n F_b_i and 𝐖_𝐁,𝐪(∅) =1.



Then we define 𝐑𝐓_k(B,q)  
by setting 

    𝐑𝐓_k(B,q) = ∑_P ∈𝒩𝒯_k(B)𝐖_𝐁,𝐪(P)

and 

    𝐑𝐓_k(B,q) = ∑_P ∈𝒩𝒯_k(B)𝐖_𝐁,𝐪(P).

Note that because of our cancellation scheme, there is a very 
simple relationship between 𝐑𝐓_k(B,q) and 
𝐑𝐓_k(B,q) in the case where B = F(b_1, …, 
b_n). That is, in any placement P ∈𝒩𝒯_k(B), 
the empty columns have b_1, …, b_n-k uncanceled cells, 
reading from left to right, so that 

    𝐑𝐓_k(B,q) = q^∑_i=1^n-k F_b_i𝐑𝐓_k(B,q).








Let B=F(b_1, …, b_n) be  a Ferrers board and 
x be a positive integer. 
Then we let AugB_x denote the board where 
we start with B_x and add the flip of the board B about 
its baseline below the board. We shall call the 
the line that separates B from these x rows the upper bar 
and the line that separates the x rows from the flip 
of B added below the x rows the lower bar. We shall 
call the flipped version of B added below B_x the board 
B. For example, 
if B=F(2,3,4,4,5,5), then the board AugB_7 is pictured 
in Figure <ref>.

augAn example of an augmented board AugB_x.


The analogue of mixed placements in AugB_x are 
more complex than the mixed placements for B_x. We process 
the columns from left to right. 
If we are in column 1, then we can do one of the following three things. 

  i. We can put a Fibonacci tiling in  
cells in the first column in B.  Then we must 
cancel the top-most cells in each of the columns in B to its right 
so that the number of uncanceled cells in 
the columns to its right are b_1,b_2, …, b_n-1, respectively, as 
we read from left to right. This means 
that we will cancel b_i-b_i-1 at the top of column i in B 
for i=2, …, n. We also cancel the same number 
of cells at the bottom of the corresponding columns of B.

  ii. We can place a rook in any row of column 1 that 
lies between the upper bar and lower bar.  This rook 
will not cancel anything. 

  iii. We can put a flip of Fibonacci tiling 
in column 1 of B.  This tiling will not 
cancel anything. 

Next assume that when we get to column j, the 
number of uncanceled cells in the columns that have 
no tilings in B and B are b_1, …, b_k  for some k as 
we read from left to right. Suppose there are 
b_i uncanceled cells in B in column j.  
Then we can do one of three things. 

  i. We can put a Fibonacci tiling of height b_i  in the uncanceled cells in column j in B.  Then we must 
cancel top-most cells of the columns in B to its right 
so that the number of uncanceled cells in 
the columns which have no tilings up to this point 
are b_1,b_2, …, b_k-1, 
We also cancel the same number 
of cells at the bottom of the corresponding columns of B

  ii. We can place a rook in any row of column j that 
lies between the upper bar and lower bar.  This rook 
will not cancel anything. 

  iii. We can put a flip of Fibonacci tiling in the b_i 
uncanceled cells in column j of B.  This tiling will not 
cancel anything 


We let ℳ_n(AugB_x) denote set of all 
mixed rook placements on AugB_x. For any placement 
P ∈ℳ_n(AugB_x), 
we define 𝐖_𝐀𝐮𝐠𝐁_𝐱,𝐪(P) and 𝐖_𝐀𝐮𝐠𝐁_𝐱,𝐪(P)
as follows. For any column i, suppose that 
the number of uncanceled cells in B in column i is t_i. 
Then the factor 𝐖_𝐢,𝐀𝐮𝐠𝐁_𝐱,𝐪(P) that the placement 
in column i contributes to 𝐖_𝐀𝐮𝐠𝐁_𝐱,𝐪(P) is 


  * q^_t_i(T_i) if there is tiling T_i in B in 
column i, 

  * q^F_t_i+s_i-1 if there is a rook in row s_i^th row from 
the top in the x rows that lie between the upper bar and lower bar, 
and 

  * -q^_t_i(T_i) if there is a flip of a tiling T_i 
in column i of B.

Then we define 

    𝐖_𝐀𝐮𝐠𝐁_𝐱,𝐪(P) = ∏_i=1^n 𝐖_𝐢,𝐀𝐮𝐠𝐁_𝐱,𝐪(P).


Similarly, the factor 𝐖_𝐢,𝐀𝐮𝐠𝐁_𝐱,𝐪(P) 
that the tile placement 
in column i contributes to 𝐖_𝐀𝐮𝐠𝐁_𝐱,𝐪(P) is  


  * q^_t_i(T_i) if there is tiling T_i in B in 
column i, 

  * q^s_i-1 if there is a rook in row s_i^th row from 
the top in the x rows that lie between the upper bar and lower bar, 
and 

  * -q^_t_i(T_i) if there is a flip of a tiling T_i 
in column i of B.

Then we define 

    𝐖_𝐀𝐮𝐠𝐁_𝐱,𝐪(P) = ∏_i=1^n 
    𝐖_𝐢,𝐀𝐮𝐠𝐁_𝐱,𝐪(P).










For example, 
Figure <ref> pictures a mixed placement P in 
AugB_x where B = F(2,3,4,4,5,5) and x is 7 where  
_2(T_1) =0,  _4(T_4) =F_2 =1, 
and _4(T_5) =F_3 =2 where T_i is the tiling in 
column i for i ∈{1,4,5}.  The rooks columns 2 and 6 
are in row 5 and the rook in column 3 is in row 3 so that s_2 =s_6 =5 
and s_3=3. Thus 

    𝐖_𝐀𝐮𝐠𝐁_𝐱,𝐪(P)    =    -q^0+(4+F_2)+(2+F_3)+1+2+(4+F_4) = -q^19 
    𝐖_𝐀𝐮𝐠𝐁_𝐱,𝐪(P)    =    
    -q^0+(4)+(2)+1+2+(4) = -q^13
 


aug2A mixed rook placement.


Our next theorem results from counting 
∑_P ∈ℳ_n(AugB_x)𝐖_𝐀𝐮𝐠𝐁_𝐱,𝐪(P) 
in two different ways. 

 
Let B =F(b_1, …, b_n) be a Ferrers 
board where 0 ≤ b_1 ≤⋯≤ b_n and b_n > 0 
and x ∈. Then 

    [x]_q^n =
    ∑_k=0^n 𝐑𝐓_n-k(B,q) 
    ([x]_q-[F_b_1]_q) ([x]_q-[F_b_2]_q)⋯ 
    ([x]_q-[F_b_k]_q).



Fix x to be a positive integer and consider 
the sum S=∑_P ∈ℳ_n(AugB_x)𝐖_𝐀𝐮𝐠𝐁_𝐱,𝐪(P). 
First we consider the contribution of each column as we proceed 
from left to right. Given our three choices 
in column 1, the contribution of our choice of the tilings of 
height b_1 in column 1 of B is [F_b_1]_q, the choice 
of placing a rook in between the upper  bar and the lower is [x]_q, 
and the contribution of our choice of the tilings of 
height b_1 in column 1 of B is -[F_b_1]_q. 
Thus the contribution of our choices in 
column 1 to S is [F_b_1]_q+[x]_q -[F_b_1]_q = [x]_q. 

In general, after we have processed our choices in 
the first j columns, our cancellation scheme ensures 
that the number of uncanceled cells in B and B in 
the j-th column is b_i for some i ≤ j. 
Thus given our three choices 
in column j, the contribution of our choice of the tilings of 
height b_i in column j of B is [F_b_i]_q, the choice 
of placing a rook in between the upper  bar and the lower is [x]_q, 
and the contribution of our choice of the tilings of 
height b_i in column j of B is -[F_b_i]_q. 
Thus the contribution of our choices in 
column j to S is [F_b_i]_q+[x]_q -[F_b_i]_q = [x]_q.
It follows that S = [x]_q^n. 


On the other 
hand, suppose that we fix a Fibonacci rook placement  
P ∈𝒩𝒯_n-k(B). 
Then we want to compute the S_P = ∑_Q ∈ℳ_n(AugB_x), 
Q ∩ B = P𝐖_𝐀𝐮𝐠𝐁_𝐱,𝐪(P) which is the sum 
of 𝐖_𝐀𝐮𝐠𝐁_𝐱,𝐪(P) over all 
mixed placements Q such that Q intersect B equals P. 
Our cancellation scheme ensures that the number of uncanceled cells in B and B 
in the k columns that do not contain tilings in P is 
b_1, …, b_k as we read from right to left. 
For each such 1 ≤ i ≤ k, the factor that 
arises from either choosing a rook to be placed 
in between the upper bar and lower bar or a flipped 
Fibonacci tiling of height b_i in B is 
[x]_q-[F_b_i]_q. It follows that 

    S_P =𝐖_𝐁,𝐪(P) ∏_i=1^k [x]_q-[F_b_i]_q.

Hence it follows that 

    S    =   ∑_k=0^n ∑_P ∈𝒩𝒯_n-k(B) S_P 
       =   ∑_k=0^n (∏_i=1^k [x]_q-[F_b_i]_q)  
    ∑_P ∈𝒩𝒯_k(B)𝐖_𝐁,𝐪(P) 
       =   ∑_k=0^n 𝐑𝐓_n-k(B,q) ( ∏_i=1^k [x]_q-[F_b_i]_q).



 

 
Let B =F(b_1, …, b_n) be a Ferrers 
board where 0 ≤ b_1 ≤⋯≤ b_n and b_n > 0 
and x ≥ b_n. Then 

    [x]_q^n =
    ∑_k=0^n 𝐑𝐓_n-k(B,q) [x-F_b_1]_q [x-F_b_2]_q⋯ 
    [x-F_b_k]_q.



It is easy to see from our cancellation scheme that 

    𝐑𝐓_n-k(B,q) =q^F_b_1+ ⋯ + F_b_k𝐑𝐓_n-k(B,q).

Thus it follows from (<ref>) 
that 

    [x]_q^n =
    ∑_k=0^n 𝐑𝐓_n-k(B,q)q^-(F_b_1+ ⋯ + F_b_k) 
    ([x]_q-[F_b_1]_q) ([x]_q-[F_b_2]_q)⋯ 
    ([x]_q-[F_b_k]_q).

However since x ≥ F_b_i for every i, 

    [x]_q-[F_b_i]_q = q^F_b_i[x-F_b_i]_q

so that 

    [x]_q^n =
    ∑_k=0^n 𝐑𝐓_n-k(B,q)  
    [x-F_b_1]_q [x-F_b_2]_q⋯ 
    [x-F_b_k]_q.


 









Now consider the special case of the previous three  theorems 
when B_n = F(0,1,2, …, n-1). Then (<ref>) implies 
that 

    𝐑𝐓_n+1-k(B_n+1,q) = q^F_k-1𝐑𝐓_n+1-k(B_n,q) + 
    [F_k]_q 𝐑𝐓_n-k(B_n,q).

Similarly (<ref>) implies 
that 

    𝐑𝐓_n+1-k(B_n+1,q) = 
    𝐑𝐓_n+1-k(B_n,q) + 
    [F_k]_q 𝐑𝐓_n-k(B_n,q).


It then easily follows that for all 0 ≤ k ≤ n, 

    𝐒𝐅_n,k(q) = 𝐑𝐓_n-k(B_n,q)

and 

    𝐒𝐅_n,k(q) = 𝐑𝐓_n-k(B_n,q).




Note that 𝐒𝐅_n,0(q) = 𝐒𝐅_n,0(q) =0 
for all n ≥ 1 since there are no Fibonacci rook placements in 
𝒩𝒯_n(B_n) since there are only n-1 non-zero columns. 
Moreover such a situation, we see that (<ref>) 
implies that for x ≥ n, 

    [x]_q^n = 
    ∑_k=1^n 𝐒𝐅_n,k(q) [x]_q [x-F_1]_q [x-F_2]_q⋯ 
    [x-F_k-1]_q

Thus we have given a combinatorial proof of 
(<ref>). 
Similarly, (<ref>) 
implies that for x ≥ n, 

    [x]_q^n = 
    ∑_k=1^n 𝐒𝐅_n,k(q) [x]_q ([x]_q-[F_1]_q) 
    ([x]_q-[F_2]_q)⋯ 
    ([x]_q-[F_k-1]_q)

Thus we have given a combinatorial proof of 
(<ref>). 





§ IDENTITIES FOR 𝐒𝐅_N,K(Q) AND 
𝐜𝐅_N,K(Q)



In this section, we shall derive various identities and special values 
for the Fibonacci analogues of the Stirling numbers 𝐒𝐅_n,k(q), 
𝐒𝐅_n,k(q), 𝐜𝐅_n,k(q),  and 
𝐜𝐅_n,k(q). 

Note that by (<ref>), 

    𝐒𝐅_n,k(q) = q^∑_i=1^k-1 F_i𝐒𝐅_n,k(q).


Then we have the following theorem. 
 


  * 𝐒𝐅_n,n(q) =1 and 
𝐒𝐅_n,n(q) =q^∑_i=1^n-1 F_i.


  * 𝐒𝐅_n,n-1(q) =∑_i=1^n-1 [F_i]_q 
and 
𝐒𝐅_n,n-1(q) =q^∑_i=1^n-2 F_i∑_i=1^n-1 [F_i]_q.

  * 𝐒𝐅_n,n-2(q) =∑_i=1^n-2 [F_i]_q(∑_j=i^n-2 
[F_j]_q)
and 
𝐒𝐅_n,n-2(q) =q^∑_i=1^n-3 F_i∑_i=1^n-2 [F_i]_q(∑_j=i^n-2 
[F_j]_q).



  * 𝐒𝐅_n,1(q) =1 and 
𝐒𝐅_n,1(q) =1.


  * 𝐒𝐅_n,2(q) =(n-1) and 
𝐒𝐅_n,2(q) =q(n-1).


  * 𝐒𝐅_n,3(q) =(1+q)^n-1 -(q(n-1)+1)/q^2 and 
𝐒𝐅_n,3(q) =(1+q)^n-1 -(q(n-1)+1).
 




For (1), it is easy to see that 𝐒𝐅_n,n(q) =1 since 
the only placement in ℱ𝒯_n-n(B_n) is the empty placement. 
The fact that 𝐒𝐅_n,k(q) =q^∑_i=1^n-1 F_i then 
follows from (<ref>). 


For (2), we can see that 𝐒𝐅_n,n-1(q) =
∑_i=1^n-1[F_i]_q because placements in 
ℱ𝒯_n-(n-1)(B_n) = ℱ𝒯_1(B_n) 
have exactly one column which 
is filled with a Fibonacci tiling.  If that column is 
column i+1, then i ≥ 1 and 
the sum of the weights of the possible tilings 
in column i is [F_i]_q.  The fact that 
𝐒𝐅_n,n-1(q) =q^∑_i=1^n-2 F_i∑_i=1^n-1 [F_i]_q then follows from (<ref>).


For (3), we can classify the placements in 
ℱ𝒯_n-(n-2)(B_n) = ℱ𝒯_2(B_n) by the left-most 
column which contains a tiling.  If that column is 
column i+1, then i ≥ 1 and 
the sum of the weights of the possible tilings 
in column i is [F_i]_q.  Moreover, any tiling 
in column i cancels one cell in the remaining columns so 
that number of uncanceled cells in the columns to the right 
of column i+1 will be i, …, n-2, reading from right to left. 
It then follows that 

    𝐒𝐅_n,n-2(q) =∑_i=1^n-2 [F_i]_q(∑_j=i^n-2 
    [F_j]_q).
 
The fact that 

    𝐒𝐅_n,n-1(q) =q^∑_i=1^n-3 F_i∑_i=1^n-2 [F_i]_q(∑_j=i^n-2 
    [F_j]_q)
 
then follows from (<ref>).




For (4), note that the elements in ℱ𝒯_n-1(B_n) have 
a tiling in every column. Given our cancellation scheme, there is 
exactly one such configuration. For example, the unique 
element of ℱ𝒯_5(B_6) is pictured in Figure 
<ref> where we have placed is in the cells 
canceled by the tiling in column i. Thus the unique 
element of ℱ𝒯_n-1(B_n) is just the Fibonacci 
rook placement where there is tiling of height one in each column. Thus 
𝐒𝐅_n,1(q) =𝐒𝐅_n,1(q) =1 since 
the rank of each tiling height 1 is 0. 

fullrooksThe Fibonacci rook tiling in ℱ𝒯_5(B_6).


For (5), note that the elements in ℱ𝒯_n-2(B_n) 
have exactly one column i ≥ 2 which does not 
have a tiling.  Given our cancellation scheme, if the column 
with out a tiling is column i ≥ 2, then any non-empty 
column to the left of column i will be filled with a tiling 
of height 1 and every column to the right of column i will 
be filled with a tiling of height 2.  For example, the unique 
element of ℱ𝒯_6(B_8) is pictured in Figure 
<ref> where we have placed is in the cells 
canceled by the tiling in column i. Since the ranks of 
the tilings of heights 1 and 2 are 0, it follows 
that 𝐒𝐅_n,2(q) =n-1.  
The fact that 
𝐒𝐅_n,n-1(q) =q(n-1)  
then follows from (<ref>). 

2fullrooksA Fibonacci rook tiling in ℱ𝒯_6(B_8).


For (6), we proceed by induction. Note that we have 
proved 

    𝐒𝐅_3,3(q) =q^F_1+F_2=q^2 = (1+q)^2-(2q +1).

Now assume that n ≥ 3 and 
𝐒𝐅_n,3(q)=(1+q)^n-1-((n-1)q+1). Then 

    𝐒𝐅_n+1,3(q)    =   q^F_2𝐒𝐅_n,2(q)+[F_3]_q
    𝐒𝐅_n,3(q)
       =    q(q(n-1) )+(1+q)((1+q)^n-1-((n-1)q+1)) 
       =    q^2(n-1) +(1-q)^n -(n-1)q-(n-1)q^2 -q -1
       =    (1-q)^n -(nq+1).

The fact that 𝐒𝐅_n,3(q)=
(1+q)^n-1-((n-1)q+1)/q^2 then follows from (<ref>). 








Next we define  

    𝕊𝔽_k(q,t) := 
    ∑_n ≥ k𝐒𝐅_n,k(q) t^n
 
for k ≥ 1
It follows from Theorem <ref> that 

    𝕊𝔽_1(q,t) = 
    ∑_n ≥ 1𝐒𝐅_n,1(q) t^n = 
    ∑_n ≥ 1 t^n = t/1-t.

Then for k > 1,  

    𝕊𝔽_k(q,t)    =   ∑_n ≥ k𝐒𝐅_n,k(q) t^n 
       =    t^k + ∑_n > k𝐒𝐅_n,k(q) t^n 
       =    t^k + t ∑_n > k( 𝐒𝐅_n-1,k-1(q) +
    [F_k]_q 𝐒𝐅_n-1,k-(p,q)) t^n-1
       =    t^k + t(∑_n > k𝐒𝐅_n-1,k-1(q) t^n-1) 
    + [F_k]_qt(∑_n > k𝐒𝐅_n-1,k(q) t^n-1) 
       =    t^k + t(𝕊𝔽_k-1(q,t) -t^k-1) +  [F_k]_qt 
    𝕊𝔽_k(q,t).

It follows 
that 

    𝕊𝔽_k(q,t) = t/(1 - [F_k]_qt)𝕊𝔽_k-1(q,t).

The following theorem easily follows from (<ref>) and 
(<ref>).
 For all k ≥ 1, 

    𝕊𝔽_k(q,t)= t^k/(1-[F_1]_qt) (1-[F_2]_qt)⋯ 
    (1-[F_k]qt).



Note that it follows from (<ref>) and  Theorem <ref> that 

    𝕊𝔽_k(q,t)= ∑_n ≥ k𝐒𝐅_n,k(q) t^n = 
    q^∑_i=1^k-1F_i t^k/(1-[F_1]_qt) (1-[F_2]_qt)⋯ 
    (1-[F_k]qt).







For any formal power series in f(x) = ∑_n ≥ 0f_n x^n, 
we let f(x)|_x^n = f_n denote the coefficient of x^n in 
f(x). Our next result will give formulas for 
𝐒𝐅_n,k(q)|_q^s for s =0,1,2. 


 

  * For all n ≥ k ≥ 1, 𝐒𝐅_n,k(q)|_q^0 = n-1k-1. 


  * For all n > k ≥ 2, 𝐒𝐅_n,k(q)|_q = (k-2)n-1k.



  * For all 
n ≥ s, 𝐒𝐅_n,3(q)|_q^s=  n-1s+2.



  * For all n ≥ k ≥ 3,  𝐒𝐅_n,k(q)|_q^2 = (k-3)n-1k 
+ k-12n-1k+1. 


  * for all n ≥ k ≥ 4,

    𝐒𝐅_n,k(q)|_q^3 = (k-4)n-1k 
    + (k-12 +k-22 -1)n-1k+1 
    +k3n-1k+2.






  * For all 
n ≥ k ≥ 4,

    𝐒𝐅_n,k(q)|_q^4   =    (k-4)n-1k 
    + (k-12 +k-22 +k-32-3)n-1k+1 +
          (2k3+k-13 -k +1) n-1k+2 + k+14n-1k+3.




 

For (1), note that a placement P in ℱ𝒯_n-k(B_n) must 
have k-1 empty columns among columns 2, …, n.  If 
WF(P) =1, then it must be the case that all the tilings  
in the columns which contain tilings in P must have rank 0 so that 
the tiling must contain only tiles of height 1.  Thus P is 
completely determined by the choice of the k-1 empty columns among 
columns 2, …, n.  Thus 
𝐒𝐅_n,k(p,q)|_q^0 = n-1k-1. 

For (3), note that by part 6 of Theorem <ref>, 
we have that for any s ≥ 0, 

    𝐒𝐅_n,3(q)|_q^s   =   𝐒𝐅_n,3(q)|_q^s+2 = (1+q)^n-1-((n-1)q+1)|_q^s+2
       =   n-1s+2.





For (2), note that 
𝐒𝐅_n,2(q)|_q = 0 since 
𝐒𝐅_n,k(q) = (n-1) by part 5 of Theorem <ref>.
By (3), 𝐒𝐅_n,3(q)|_q= n-13. 
Thus our formula holds for n =2 and n=3. 


Next fix k ≥ 4 and assume by induction that 
𝐒𝐅_n,k-1(q)|_q= (k-3) n-1k-1 
for all n ≥ k-1.  Then we shall prove by induction on 
n that 𝐒𝐅_n,k(q)|_q= (k-2) n-1k.
The base case n =k holds since 
𝐒𝐅_k,(q)=1. But then assuming that 
𝐒𝐅_n,k(q)|_q= (k-2) n-1k, 
we see that 

    𝐒𝐅_n+1,k(q)|_q   =   𝐒𝐅_n,k-1(q)|_q+ 
    ((1+q+q^2+ ⋯ + q^F_k-1)𝐒𝐅_n,k(q))|_q
       =    (k-3) n-1k-1 + 𝐒𝐅_n,k(q))|_q^0 +
    𝐒𝐅_n,k(q))|_q
       =    (k-3) n-1k-1 + n-1k-1 +(k-2) n-1k
       =    (k-2) nk.




Parts (4), (5), and (6) can easily be proved by induction.  

For example, by (3),

    𝐒𝐅_n,3(q)|_q^2 = n-14
 
so that our formula holds for k=3.  Now suppose 
that k ≥ 4 and our formula holds for k-1. That is, 

    𝐒𝐅_n,k-1(q)|_q^2 = (k-4) 
    n-1k-1 + k-22n-1k.

Next observe that 𝐒𝐅_k,k(q)|_q^2 =0 since 
 𝐒𝐅_k,k(q) =1 so that our formula 
holds for n =k.  Note also that for 
k ≥ 4, F_k ≥ 3. 
But then for n ≥ k ≥ 4,  

    𝐒𝐅_n+1,k(q)|_q^2   =   𝐒𝐅_n,k-1(q)|_q^2+([F_k]_q 
    𝐒𝐅_n,k(q))|_q^2
       =   𝐒𝐅_n,k-1(q)|_q^2+((1+q+q^2)
    𝐒𝐅_n,k(q))|_q^2
       =   𝐒𝐅_n,k-1(q)|_q^2+
    𝐒𝐅_n,k(q)|_q^0+
    𝐒𝐅_n,k(q)|_q+𝐒𝐅_n,k(q)|_q^2
       =    (k-4) n-1k-1 + k-22n-1k 
    + n-1k-1 + (k-2)n-1k
          + 𝐒𝐅_n,k(q)|_q^2
       =    (k-3) n-1k-1 + k-12n-1k + 𝐒𝐅_n,k(q)|_q^2.

This gives us a recursion for 𝐒𝐅_n+1,k(q)|_q^2 in terms of 𝐒𝐅_n,k(q)|_q^2 which we can iterate 
to prove that 

    𝐒𝐅_n,k(q)|_q^2 = 
    (k-3) n-1k + k-12n-1k+1.



For (5), we first have to establish the base case k=4. 


    𝐒𝐅_n+1,4(q)|_q^3   =   𝐒𝐅_n,3(q)|_q^3+([F_4]_q 
    𝐒𝐅_n,4(q))|_q^3
       =   𝐒𝐅_n,3(q)|_q^3+((1+q+q^2)
    𝐒𝐅_n,4(q))|_q^3
       =   𝐒𝐅_n,k-1(q)|_q^3+
    𝐒𝐅_n,k(q)|_q+
    𝐒𝐅_n,k(q)|_q^2+𝐒𝐅_n,k(q)|_q^3
       =   n-15 + 2n-14 +
    (n-14+3n-15)
    +𝐒𝐅_n,k(q)|_q^3
       =    3n-14 + 4n-15 +𝐒𝐅_n,4(q)|_q^3.
 
This gives us a recursion for 𝐒𝐅_n+1,4(q)|_q^3 in terms of 𝐒𝐅_n,4(q)|_q^3 which we can iterate 
to prove that 

    𝐒𝐅_n,4(q)|_q^3 = 
    3n-15 + 4n-16.

Thus our formula for (5) holds for k =4. 


Next assume that k ≥ 5. First we note that
𝐒𝐅_k,k(q)|_q^3 =0 since 
 𝐒𝐅_k,k(q) =1 so that our formula 
holds for n =k.  Note also that for 
k ≥ 5, F_k ≥ 5. 
Now suppose our formula holds for k-1. That is, 

    𝐒𝐅_n,k-1(q)|_q^3 = 
    
    (k-5) 
    n-1k-1 + (k-22+k-32-1)
    n-1k +k-13n-1k+1.

Next observe that 𝐒𝐅_k,k(q)|_q^3 =0 since 
 𝐒𝐅_k,k(q) =1 so that our formula 
holds for n =k.  Note also that for 
k ≥ 4, F_k ≥ 3. 
But then for n ≥ k ≥ 5,  

    𝐒𝐅_n+1,k(q)|_q^3   =   𝐒𝐅_n,k-1(q)|_q^3+([F_k]_q 
    𝐒𝐅_n,k(q))|_q^3
       =   𝐒𝐅_n,k-1(q)|_q^2+((1+q+q^2+q^3)
    𝐒𝐅_n,k(q))|_q^3
       =   𝐒𝐅_n,k-1(q)|_q^3+
    𝐒𝐅_n,k(q)|_q^0+
    𝐒𝐅_n,k(q)|_q+
    𝐒𝐅_n,k(q)|_q^2+ 
    𝐒𝐅_n,k(q)|_q^3
       =    (k-5) 
    n-1k-1 + (k-22+k-32-1)
    n-1k +k-13n-1k+1
           + n-1k-1+(k-2)n-1k+ 
    (k-3)n-1k + k-12n-1k+1
          + 𝐒𝐅_n,k(q)|_q^3
       =    (k-4) n-1k-1 + (k-12+k-22-1)
    n-1k + k3n-1k+1
          + 𝐒𝐅_n,k(q)|_q^3.

This gives us a recursion for 𝐒𝐅_n+1,k(q)|_q^3 in terms of 𝐒𝐅_n,k(q)|_q^3 which we can interate 
to prove that 

    𝐒𝐅_n,k(q)|_q^3 = 
    (k-4) n-1k + (k-12+k-22-1)
    n-1k+1 + k3n-1k+2.



For (6), again, we first have to establish the base case k=4. 


    𝐒𝐅_n+1,4(q)|_q^4   =   𝐒𝐅_n,3(q)|_q^4+([F_4]_q 
    𝐒𝐅_n,4(q))|_q^4
       =   𝐒𝐅_n,3(q)|_q^4+((1+q+q^2)
    𝐒𝐅_n,4(q))|_q^4
       =   𝐒𝐅_n,k-1(q)|_q^4+
    𝐒𝐅_n,k(q)|_q^2+
    𝐒𝐅_n,k(q)|_q^3+𝐒𝐅_n,k(q)|_q^4
       =   n-16 + n-14 +3n-15 + 
    3n-15+4n-16 + 
          𝐒𝐅_n,k(q)|_q^4
       =   n-14 + 6n-15 + 5n-16+𝐒𝐅_n,4(q)|_q^3.
 
This gives us a recursion for 𝐒𝐅_n+1,4(q)|_q^4 in terms of 𝐒𝐅_n,4(q)|_q^4 which we can iterate 
to prove that 

    𝐒𝐅_n,4(q)|_q^4 =  n-15 + 6n-16 + 5n-17.

Thus our formula for (6) holds for k =4. 


Next assume that k ≥ 5. First we note that
𝐒𝐅_k,k(q)|_q^4 =0 since 
 𝐒𝐅_k,k(q) =1 so that our formula 
holds for n =k.  Note also that for 
k ≥ 5, F_k ≥ 5. 
Now suppose our formula holds for k-1. That is, 

    𝐒𝐅_n,k-1(q)|_q^4   =    (k-5) 
    n-1k-1 + (k-22+k-32+k-32-3)
    n-1k +
          (k-13+k-23-(k-1)+1)n-1k+1 + k4n-1k+2.

Next observe that 𝐒𝐅_k,k(q)|_q^5 =0 since 
 𝐒𝐅_k,k(q) =1 so that our formula 
holds for n =k.  Note also that for 
k ≥ 5, F_k ≥ 5. 
But then for n ≥ k ≥ 5,  

    𝐒𝐅_n+1,k(q)|_q^4   =   𝐒𝐅_n,k-1(q)|_q^4+([F_k]_q 
    𝐒𝐅_n,k(q))|_q^4
       =   𝐒𝐅_n,k-1(q)|_q^4+((1+q+q^2+q^3+q^4)
    𝐒𝐅_n,k(q))|_q^4
       =   𝐒𝐅_n,k-1(q)|_q^3+
    𝐒𝐅_n,k(q)|_q^0+
    𝐒𝐅_n,k(q)|_q+
    𝐒𝐅_n,k(q)|_q^2+ 
    𝐒𝐅_n,k(q)|_q^3
           + 𝐒𝐅_n,k(q)|_q^4
       =    (k-5) 
    n-1k-1 + (k-22+k-32+k-32-3)
    n-1k +
          + (k-13+k-23-(k-1)+1)n-1k+1 + 
    k4n-1k+2
          + n-1k-1+(k-2)n-1k+ 
    (k-3)n-1k + k-12n-1k+1
          + (k-4) n-1k + (k-12+k-22-1)
    n-1k+1 + k3n-1k+2
          + 𝐒𝐅_n,k(q)|_q^4
       =    (k-4)n-1k-1 + 
    ( k-12+k-22+k-32-3 )
    n-1k
          +(k3+k-13-k+1) n-1k+1 +
    k+14n-1k+2
          +𝐒𝐅_n,k(q)|_q^4.

This gives us a recursion for 𝐒𝐅_n+1,k(q)|_q^4 in terms of 𝐒𝐅_n,k(q)|_q^4 which we can iterate 
to prove that 

    𝐒𝐅_n,k(q)|_q^4   =    
    (k-4) n-1k + (k-12+k-22+k-22-3)n-1k+1 + 
          (2k3+k-12-k+1)
    n-1k+2+ k+14n-1k+3.



A sequence of real numbers a_0, …, a_n is is said to be unimodal 
if there is a 0 ≤ j ≤ n such that 
a_0 ≤⋯≤ a_j ≥ a_j+1≥⋯≥ a_n and 
is said to be log-concave if for 0 ≤ i ≤ n, 
a_i^2- a_i-1a_i+1≥ 0 where we set a_-1 = a_n+1 =0.
If a sequence is log-concave, then 
it is unimodal. A polynomial P(x) = ∑_k=0^n a_k x^k is said to be unimodal if a_0, …, a_n is a unimodal sequence and is said 
to be log-concave if a_0, …, a_n is log concave. 

It is easy to see from Theorem <ref> that 
𝐒𝐅_n,k(q) is unimodal for 
all n ≥ k when k ∈{1,2,3}.  
Computational evidence suggests that 
𝐒𝐅_n,4(q) is unimodal for all n ≥ 4 and that 
𝐒𝐅_n,5(q) is unimodal for all n ≥ 5. 
However, it is not the case that  
𝐒𝐅_n,6(q) is unimodal for all n ≥ 6. 
For example, one can use part 3 of Theorem <ref> to compute 

    𝐒𝐅_8,6(q) = 
    
    21 +28q +31q^2+29q^3+30q^4+25q^5+23q^2+22q^7+15q^8+10q^9+7q^10+5q^11+3q^12+2q^13+q^14.



It is not difficult to see that for any Ferrers board 
B =F(b_1,…,b_n), the coefficients that appear 
in the polynomials 𝐅𝐓_k(B,q) and 𝐅𝐓_k(B,q) 
are essentially the same. That is, we have the following theorem. 


Let B=F(b_1, …, b_n) be a skyline board. Then 

    𝐅𝐓_k(B,q) = ( ∏_i=1^n (1+[F_b_i]_qz))|_z^k

and 

    𝐅𝐓_k(B,q) = 
    ( ∏_i=1^n (q^F_b_i+[F_b_i]_qz))|_z^k =
    q^∑_i=1^n F_b_i( ∏_i=1^n (1+1/q[F_b_i]_1/qz))|_z^k



It is easy to see that if we are creating a Fibonacci file tiling 
in ℱ𝒯_k(B), then in column i, we have two choices, 
namely, we can leave the column empty or put a Fibonacci tiling 
of height b_i.  For 𝐅𝐓_k(B,q), 
the weight of an empty column is 1 and the sum of weights of 
the Fibonacci tilings of height b_i is [F_b_i]_q.  Thus 
( ∏_i=1^n (1+[F_b_i]_qz))|_z^k is equal 
to the sum over all Fibonacci file tilings where exactly k columns 
have tiling which is equal to 𝐅𝐓_k(B,q).


Similarly, For 𝐅𝐓_k(B,q), 
the weight of an empty column i when it is empty is 
q^F_b_i and the sum of weights of 
the Fibonacci tilings of height b_i is [F_b_i]_q.  Thus 
( ∏_i=1^n (q^F_b_i+[F_b_i]_qz))|_z^k is equal 
to the sum over all Fibonacci file tilings where exactly k columns 
have tiling which is equal to 𝐅𝐓_k(B,q).



It follow that for any n, the coefficient of q^n in 
( ∏_i=1^n (1+[F_b_i]_qz))|_z^k is equal to 
the coefficient of 1/q^n+k in 
( ∏_i=1^n (1+1/q[F_b_i]_1/qz))|_z^k.
It follows that 

    𝐅𝐓_k(B,q)_q^n = 
    𝐅𝐓_k(B,q)|_q^-n-k+∑_i=1^n F_b_i.



It is easy to see from (<ref>) that 

    𝐜𝐅_n,n-1(q) = ∑_i=1^n-1 [F_i]_q

so that coefficient of q^k in 𝐜𝐅_n,n-1(q) weakly decreases as k goes from 0 to F_n-1-1. It follows that the coefficient 
of q^k in 𝐜𝐅)_n,n-1(q) weakly increase. Similarly, 
it is easy to see that 

    𝐜𝐅_n,1(q) = ∏_i=1^n-1[F_i]_q

so that 𝐜𝐅_n,1(q) is just the rank generating 
function of a product of chains which is know to be symmetric and unimodal, 
see <cit.>.




From our computational evidence, it seems that the polynomials 
𝐜𝐅_n,2(q) 
are unimodal. However, it is not the case 
𝐜𝐅_n,k(q) are unimodal for all n and k.
For example, 𝐜𝐅_9,7(q) starts out 

    28+42q+50q^2+53q^3+58q^4+57q^5+58q^6+60q^7+ … .




Finally, our results show that the matrices  
||(-1)^n-k𝐜𝐅_n,k(q)|| and 
||𝐒𝐅_n,k(q)|| are inverses of each other. One can give a combinatorial proof of this fact.  Indeed, the combinatorial 
proof of <cit.> which shows that matrices 
||(-1)^n-k𝐜𝐟_n,k(q)|| and 
||𝐒𝐟_n,k(q)|| are inverses of each other can also be applied to show 
that the matrices  
||(-1)^n-k𝐜𝐅_n,k(q)|| and 
||𝐒𝐅_n,k(q)|| are inverses of each other.





























20



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