Available at
http://pvamu.edu/aam
Appl. Appl. Math.
ISSN: 1932-9466
Applications and Applied
Mathematics:
An International Journal
(AAM)
Vol. 7, Issue 2 (December 2012), pp. 508 - 533
Generalizations of Two Statistics on Linear Tilings
Toufik Mansour & Mark Shattuck
Department of Mathematics
University of Haifa
31905 Haifa, Israel
tmansour@univ.haifa.ac.il; maarkons@excite.com
Received:April 28, 2012;Accepted: September 11, 2012
Abstract
In this paper, we study generalizations of two well-known statistics on linear square-and-domino
tilings by considering only those dominos whose right half covers a multiple of , where is a
fixed positive integer. Using the method of generating functions, we derive explicit expressions
for the joint distribution polynomials of the two statistics with the statistic that records the
number of squares in a tiling. In this way, we obtain two families of q -generalizations of the
Fibonacci polynomials. When
1, our formulas reduce to known results concerning previous
statistics. Special attention is payed to the case
2. As a byproduct of our analysis, several
combinatorial identities are obtained.
Keywords: Tilings, Fibonacci numbers, Lucas numbers, polynomial generalization
MSC 2010 No.: 11B39, 05A15, 05A19.
1. Introduction
Let Fn be the Fibonacci number defined by the recurrence Fn = Fn 1  Fn  2 if n  2 , with initial
conditions F0 = 0 and F1 = 1 . Let Ln be the Lucas number satisfying the same recurrence, but
with L0 = 2 and L1 = 1 . See, for example, sequences A000045 and A000032 in [Sloan (2010)].
Let Gn = Gn (t ) be the Fibonacci polynomial defined by Gn = tGn 1  Gn  2 if n  2 , with G0 = 0
and G1 = 1 ; note that Gn (1) = Fn for all n . See, for example, [Benjamin and Quinn (2003) p.
508
AAM: Intern. J., Vol. 7, Issue 2 (December 2012)
509
[ m ]q !
 m
141]. Finally, let   denote the q -binomial coefficient given by
if 0  j  m ,
[ j ]q ![m  j ]q !
 j q
where [m]q != i =1[i ]q if m  1 denotes the q -factorial and [i ]q = 1  q    q i 1 if i  1 denotes
m
 m
the q -integer (with [0]q != 1 and [0]q = 0 ). We will take   to be zero if 0  m < j or if
 j q
j < 0. Polynomial generalizations of Fn have arisen in connection with statistics on binary words
Carlitz (1974), lattice paths [Cigler (2004)], Morse code sequences [Cigler(2003)], and linear
domino arrangements [Shattuck and Wagner ( 2005, 2007)]. Let us recall now two statistics
related to domino arrangements. If n  1, then let  n denote the set of coverings of the numbers
1,2,  , n , arranged in a row by indistinguishable dominos and indistinguishable squares, where
pieces do not overlap, a domino is a rectangular piece covering two numbers, and a square is a
piece covering a single number. The members of  n are also called (linear) tilings or domino
arrangements. (If n = 0 , then  0 consists of the empty tiling having length zero.)
Note that such coverings correspond uniquely to words in the alphabet {d , s} comprising i d 's
and n  2i s 's for some i , 0  i  n / 2 .
In what follows, we will frequently identify tilings c by such words c1c2  . For example, if
n = 4 , then  4 = {dd , dss, sds, ssd , ssss} . Note that |  n |= Fn 1 for all n . Given    n , let
 ( ) denote the number of dominos in  and let  ( ) denote the sum of the numbers covered
by the left halves of dominos in  . For example, if n = 16 and  = s d s d s d d s s d s  16 (see
Figure 1 below), then  ( ) = 5 and  ( ) = 2  5  8  10  14 = 39 .
1
2 3
4
5
6
7
8
9
10 11
12
13
14 15 16
Figure 1. The tiling  = s d s d s d d s s d s  16 has  ( ) = 39 .
The following results concerning the distribution of the  and  statistics on  n are wellknown; see, e.g., [Shattuck and Wagner ( 2005)] or [Shattuck and Wagner (2007)], respectively:
n
n i
 ( )

q
q i 
=


  n
i =0
 i 
(1)
and
510
T. Mansour & M. Shattuck
n
2 n i
 ( )
 .
q
q i 
=


i =0
  n
 i q
(2)
Note that both polynomials reduce to Fn 1 when q = 1 .
We remark that the polynomial in (2) first arose in a paper of Carlitz (1974), where he showed
that it gives the distribution of the statistic a1  2 a2    ( n  1) an 1 on the set of binary words
a1a2  an 1 with no consecutive ones. To see that this statistic is equivalent to the  statistic on
 n , simply append a 0 to any binary word of length n  1 having no two consecutive 1's and
identify occurrences of 1 followed by a 0 as dominos and any remaining 0 's as squares. The
polynomials (2) or close variants thereof also appear in [Carlitz (1974, 1975), Cigler (2004)].
In this paper, we study generalizations of the  and  statistics obtained by considering only
those dominos whose right half covers a multiple of k , where k is a fixed positive integer. More
precisely, let  k record the number of dominos whose right half covers a multiple of k and let
 k record the sum of the numbers of the form ik  1 covered by the left halves of dominos
within a member of  n . The  k and  k statistics reduce to  and  when k = 1 . We remark
that the  k statistic is related to a special case of the recurrence
Qm = a j Qm 1  b j Qm 2 ,
m  j (mod k ),
with Q0 = 0 and Q1 = 1 , which was considered in [Petronilho (2012)] from a primarily algebraic
standpoint through the use of orthogonal polynomials.
In the second and third sections, respectively, we consider the  k and  k statistics and obtain
explicit formulas for their distribution on  n (see Corollary 2.5 and Theorem 3.2 below), using
the method of generating functions. Our formulas reduce to (1) and (2) when k = 1 and involve
q k -binomial coefficients in the latter case. By taking  k and  k jointly with the statistic that
records the number of squares within a tiling, we obtain q -generalizations of the Fibonacci
polynomials Gn defined above. As a consequence of our analysis, several identities involving
Gn are obtained. Special attention is payed to the case k = 2 , where some further combinatorial
results may be given. Note that  2 records the number of dominos whose left half covers an odd
number and  2 records the sum of the odd numbers covered by the left halves of these dominos.
2. A Generalization of the Statistic
Suppose k is a fixed positive integer. Given    n , let s( ) denote the number of squares of
 and let  k ( ) denote the number of dominos of  that cover numbers ik  1 and ik for some
i , i.e., the number of dominos whose right half covers a multiple of k . For example, if n = 24 ,
AAM: Intern. J., Vol. 7, Issue 2 (December 2012)
511
k = 3 , and  = s d s d s s d d s d s s d s d s s   24 (see Figure 2 below), then s ( ) = 10 and
 3 ( ) = 4 .
1
2 3 4
5 6 7
8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
•
•
•
•
Figure 2. The tiling  = s d s d s s d d s d s s d s d s s   24 has  3 ( ) = 4 .
If q and t are indeterminates, then define the distribution polynomial an( k ) (q, t ) by
an(k ) (q, t ) :=
q


 k ( ) s ( )
t
,
n  1,
n
with a0( k ) ( q, t ) := 1 . For example, if n = 6 and k = 3 , then
a6(3) (q, t ) = t 2 (t 2  1)(t 2  2)  q (t 2  1)(2t 2  1)  q 2t 2 .
Note that an( k ) (1, t ) = Gn 1 for all k and n .
In this section, we derive explicit formulas for the polynomials an( k ) (q, t ) and consider
specifically the case k = 2 .
2.1. Preliminary Result
To establish our formulas for an( k ) (q, t ) , we will need the following preliminary result, which was
shown in (Shattuck). [See also (Petronilho (2012)] for an equivalent, though more complicated,
formula involving determinants and Yayenie (2011) for the case k = 2 .) Given indeterminates
x1 , x2 , , xk and y1 , y 2 , , yk , let pn be the sequence defined by
if n  2 (mod k );
 x1 pn 1  y1 pn  2 ,
 x p y p ,
if n  3 (mod k );
2 n2
 2 n 1
p0 = 0, p1 = 1, pn = 

 x p  y p , if n  0 (mod k );
k 1 n  2
 k 1 n 1
 xk pn 1  yk pn 2 ,
if n  1 (mod k ),
Let
(n  2).
pn* be the generalized Fibonacci sequence defined by
p0* = 0 ,
(3)
p1* = 1 , and
pn* = xi pn*1  yi pn* 2 if n  2 and n  i ( mod k ) . The sequence pn then has the following Binetlike formula.
512
T. Mansour & M. Shattuck
Theorem 2.1.
If m  0 and 1  r  k , then
  m 1   m 1 
m   m 
 pr ,
 pk  r   
pmk  r = 
   
   
(4)
where  and  are the roots of the quadratic equation x 2  Lx   = 0 , L = pk 1  y1 pk*1 , and
 = (1) k 1  j =1y j .
k
2.2. General Formulae
For ease of notation, we will often suppress arguments and write a n for an( k ) (q, t ) . Using
Theorem 2.1, one can give a Binet-like formula for a n .
Theorem 2.2.
If m  0 and 0  r  k  1 , then
  m 1   m 1 
 m   m 
ar ,
ak  r  (1) k 1 q
amk  r = 
   
   
(5)
where  and  are the roots of the quadratic equation
x 2  (qGk 1 (t )  Gk 1 (t )) x  (1) k q = 0.
Proof:
Considering whether the last piece within a member of  n is a square or a domino yields the
recurrence
an = ta n 1  qa n  2 ,
n  2,
(6)
if n is divisible by k , and the recurrence
an = ta n 1  an  2 ,
n  2,
(7)
if n is not, with the initial conditions a0 = 1 and a1 = t . By induction, recurrences (3), (6), and
(7) together show that
AAM: Intern. J., Vol. 7, Issue 2 (December 2012)
n  0,
an = pn 1 ,
where
denotes
pn
513
(8)
here
the
sequence
defined
by
(3)
with
x1 = x2 =  = xk = t ,
y1 = y 2 =  = y k 1 = 1 , and y k = q . Thus, we have  = (1) k 1  j =1y j = (1) k 1 q and
k
L = pk 1  y1 pk*1 = ak  pk*1 = tak 1  qak  2  pk*1
= tGk (t )  qGk 1 (t )  Gk 1 (t ) = qGk 1 (t )  Gk 1 (t ),
since ai = Gi 1 (t ) and pi* = Gi (t ) if i < k , as there is no domino whose right half covers a
multiple of k . Formula (5) follows from writing amk  r = pmk  r 1 and using (4), which completes
the proof.
In determining our next formula for a n , we will need the generating function for the sequence
pn given by (3).
Lemma 2.3.
If pn is defined as above, then
p x
n0
n
n
k 1
k 1
r =0
r =0
 pr x r  ( pk r  Lpr ) x k r
=
1  Lx k  x 2 k
,
(9)
where L and  are given in Theorem 2.1.
Proof:
From (4), we have
p x
n0
=
=
=
n
n
k 1
=   pmk  r x mk  r
r = 0 m0
k 1

1
pr 
m mk  r

  x
  m0
 
1
 
 p
1
 
 1

 1  x k

 p
1
 
 1

k
 1  x
1
pr r
x 

1  x k
r =0 
k r
r =0
1

 

k 1
r =0
k 1

k r

k 1

 p
r =0
1
pr  r
x 
1  x k
 

k 1
r =0

 k 1

(   ) x k
 p xr .

k
k  k  r
 (1  x )(1  x )  r = 0


 p
pr r 
x 
r =0 

k 1
k r

k r
pr 
  m x mk  r
 m0

pr  r 
x 
  
514
T. Mansour & M. Shattuck
Note that

 (1  x k )


 (1  x k )
=
 (1  x k )   (1   ) x k
 (1  x k )(1  x k )
    (   )(   ) x k
(1  x k )(1   x k )
=
since  =  . Thus, the first two sums in the last expression for
k 1
p x
n
=
n
n0
=
(1  (   ) x
k
) pr x r
r =0
(1  x )(1  x )
k
k 1
k 1
r =0
r =0
k
k 1

p
k r

n0
pn x n combine to give
x k r
r =0
(1  x k )(1   x k )
 pr x r  ( pk  r  Lpr ) x k  r
1  Lx k  x 2 k
since L =    , which completes the proof.

The generating function for the sequence a n may be given explicitly as follows.
Theorem 2.4.
We have
k 1
a x
n0
n
n
G
r 1
=
r =0
k 1
x r  (1) r 1 Gk  r 1 x k  r
r =0
1  (qGk 1  Gk 1 ) x k  (1) k qx 2 k
Proof:
By (8) and (9), we have
.
(10)
AAM: Intern. J., Vol. 7, Issue 2 (December 2012)
a x
n0
n
n
=
=
=  pn 1 x n =
k 1
k 1
r =1
r =0
 pr x r 1  ( pk  r  Lpr ) x k  r 1
n 0
k
k 1
r =1
r =1
515
1  Lx k  x 2 k
ar 1 x r 1  (ak  r 1  (qGk 1  Gk 1 )ar 1 ) x k  r 1
1  (qGk 1  Gk 1 ) x k  (1) k qx 2 k
k 1
k 2
r =0
r =0
ar x r  (ak  r  (qGk 1  Gk 1 )ar ) x k r
1  (qGk 1  Gk 1 ) x k  (1) k qx 2 k
,
by p0 = 0 and the expressions for L and  given in the proof of Theorem 2.2 above. If
0  r  k  2 , then ar = Gr 1 and
ak  r = qa k  2 ar  ak 1ar 1 = qGk 1Gr 1  Gk Gr  2 ,
the first relation upon considering whether or not the numbers k  1 and k are covered by a
single domino within a member of  k  r . Thus,
ak  r  (qGk 1  Gk 1 )ar
= qGk 1Gr 1  Gk Gr  2  (qGk 1  Gk 1 )Gr 1
= Gk Gr  2  Gk 1Gr 1
= (1) r 1 Gk  r 1 ,
the last equality by the identity (1) m Gn  m = Gm 1Gn  GmGn 1 , 0  m  n , which can be shown
by induction (see [Benjamin and Quinn (2003), p. 30, Identity 47] for the case when t = 1 ).
Substituting this into the last expression above for n0an x n , and noting G0 = 0 , completes the
proof.

Corollary 2.5.
If m  0 and 0  s  k  1 , then
amk  s
m
2
m 
= Gs 1 (1) ( k 1) j q j 
j =0
 j
m 1
2
j
(qGk 1  Gk 1 ) m  2 j

 m 1
 (1) s 1 Gk  s 1 (1) ( k 1) j q j 
j
j =0

Proof:
By (10), we have
j
(qGk 1  Gk 1 ) m 1 2 j .

(11)
516
T. Mansour & M. Shattuck
a x
n0
n
n
=
k 1
k 1
r =0
r =0
Gr 1 x r  (1) r 1 Gk r 1 x k r
1  x k (qGk 1  Gk 1  (1) k qx k )
k 1
 k 1

=  Gr 1 x r  (1) r 1 Gk  r 1 x k  r x jk (qGk 1  Gk 1  (1) k qx k ) j
r =0
 r =0
 j 0
j
k 1
 j
 k 1

=  Gr 1 x r  (1) r 1 Gk  r 1 x k  r  (qGk 1  Gk 1 ) j i (1)i ( k 1) q i x ik  jk .
r =0
 r =0
 j 0 i = 0  i 
Since each power of x in the infinite double sum on the right side of the last expression is a
multiple of k for all i and j , only one term from each of the two finite sums on the left
contributes towards the coefficient of x mk  s , namely, the r = s term. Thus, the coefficient to
x mk  s in the last expression is given by
m
 j
Gs 1 ((1) k 1 q) m  j 
j =0
m 

(qGk 1  Gk 1 ) 2 j  m
j 
m 1
j


(qGk 1  Gk 1 ) 2 j 1m .
 (1) s 1 Gk  s 1 ((1) k 1 q) m  j 1 
j =0
 m  j  1
Replacing j by m  j in the first sum and j by m 1  j in the second gives (11).

Taking k = 1 in (11) implies
n
n 
an(1) (q, t ) = q j t n  2 j 
j =0
 j
j
,

n  0,
(12)
which is well-known (see, e.g., [Benjamin and Quinn (2003), Shattuck and C. Wagner (2005)].
Taking k = 2 in (11) implies
if n = 2m;
Q(m)  Q(m  1),
an(2) (q, t ) = 
if n = 2m  1,
tQ(m),

where
m
2
m 
Q(m) = (1) j q j 
j =0
 j
Taking k = 3 in (11) implies
j 2
(t  q  1) m  2 j .

(13)
AAM: Intern. J., Vol. 7, Issue 2 (December 2012)
517
if n = 3m;
 R(m)  tR(m  1),

a (q, t ) = tR(m)  R(m  1), if n = 3m  1;
 (t 2  1) R(m),
if n = 3m  2,

(3)
n
(14)
where
m
2
m 
R(m) = q j 
j =0
 j
j 3
(t  (2  q)t ) m  2 j .

Let H n = H n (t ) denote the Lucas polynomial defined by the recurrence H n = tH n 1  H n  2 if
n  2 , with H 0 = 2 and H 1 = t , or, equivalently, by H n = Gn 1  Gn 1 if n  1.
Corollary 2.6.
If m  0 and 0  s  k  1 , then
Gmk  s 1
m
2
m 
= Gs 1 (1) ( k 1) j 
j =0
 j
m 1
2
j  m2 j
 H k

 m 1
 (1) s 1 Gk  s 1  (1) ( k 1) j 
j
j =0

(15)
j  m 1 2 j
 H k
.

In particular, we have
m 1
2
 m 1 
Gmk = Gk (1) ( k 1) j 
j
j =0

j  m 1 2 j
 H k
,

m  0.
(16)
Proof:
Taking q = 1 in (11) and noting an( k ) (1, t ) = Gn 1 (t ) for all k gives (15). Furthermore, if s = k  1
in (15), then the second sum drops out since G0 = 0 , which yields (16).
We were unable to find formulas (15) or (16) in the literature, though the t = 1 case of (16) is
similar in form to Identities V82 and V83 in (Benjamin and Quinn (2003), p. 145).
2.3. The Case
.
We consider further the case when k = 2 . Note that an(2) (q, t ) is the joint distribution polynomial
on  n for the statistics recording the number of squares and the number of dominos whose right
518
T. Mansour & M. Shattuck
half covers an even number. The next result follows from taking k = 2 in formula (10), though
we provide another derivation here. Let an(2) = an(2) (q, t ) and a( x; q, t ) = n0an(2) (q, t ) x n , which
we'll often denote by a (x) .
Proposition 2.7.
We have
a ( x; q, t ) =
1  tx  x 2
.
1  (1  q  t 2 ) x 2  qx 4
(17)
Proof:
Considering whether or not a tiling ends in a square yields the recurrences
a2(2)n = ta2(2)n 1  qa2(2)n  2 ,
n  1,
and
a2(2)n 1 = ta2(2)n  a2(2)n 1 ,
n  1,
with a0(2) = 1 and a1(2) = t . Multiplying the first recurrence by x 2 n and the second by x 2 n 1 ,
summing both over n  1, and adding the two equations that result implies
 a ( x)  a ( x) 
2  a ( x)  a ( x) 
a ( x)  tx  1 = tx (a ( x)  1)  x 2 
,
  qx 
2
2




which may be rewritten as
(2  2tx  (1  q ) x 2 ) a ( x ) = 2  x 2 ( q  1) a (  x ).
(18)
Replacing x with  x in (18) gives
(2  2tx  (1  q ) x 2 ) a ( x ) = 2  x 2 ( q  1) a ( x ),
(19)
and solving the system of equations (18) and (19) in a (x) and a ( x ) yields
a( x) =
as desired.
1  tx  x 2
,
1  (1  q  t 2 ) x 2  qx 4

AAM: Intern. J., Vol. 7, Issue 2 (December 2012)
519
We next consider some particular values of the polynomials an(2) (q, t ) .
Proposition 2.8
If n  1, then
t 2 (1  t 2 ) m1 ,
if n = 2m;
an(2) (0, t ) = 
2 m
 t (1  t ) , if n = 2m  1.
(20)
Proof:
We provide both algebraic and combinatorial proofs. Taking q = 0 in (17) implies
a ( x;0, t ) =
1  tx  x 2
= (1  tx  x 2 )  (1  t 2 ) m x 2 m
1  (1  t 2 ) x 2
m0
= 1  ((1  t 2 ) m  (1  t 2 ) m 1 ) x 2 m  t (1  t 2 ) m x 2 m 1
m 1
= 1  t (1  t )
2 m 1
2
m0
x
2m
m 1
 t (1  t ) x 2 m 1 ,
2 m
m0
from which (20) follows.
For a combinatorial proof, first let n = 2m , where m  1 . Then members  of  n having zero
 2 value are of the form
a
a
a
 = ( sd 1 s)(sd 2 s) (sd  s)
for some  , where ai  0 for each i  [] = {1,2,  , } .
Note that the sequence (a1  1, a2  1, , a  1) is a composition of m .
Thus, the polynomial an(2) (0, t ) may be viewed as the weighted sum of compositions of m , where
 m  1
 compositions
the weight of a composition having exactly  parts is t 2  . Since there are 
  1 
of m having  parts, we have
m m 1
 2  m 1 m  1 2   2 2

t
t = 
= t (1  t 2 ) m 1 ,
an(2) (0, t ) = 
 =1    1 
 =0   
which gives the even case.
520
T. Mansour & M. Shattuck
If n = 2m  1 , then the weighted sum of tilings s , where    n has zero  2 value, is given by
t 2 (1  t 2 ) m , by the even case. Dividing this by t (to account for the square that was added at the
end) gives the odd case and completes the proof.
Proposition 2.9.
If n  1, then
G (t 2 )  Gm (t 2 ),
if n = 2m;
an(2) (1, t ) =  m1
2
tGm 1 (t ),
if n = 2m  1.

(21)
Proof:
We provide both algebraic and combinatorial proofs of this result. Taking q = 1 in (17) and
replacing x with x 2 and t with t 2 in
G
m0
m 1
(t ) x m =
1
1  tx  x 2
implies
a ( x;1, t ) =
1  tx  x 2
= (1  tx  x 2 ) Gm 1 (t 2 ) x 2 m
2 2
4
1 t x  x
m 0
= tGm 1 (t 2 ) x 2 m 1  (Gm 1 (t 2 )  Gm (t 2 )) x 2 m ,
m0
m0
which gives the result.
We provide a bijective proof of (21) in the case when t = 1 , the general case being similar, and
show
if n = 2m;
F ,
an(2) (1,1) =  m 1
(22)
 Fm 1 , if n = 2m  1.
 ( )
To do so, define the sign of    n by sgn( ) = (1) 2 , and let  ne and  no denote the subsets
of  n whose members have positive and negative sign, respectively. Then
an(2) (1,1) =|  ne |  |  no | and it suffices to identify a subset  n* of  ne having cardinality Fm 1 or
Fm 1 , along with a sign-changing involution of  n   n* .
Let n = 2m and  n   n consist of those coverings  = 12  such that 2i 1 = 2i for all i . If
AAM: Intern. J., Vol. 7, Issue 2 (December 2012)
521
   n   n , then let io denote the smallest index i such that 2i 1  2i , i.e., 2i 12i = ds or sd
. Let f ( ) denote the covering that is obtained from  by exchanging the positions of the
(2io  1) -st and (2io ) -th pieces of  , leaving all other pieces undisturbed. Then the mapping f
is seen to be a sign-changing involution of  n   n .
We now define an involution of  n . Let  n*   n consist of those members containing an even
number of pieces and ending in a domino. Note that  n*   ne and that |  n* |= Fm 1 since
members of  n* are synonymous with members of  m ending in a domino, upon halving.
Observe further that if    n   n* has an odd number of pieces, then  ends in a domino since
n is even, while if  has an even number of pieces, it must end in two squares. If    n   n* ,
then let g ( ) be obtained from  by either changing the final domino to two squares or
changing the final two squares to a domino. Then g is seen to be a sign-changing involution of
 n   n* . Combining the two mappings f and g yields a sign-changing involution of  n   n* ,
as desired.
If n = 2m  1 , then apply the mapping f defined above to  n . Note that the set of survivors has
cardinality Fm 1 , upon halving, since they are of the form  = 12  2  2 1 for some  , with
2i 1 = 2i for each i  [] and 2 1 = s . This completes the proof of (22).
Let t n ( 2 ) denote the sum of the  2 values taken over all of the members of  n .
Proposition 2.10 .
If n  1, then
 nLn  4 Fn
,
if n is even;

10

t n ( 2 ) = 
 (n  1) Ln  2 Fn 1
, if n is odd.

10

(23)
Proof:
To find t n ( 2 ) , we consider the contribution of the dominos that cover the numbers 2i  1 and 2i
for some i fixed within all of the members of  n . Let n = 2m  1 .
Note that there are F2 i 1 F2 m  2  2 i dominos that cover the numbers 2i  1 and 2i within all of the
members of  n .
Summing over all i , we have
522
T. Mansour & M. Shattuck
m
m 1
i =1
i =0
t n ( 2 ) = F2i 1 F2 m  2i  2 = F2i 1 F2 m  2i .
To simplify this sum, we recall the Binet formulas Fn =
n n
and Ln =  n   n , n  0, where
5
 and  denote the positive and negative roots, respectively, of the equation x 2  x  1 = 0 .
Then for m even, we have
m 1
m 1
5F2i 1 F2 m  2i
= ( 2i 1   2i 1 )( 2 m  2i   2 m  2i )
i =0
i=0
m 1
m
1
2
i =0
i =0
= ( 2 m 1   2 m 1 )   ( 2 m  4i 1   2 m  4i 1 )
m 1
 ( 4i  2 m 1   4i  2 m 1 )
i=
m 1
m
2
m
1
2
m
1
2
i =0
i=0
= L2 m 1  L2 m  4i 1  L2 m  4i 3
i =0
m
1
2
m
1
2
i =0
i =0
= mL2 m 1  L2 m  4i  2 = mL2 m 1   ( F2 m  4i 1  F2 m  4i 3 )
= mL2 m 1  Fm Fm 1  Fm 1 Fm = mL2 m 1  Fm Lm
= mL2 m 1  F2 m ,
by Identities 28, 26 and 33 in Benjamin and Quinn (2003) and since Lm = Fm 1  Fm 1 . Substituting
n 1
gives the second formula when n  1 ( mod 4) . A similar calculation gives the same
m=
2
formula when n  3 (mod 4) .
If n = 2m and m is odd, then similar reasoning shows that
AAM: Intern. J., Vol. 7, Issue 2 (December 2012)
523
m 1
5t n ( 2 ) = 5F2i 1 F2 m  2i 1
i =0
m 1
m 1
2
i =0
i =0
m 1
= ( 2 m   2 m )   ( 2 m  4i  2   2 m  4i  2 ) 
 (
i=
4i  2 m  2
  4i  2 m  2 )
m 1
2
m 3
2
= mL2 m  2  2  L2 m  4i  2
i =0
m 1
2
m 3
2
i =0
i =0
= mL2 m  2 F2 m  4i 1  2  F2 m  4i 3
= mL2 m  2 Fm Fm 1  2 Fm 1 Fm = mL2 m  2 F2 m ,
and the first formula in (23) follows when n  2 ( mod 4) , upon replacing m with
calculation gives the same formula when n  0 (mod 4) .
n
. A similar
2
We close this section with a general formula for an(2) (q, t ) .
Theorem 2.11.
If n  0 , then
 m m 1 i  m  i  j  m  j  1 j 2 m  2i
q t

,
if n = 2m;
q  
j
i
j

i
=
0
j
=
0





an(2) (q, t ) = 
m i

 m  i  j  m  j  j 2 m  2i 1
q t

,
if n = 2m  1.
 
j
 i = 0 j = 0 
 i  j 
(24)
Proof:
We will refer to a domino whose left half covers an odd (resp., even) number as odd-positioned
(resp., even-positioned). First suppose n = 2m is even. If    n contains no squares, then it
consists of m odd-positioned dominos, whence the q m term. So suppose that  contains i
dominos, where 0  i  m  1 , and that j of the dominos are odd-positioned. There are 2m  2i
squares and m  i  1 possible positions to insert each of the j odd-positioned dominos relative
m i  j
 choices concerning their placement. There are m  i
to the squares, whence there are 
j


possible positions to insert each of the i  j even-positioned dominos, whence there are
524
T. Mansour & M. Shattuck
 m  j  1
 m  i  j  m  j  1
 choices concerning their placement. Thus, there are 

 members

j
 i j 
 i  j 

of  n containing i dominos, j of which are odd-positioned. Summing over all i and j gives
the even case of (24). A similar argument applies to the odd case.

Remark: Setting q = 0 in (24) gives (20). Comparing the odd cases of (24) and (13) and
replacing t with
t gives the following polynomial identity in q and t :
m i 
q j t mi 


j
i =0 j =0

m
i
m
j  m  j  2
m 
 = (1) j q j 

 j
 i  j  j =0
j
(q  t  1) m2 j ,

m  0.
(25)
A similar identity can be obtained by comparing the even cases of (24) and (13). Setting q = 1
in (24), comparing with (21), and replacing t with
t gives a pair of formulas for Gm (t ) .
3. A Generalization of The Statistic
Suppose k is a fixed positive integer. Given    n , let s ( ) denote the number of squares of
 and let  k ( ) denote the sum of the numbers of the form ik  1 that are covered by the left
half of a domino. For example, if n = 24 , k = 4 , and  = s s d s d d s d s s d d s s d d   24 (see
Figure 3 below), then s ( ) = 8 and  4 ( ) = 3  11  15  23 = 52 . If q and t are indeterminates,
then define the distribution polynomial bn( k ) (q, t ) by
bn( k ) (q, t ) :=
q

 k ( ) s ( )
t
,
n  1,
 n
with b0( k ) (q, t ) := 1 . For example, if n = 6 and k = 3 , then
b6(3) (q, t ) = t 2 (t 2  1)(t 2  2)  q 2t 2 (t 2  1)  q 5 (t 2  1) 2  q 7t 2 .
Note that bn( k ) (1, t ) = Gn 1 for all k and n .
1 2 3 4 5 6 7
8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
Figure 3. The tiling  = s s d s d d s d s s d d s s d d   24 has  4 ( ) = 52 .
In what follows, we will often suppress arguments and write bn for bn( k ) (q, t ) . Considering
whether the last piece within a member of  n is a square or a domino yields the recurrence
AAM: Intern. J., Vol. 7, Issue 2 (December 2012)
bn = tbn 1  q n 1bn  2 ,
525
n  2,
if n is divisible by k , and the recurrence
bn = tbn 1  bn  2 ,
n  2,
if n is not, with initial conditions b0 = 1 and b1 = t . In [4], Carlitz studied the polynomials
(1)
bn
1 ( q, t ) from an algebraic point of view. See also the related paper by Cigler (2003).
In this section, we will derive explicit formulas for the polynomials bn( k ) (q, t ) and their
generating function, with specific consideration of the case k = 2 . Note that  2 ( ) records the
sum of the odd numbers covered by left halves of dominos in  .
3.1. General Formulas
We first establish an explicit formula for the generating function of the sequence bn .
Theorem 3.1.
We have
bn x n
n0
x
k 3
k 3
k 3
r =0
r =0
= Gr 1 x r  q k 1ck 1 (q k x k )Gr 1 x k  r  ck ( x k )Gr  2 x k  r
r =0
k 2
k 1
(26)
ck 1 ( x )  x ck ( x ),
k
k
where
j
ck 1 ( x) = 
x q
 j 1 
  j
k 
 2 
j
(G
k 1
 (1) k 1 xq ik )
i =0
j
(1  xq
j 0
ik
Gk 1 )
i=0
and
j
ck ( x) = Gk 
j 0
x q
 j 1 
  j
k 
 2 
j
(G
k 1
 (1) k 1 xq ik )
i =1
j
(1  xq
.
ik
Gk 1 )
i =0
Proof:
It is more convenient to first consider the generating function for the numbers bn := bn( k1) (q, t ) .
526
T. Mansour & M. Shattuck
Then the sequence bn has initial values b0 = 0 and b1 = 1 and satisfies the recurrences
  r = tbmk
  r 1  bmk
 r 2 ,
bmk
2rk
 1 = tbmk
  q mk 1bmk
 1 ,
bmk
m  1.
and
m  0,
(27)
with
(28)
Let
 r x m ,
cr ( x) = bmk
m 0
where r  [k ] . Then multiplying the recurrences (27) and (28) by x m , and summing the first over
m  0 and the second over m  1 , gives
cr ( x) = tcr 1 ( x)  cr  2 ( x), r = 3,4, , k ,
c2 ( x) = tc1 ( x)  xck ( x),
c1 ( x) = 1  xtck ( x)  xq k 1ck 1 (q k x).
By induction on r , we obtain
cr ( x) = Gr  Gr 1 xck ( x)  Gr xq k 1ck 1 (q k x),
1  r  k.
(29)
Taking r = k and r = k  1 in (29) gives
ck ( x ) =
Gk
xq k 1Gk
ck 1 (q k x)

1  xGk 1 1  xGk 1
and
ck 1 ( x) = Gk 1  Gk xck ( x)  Gk 1 xq k 1ck 1 (q k x)
 Gk

xq k 1Gk
ck 1 (q k x)   Gk 1 xq k 1ck 1 (q k x)
= Gk 1  Gk x

 1  xGk 1 1  xGk 1

2
2 k 1
2
G  x(Gk  Gk 1Gk 1 ) x q (Gk  Gk 1Gk 1 )  xq k 1Gk 1
ck 1 (q k x)
= k 1

1  xGk 1
1  xGk 1
=
Gk 1  x(1) k 1 xq k 1 (Gk 1  x(1) k 1 )
ck 1 (q k x),

1  xGk 1
1  xGk 1
where we have used the identity Gk2  Gk 1Gk 1 = (1) k 1 [see, e.g., (Benjamin and Quinn (2003),
AAM: Intern. J., Vol. 7, Issue 2 (December 2012)
527
Identity 246].
Iterating the last recurrence gives
Gk 1  (1) k 1 xq kj
ck 1 ( x) = 
1  xq kj Gk 1
j 0
j
=
x q
 j 1 
  j
k 
 2 
j
(G
k 1
xq ik 1 (Gk 1  (1) k 1 xq ( i 1) k )

1  xq ( i 1) k Gk 1
i =1
j
 (1) k 1 xq ik )
i =0
j
(1  xq
j 0
,
ik
Gk 1 )
i =0
which implies
j
x q
Gk
ck ( x ) =
 Gk 
1  xGk 1
j 1
 j
k    kj  j j 1
 2
(G
k 1
i=0
j
 (1) k 1 xq ( i 1) k )
(1  xq
ik
Gk 1 )
i =0
j
= Gk 
x q
 j 1 
k 
  j
 2 
j
(G
k 1
 (1) k 1 xq ik )
i =1
j
(1  xq
j 0
.
ik
Gk 1 )
i =0
Then, by (29), we have
bn x n
n0
k
k
  r x mk  r = x r cr ( x k )
=  bmk
r =1 m  0
r =1
k 2
= x r (Gr  Gr 1 x k ck ( x k )  Gr x k q k 1ck 1 (q k x k ))  x k 1ck 1 ( x k )  x k ck ( x k )
r =1
k 2
k 2
k 2
= Gr x r  ck ( x k )Gr 1 x k  r  q k 1ck 1 (q k x k )Gr x k  r
r =1
r =1
k 1
r =1
 x ck 1 ( x )  x ck ( x ),
k
k
k
where ck 1 ( x ) and ck (x ) are as given. Formula (26) now follows upon noting
b
n0
(k )
n
(q, t ) x n = bn 1 x n =
n0
1
bn x n .

x n0
One can find explicit expressions for the bn using Theorem 3.1 and the following formulas that
528
T. Mansour & M. Shattuck
involve the q -binomial coefficient [see, e.g., (Andrews (1976) or Stanley (1997)]:
a
=   x a
j
(1  xq i ) a j   q
xj
(30)
j
i=0
and
j
j
( y  xq ) = q
i
 a 1 
 2 


a =0
i =1
 j  a j a
  x y ,
 a q
(31)
where j is a non-negative integer.
Theorem 3.2.
The following formulas hold for bn . If m  0 , then
m
bmk  k 1 = Gk q
 j 1 
  j m
k 
 2 
(1)
j =0
( k 1) a
q
 a 1 

k 
 2 
 j m  a
 .
Gkj1a Gkm1a  j   
 a qk  j qk
(32)
 j  1  m  a 
 
 .
Gkj1a 1Gkm1a  j 
 a qk  j qk
(33)
a =0
If m  0 , then
m
bmk  k  2 = q
 j 1 
  j m
k 
 2 
(1)
j =0
( k 1) a
q
a
k  
2
a=0
If m  1 and 0  r  k  3 , then
 j 1 
m 1 k 
  j m 1
 2 
bmk  r = Gk Gr  2 q
(1)
j =0
j =0
q
a=0
 j 1 
m 1 k 
  j m 1
 2 
 q km1Gr 1 q
( k 1) a
(1)
a=0
( k 1) a
q
a
k  
2
 a 1 

k 
 2 
 j   m  a  1

Gkj1a Gkm1a  j 1   
j
 a qk 
qk

Gkj1a 1Gkm1a  j 1 

j  1  m  a  1
 
 ,
a  q k 
j
qk
(34)
with br = Gr 1 .
Proof:
Let n = mk  k  1 , where m  0 . Then the coefficient of x n on the right-hand side of (26) is
given by
AAM: Intern. J., Vol. 7, Issue 2 (December 2012)
529
 j 1 


 j j
k
k 1
ik 
 x j q  2 
G
xq
(
(
1)
)



k 1


i =1
[ x m ](ck ( x)) = Gk  [ x m ]
.
j
ik
j 0


(1  xq Gk 1 )



i=0


By (30) and (31), we have for each j  0 ,
a
=   x a Gka1
j k
(1  xq kiGk 1 ) a j   q
( xGk 1 ) j
j
i =0
and
j
(G
k 1
 (1)
k 1
j
xq ) = (1)
ik
( k 1) a
q
 a 1 

k 
 2 
a =0
i =1
 j  a j a
  x Gk 1 ,
 a qk
so that coefficient of x n is given by
m
Gk 
j =0
q
 j 1 
  j
k 
 2 
j
k 1
G
m
(1)
( k 1) a
q
 a 1 

k 
 2 
a =0
 j
Gkj1a  
 a  qk
m  a
 Gkm1a ,
 
j

 qk
which yields (32). Similar proofs apply to formulas (33) and (34), in the latter case, upon
extracting the coefficient of x n from two separate terms in (26).
When k = 1 in Theorem 3.2, the inner sum in (32) reduces to a single term since G0 = 0 and
gives
2 n 
b (q, t ) = q j 
j =0
 j
(1)
n
n
j  n2 j
 t ,
q
n  0,
(35)
which is well-known [see, e.g., Shattuck and Wagner (2005)] When k = 2 in Theorem 3.2, we get for
all m  0 the formulas
m
b2(2)m (q, t ) = q j
j =0
and
2
m
(1)
a =0
a
qa
2 a
 j  1  m  a 
 

(t 2  1) m a  j 
 a q2  j q2
(36)
530
T. Mansour & M. Shattuck
m
b2(2)m1 (q, t ) = t q j
j =0
2
m
(1)
a
qa
2 a
a =0
 j m  a
 .
(t 2  1) m  a  j   
 a q2  j q2
(37)
The polynomials bn(2) (q, t ) are considered in more detail below.
3.2. The Case
Let us write bn(2) for bn(2) (q, t ) . The even and odd terms of the sequence bn(2) satisfy the following
two-term recurrences.
Proposition 3.3.
If m  2 , then
b2(2)m = (q 2 m 1  t 2  1)b2(2)m  2  q 2 m 3b2( 2m) 4 ,
(38)
with b0(2) = 1 and b2(2) = t 2  q , and
b2(2)m 1 = (q 2 m 1  t 2  1)b2(2)m 1  q 2 m 1b2(2)m 3 ,
(39)
with b1(2) = t and b3(2) = t 3  (1  q )t .
Proof:
To show (39), first note that the total weight of all the members of  2 m 1 ending in d or ss is
b2(2)m1 and t 2b2(2)m1 , respectively. The weight of all members of  2 m 1 ending in ds is
q 2 m 1 (b2(2)m 1  b2(2)m 3 ) . To see this, we insert a d just before the final s in any    2 m 1 ending in
s . By subtraction, the total weight of all tilings  that end in s is b2(2)m 1  b2(2)m 3 , and the inserted
d contributes 2m  1 towards the  2 value since it covers the numbers 2m  1 and 2m . For (38),
note that by similar reasoning, the total weight of all members of  2 m ending in d , ss , and ds
is q 2 m 1b2(2)m  2 , t 2b2(2)m  2 , and b2(2)m 2  q 2 m 3b2(2)m 4 , respectively.

We were unable to find, in general, two-term recurrences comparable to (38) and (39) for the
(2)
n
(k )
sequences bmk
 r ( q, t ) , where k and r are fixed and m  0 . Let b ( x; q , t ) =  n  0bn ( q , t ) x .
Using (38) and (39), it is possible to determine explicit formulas for the generating functions
m0b2(2)m x m and m0b2(2)m1 x m and thus for b( x; q, t ) , upon proceeding in a manner analogous to
the proof of Theorem 3.1 above. The following formula results, which may also be obtained by
taking k = 2 in Theorem 3.1.
AAM: Intern. J., Vol. 7, Issue 2 (December 2012)
531
Proposition 3.4.
We have
j
x 2 j q j (1  tx  x 2 )(1  q 2i x 2 )
2
b( x; q, t ) = 
j 0
i =1
j
(1  (t
2
.
(40)
 1)q x )
2i
2
i =0
Taking q = 0 and q = 1 in (40) shows that bn(2) (0, t ) and bn(2) (1, t ) are the same as an(2) (0, t )
and an(2) (1, t ) and are thus given by Propositions 2.8 and 2.9, respectively. This is easily seen
directly since a member of  n has zero  2 value if and only if it has zero  2 value and since the
parity of the  2 and  2 values is the same for all members of  n . Comparing with (36) and (37)
when q = 1 , and replacing t with
polynomials.
t , then gives a pair of formulas for the Fibonacci
Corollary 3.5.
If m  0 , then
m m
 j  1 m  a 


Gm 1 (t )  Gm (t ) = (1) a  j (t  1) m  a  j 
j =0 a=0
 a  j 
(41)
m m
 j  m  a 
.
Gm 1 (t ) = (1) a  j (t  1) m  a  j  
j =0 a =0
 a  j 
(42)
and
Taking q = 1 in (36) and (37), and noting bn(2) (1, t ) = Gn 1 (t ) , gives another pair of formulas.
Corollary 3.6.
If m  0 , then
m m
 j  1 m  a 


G2 m 1 (t ) = (1) a (t 2  1) m  a  j 
j =0 a =0
 a  j 
(43)
m m
 j  m  a 
.
G2 m  2 (t ) = t (1) a (t 2  1) m  a  j  
j =0 a=0
 a  j 
(44)
and
532
T. Mansour & M. Shattuck
Let t n ( 2 ) denote the sum of the  2 values taken over all of the members of  n . We conclude
with the following explicit formula for t n ( 2 ) .
Proposition 3.7.
If n  0 , then
t n ( 2 ) = ( 1) n
(2n  1) Fn  2  (2n  3) Fn 1 (6n 2  2n  15) Fn 1  (2n 2  2n  5) Fn  2

.
8
40
(45)
Proof:
To find t n ( 2 ) , first note that

t ( 2 ) x n =
n0 n
d
b( x; q,1) |q =1 . By Proposition 3.4 and partial
dq
fractions, we have
d
n 2 x 2 n (1  x 2 ) n x 2 (1  x  x 2 )
n(n  1) x 2 n (1  x 2 ) n

b( x; q,1) |q =1 = (1  x  x 2 )

2 n 1
1 x2
(1  2 x 2 ) n 1
dq
n  0 (1  2 x )
n 0

2n(n  1) x 2 n (1  x 2 ) n
x 2 (1  x  x 2 )

1 2x2
(1  2 x 2 ) n 1
n0
=

1
7 x
3  4x
9  7x
3  3x
4  7x

.




2
2 2
2
2 2
2 3 
2(1  x  x ) 
x  8(1  x  x ) 4(1  x  x ) 8(1  x  x ) (1  x  x )
Note that
1
= Fn 1 ,
1 x  x2
1
(n  1) Fn  2  2(n  2) Fn 1
[xn ]
=
,
2 2
(1  x  x )
5
1
(5n  16)(n  1) Fn  2  (5n  17)(n  2) Fn 1
[xn ]
=
;
2 3
(1  x  x )
50
[xn ]
see sequences A000045, A001629, and A001628, respectively, in Sloane (2010). Thus, the
d
coefficient of x n in
b( x; q,1) |q =1 is given by
dq
( 1) n
(2n  1) Fn  2  (2n  3) Fn 1 (6n 2  2n  15) Fn 1  (2n 2  2n  5) Fn  2

,
8
40
which completes the proof.

AAM: Intern. J., Vol. 7, Issue 2 (December 2012)
533
4. Conclusion
In this paper, we have studied two statistics on square-and-domino tilings that generalize
previous ones by considering only those dominos whose right half covers a multiple of k , where
k is a fixed positive integer. We have derived explicit formulas for all k for the joint
distribution polynomials of the two statistics with the statistic that records the number of squares
in a tiling. This yields two infinite families of q -generalizations of the Fibonacci polynomials.
When k = 1 , our formulas reduce to prior results. Upon noting some special cases, several
combinatorial identities were obtained as a consequence. Finally, it seems that other statistics on
square-and-domino tilings could possibly be generalized. Perhaps one could also modify
statistics on permutations and set partitions by introducing additional requirements concerning
the positions, mod k , of various elements.
References
Andrews, G. (1976). The Theory of Partitions, in Encyclopedia of Mathematics and
Applications, (No. 2), Addison-Wesley.
Benjamin, A. T. and Quinn, J. J. (2003). Proofs that Really Count: The Art of Combinatorial
Proof, Mathematical Association of America.
Carlitz, L. (1974). Fibonacci notes 3: q -Fibonacci numbers, Fibonacci Quart, 12, 317—322.
Carlitz, L. (1975). Fibonacci notes 4: q -Fibonacci polynomials, Fibonacci Quart, 13, 97—102.
Cigler, J. (2003). q -Fibonacci polynomials, Fibonacci Quart, 41, 31—40.
Cigler, J. (2003). Some algebraic aspects of Morse code sequences, Discrete Math. Theor.
Comput. Sci. 6, 55—68.
Cigler, J. (2004). q -Fibonacci polynomials and the Rogers-Ramanujan identities, Ann. Comb.
8, 269—285.
Petronilho, J. (2012). Generalized Fibonacci sequences via orthogonal polynomials, Appl. Math.
Comput. 218, 9819—9824.
Shattuck, M. Some remarks on an extended Binet formula, pre-print.
Shattuck, M. and Wagner, C. (2005). Parity theorems for statistics on domino arrangements,
Electron. J. Combin. 12, #N10.
Shattuck, M. and Wagner, C. (2007). Some generalized Fibonacci polynomials, J. Integer Seq.
10, Article 07.5.3.
Sloane, N. J. (2010). The On-Line Encyclopedia of Integer Sequences, published electronically
at http://oeis.org.
Stanley, R. P. (1997). Enumerative Combinatorics, Vol. I, Cambridge University Press.
Yayenie, O. (2011). A note on generalized Fibonacci sequences, Appl. Math. Comput. 217,
5603—5611.