Generalizations of Two Statistics on Linear Tilings

508

Available at http://pvamu.edu/aam

Appl. Appl. Math.

ISSN: 1932-9466

Vol. 7, Issue 2 (December 2012), pp. 508 - 533

Applications and Applied Mathematics:

An International Journal (AAM)

Generalizations of Two Statistics on Linear Tilings

Toufik Mansour & Mark Shattuck

Department of Mathematics University of Haifa 31905 Haifa, Israel

[email protected]; [email protected]

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 nF be the Fibonacci number defined by the recurrence 21= nnn FFF if 2n , with initial

conditions 0=0F and 1=1F . Let nL be the Lucas number satisfying the same recurrence, but

with 2=0L and 1=1L . See, for example, sequences A000045 and A000032 in [Sloan (2010)].

Let )(= tGG nn be the Fibonacci polynomial defined by 21= nnn GtGG if 2n , with 0=0G

and 1=1G ; note that nn FG =(1) for all n . See, for example, [Benjamin and Quinn (2003) p.

AAM: Intern. J., Vol. 7, Issue 2 (December 2012) 509

141]. Finally, let q

j

m

denote the q -binomial coefficient given by

!][!][

!][

qq

q

jmj

m

if mj 0 ,

where q

m

iq im ][=!][1= if 1m denotes the q -factorial and 11=][ i

q qqi if 1i denotes

the q -integer (with 1=![0]q and 0=[0]q ). We will take q

j

m

to be zero if jm <0 or if

< 0.j Polynomial generalizations of nF 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 1n , then let n denote the set of coverings of the numbers

n,1,2, , 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 0=n , then 0 consists of the empty tiling having length zero.)

Note that such coverings correspond uniquely to words in the alphabet },{ sd comprising i d 's

and in 2 s 's for some i , 2/0 ni . In what follows, we will frequently identify tilings c by such words 21cc . For example, if

4=n , then },,,,{=4 ssssssdsdsdssdd . Note that 1|=| nn F 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 16=n and 16= sdssddsdsds (see

Figure 1 below), then 5=)( and 39=1410852=)( . 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 Figure 1. The tiling 16= sdssddsdsds has 39=)( . The following results concerning the distribution of the and statistics on n are well-

known; see, e.g., [Shattuck and Wagner ( 2005)] or [Shattuck and Wagner (2007)], respectively:

i

inqq i

n

in 0=

)( =

(1) and

510 T. Mansour & M. Shattuck

.=2

0=

)(

q

in

ini

inqq

(2)

Note that both polynomials reduce to 1nF when 1=q .

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 121 1)(2 nanaa on the set of binary words

121 naaa 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 1n 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 1ik covered by the left halves of dominos

within a member of n . The k and k statistics reduce to and when 1=k . We remark

that the k statistic is related to a special case of the recurrence

), mod( ,= 21 kjmQbQaQ mjmjm

with 0=0Q and 1=1Q , 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 1=k and involve kq -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 nG defined above. As a consequence of our analysis, several identities involving

nG are obtained. Special attention is payed to the case 2=k , 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 1ik and ik for some

i , i.e., the number of dominos whose right half covers a multiple of .k For example, if ,24=n


,3=k and 24= ssdsdssdsddssdsds (see Figure 2 below), then 10=)(s and

4=)(3 .

• • • • Figure 2. The tiling 24= ssdsdssdsddssdsds has 4=)(3 .

If q and t are indeterminates, then define the distribution polynomial ),()( tqa k

n by

1,,:=),( )()()(

ntqtqa sk

n

kn

with 1:=),()(

0 tqa k . For example, if 6=n and 3=k , then

.1)1)(2(2)1)((=),( 2222222(3)

6 tqttqttttqa

Note that 1

)( =)(1, nk

n Gta for all k and n .

In this section, we derive explicit formulas for the polynomials ),()( tqa k

n and consider

specifically the case 2=k . 2.1. Preliminary Result To establish our formulas for ),()( tqa k

n , 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 2=k .) Given indeterminates

kxxx ,,, 21 and kyyy ,,, 21 , let np be the sequence defined by

2).(

), mod( 1 if,

); mod( 0 if,

); mod( 3 if,

); mod( 2 if,

= 1,= 0,=

21

2111

2212

2111

10

n

knpypx

knpypx

knpypx

knpypx

ppp

nknk

nknk

nn

nn

n (3)

Let *

np be the generalized Fibonacci sequence defined by 0=*0p , 1=*

1p , and *

2*

1* = ninin pypxp if 2n and ) mod( kin . The sequence np then has the following Binet-

like formula.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24


Theorem 2.1. If 0m and kr 1 , then

,=11

r

mm

rk

mm

rmk ppp

(4)

where and are the roots of the quadratic equation 0=2 Lxx , *

111= kk pypL , and

j

k

j

k y1=

11)(= .

2.2. General Formulae For ease of notation, we will often suppress arguments and write na for ),()( tqa k

n . Using

Theorem 2.1, one can give a Binet-like formula for na .

Theorem 2.2. If 0m and 10 kr , then

,1)(=11

1r

mmk

rk

mm

rmk aqaa

(5)

where and are the roots of the quadratic equation

0.=1)())()(( 112 qxtGtqGx k

kk

Proof: Considering whether the last piece within a member of n is a square or a domino yields the

recurrence

2,,= 21 nqataa nnn (6)

if n is divisible by k , and the recurrence

2,,= 21 nataa nnn (7)

if n is not, with the initial conditions 1=0a and ta =1 . By induction, recurrences (3), (6), and

(7) together show that


0,,= 1 npa nn (8)

where np denotes here the sequence defined by (3) with txxx k ==== 21 ,

1==== 121 kyyy , and qyk = . Thus, we have qy kj

k

j

k 1

1=

1 1)(=1)(= and

),()(=)()()(=

===

1111

*121

*1

*111

tGtqGtGtqGttG

pqatapapypL

kkkkk

kkkkkkk

since )(= 1 tGa ii and )(=* tGp ii if ki < , as there is no domino whose right half covers a

multiple of k . Formula (5) follows from writing 1= rmkrmk pa and using (4), which completes

the proof. In determining our next formula for na , we will need the generating function for the sequence

np given by (3).

Lemma 2.3. If np is defined as above, then

,1

)(=

2

1

0=

1

0=

0kk

rkrrk

k

r

rr

k

rnn

n xLx

xLppxpxp

(9)

where L and are given in Theorem 2.1. Proof: From (4), we have

.))(1(1

)(1

1

1

1

11=

1

1

1

11=

11=

=

1

0=

1

0=

1

0=

1

0=

1

0=

0

1

0=0

1

0=

0

1

0=0

rrk

k

rkk

k

rrk

rk

rrk

rk

rrrk

k

rk

rrrk

k

rk

rmkm

m

rrk

k

r

rmkm

m

rrk

k

r

rmkrmk

m

k

r

nn

n

xpxx

x

xp

xx

p

x

xp

px

xp

px

xp

pxp

p

xpxp


Note that

))(1(1

))((=

))(1(1

)(1)(1=

)(1)(1

kk

k

kk

kk

kk

xx

x

xx

xx

xx

since = . Thus, the first two sums in the last expression for n

nnxp 0

combine to give

kk

rkrrk

k

r

rr

k

r

kk

rkrk

k

rkk

rr

kk

rnn

n

xLx

xLppxp

xx

xp

xx

xpxxp

2

1

0=

1

0=

1

0=

1

0=

0

1

)(=

))(1(1))(1(1

))((1=

since =L , which completes the proof. The generating function for the sequence na may be given explicitly as follows.

Theorem 2.4. We have

.1)()(1

1)(=

211

11

1

0=1

1

0=

0kkk

kk

rkrk

rk

r

rr

k

rnn

n qxxGqG

xGxGxa

(10)

Proof: By (8) and (9), we have


,1)()(1

))((=

1)()(1

))((=

1

)(==

211

11

2

0=

1

0=

211

11111

1

1=

11

1=

2

11

0=

11

1=1

00

kkkkk

rkrkkrk

k

r

rr

k

r

kkkkk

rkrkkrk

k

r

rr

k

r

kk

rkrrk

k

r

rr

k

rnn

n

nn

n

qxxGqG

xaGqGaxa

qxxGqG

xaGqGaxa

xLx

xLppxpxpxa

by 0=0p and the expressions for L and given in the proof of Theorem 2.2 above. If

20 kr , then 1= rr Ga and

,== 211112 rkrkrkrkrk GGGqGaaaqaa

the first relation upon considering whether or not the numbers 1k and k are covered by a single domino within a member of rk . Thus,

,1)(=

=

)(=)(

11

112

11121111

rkr

rkrk

rkkrkrkrkkrk

G

GGGG

GGqGGGGqGaGqGa

the last equality by the identity 11=1)( nmnmmn

m GGGGG , nm 0 , which can be shown

by induction (see [Benjamin and Quinn (2003), p. 30, Identity 47] for the case when 1=t ). Substituting this into the last expression above for n

nnxa 0

, and noting 0=0G , completes the

proof. Corollary 2.5. If 0m and 10 ks , then

.)(1

1)(1)(

)(1)(=

2111

1)(2

1

0=1

1

211

1)(2

0=1

jmkk

jjk

m

jsk

s

jmkk

jjk

m

jssmk

GqGj

jmqG

GqGj

jmqGa

(11)

Proof: By (10), we have


.1)()(1)(=

)1)((1)(=

)1)((1

1)(=

1)(11

0=01

11

0=1

1

0=

110

11

1

0=1

1

0=

11

11

1

0=1

1

0=

0

jkikikiijkk

j

ij

rkrk

rk

r

rr

k

r

jkkkk

jk

j

rkrk

rk

r

rr

k

r

kkkk

k

rkrk

rk

r

rr

k

rnn

n

xqGqGi

jxGxG

qxGqGxxGxG

qxGqGx

xGxGxa

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 smkx , namely, the sr = term. Thus, the coefficient to smkx in the last expression is given by

.)(1

)1)((1)(

)()1)((

1211

111

0=1

1

211

1

0=1

mjkk

jmkm

jsk

s

mjkk

jmkm

js

GqGjm

jqG

GqGjm

jqG

Replacing j by jm in the first sum and j by jm 1 in the second gives (11). Taking 1=k in (11) implies

0,,=),( 2

0=

(1)

nj

jntqtqa jnj

n

jn (12)

which is well-known (see, e.g., [Benjamin and Quinn (2003), Shattuck and C. Wagner (2005)]. Taking 2=k in (11) implies

1,2= if),(

;2= if1),()(=),((2)

mnmtQ

mnmQmQtqan (13)

where

.1)(1)(=)( 222

0=

jmjj

m

j

qtj

jmqmQ

Taking 3=k in (11) implies


2,3= if),(1)(

1;3= if1),()(

;3= if1),()(

=),(2

(3)

mnmRt

mnmRmtR

mnmtRmR

tqan (14)

where

.))(2(=)( 232

0=

jmj

m

j

tqtj

jmqmR

Let )(= tHH nn denote the Lucas polynomial defined by the recurrence 21= nnn HtHH if

2n , with 2=0H and tH =1 , or, equivalently, by 11= nnn GGH if 1n .

Corollary 2.6. If 0m and 10 ks , then

.1

1)(1)(

1)(=

211)(2

1

0=1

1

21)(2

0=11

jmk

jk

m

jsk

s

jmk

jk

m

jssmk

Hj

jmG

Hj

jmGG

(15)

In particular, we have

0.,1

1)(= 211)(2

1

0=

mHj

jmGG jm

kjk

m

jkmk (16)

Proof: Taking 1=q in (11) and noting )(=)(1, 1

)( tGta nk

n for all k gives (15). Furthermore, if 1= ks

in (15), then the second sum drops out since 0=0G , which yields (16).

We were unable to find formulas (15) or (16) in the literature, though the 1=t 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 2=k . Note that ),((2) tqan is the joint distribution polynomial

on n for the statistics recording the number of squares and the number of dominos whose right


half covers an even number. The next result follows from taking 2=k in formula (10), though we provide another derivation here. Let ),(= (2)(2) tqaa nn and n

nnxtqatqxa ),(=),;( (2)

0 , which

we'll often denote by )(xa . Proposition 2.7. We have

.)(11

1=),;(

422

2

qxxtq

xtxtqxa

(17)

Proof: Considering whether or not a tiling ends in a square yields the recurrences

1,,= (2)22

(2)12

(2)2 nqataa nnn

and

1,,= (2)12

(2)2

(2)12 nataa nnn

with 1=(2)

0a and ta =(2)1 . Multiplying the first recurrence by nx2 and the second by 12 nx ,

summing both over 1n , and adding the two equations that result implies

,2

)()(

2

)()(1))((=1)( 22

xaxa

qxxaxa

xxatxtxxa

which may be rewritten as

).(1)(2=)())(12(2 22 xaqxxaxqtx (18) Replacing x with x in (18) gives

),(1)(2=)())(12(2 22 xaqxxaxqtx (19) and solving the system of equations (18) and (19) in )(xa and )( xa yields

,)(11

1=)(

422

2

qxxtq

xtxxa

as desired.


We next consider some particular values of the polynomials ),((2) tqan .

Proposition 2.8 If 1n , then

1.2= if,)(1

;2= if,)(1=)(0,

2

122(2)

mntt

mnttta

m

m

n (20)

Proof: We provide both algebraic and combinatorial proofs. Taking 0=q in (17) implies

,)(1)(11=

)(1))(1)((11=

)(1)(1=)(11

1=);0,(

122

0

2122

1

122

0

2122

1

22

0

222

2

mm

m

mm

m

mm

m

mmm

m

mm

m

xttxtt

xttxtt

xtxtxxt

xtxtxa

from which (20) follows. For a combinatorial proof, first let mn 2= , where 1m . Then members of n having zero

2 value are of the form

)())((= 21 ssdssdssdaaa

for some , where 0ia for each },{1,2,=][ i .

Note that the sequence 1),1,1,( 21 aaa is a composition of m . Thus, the polynomial )(0,(2) tan may be viewed as the weighted sum of compositions of m , where

the weight of a composition having exactly parts is 2t . Since there are

1

1

m

compositions

of m having parts, we have

,)(1=1

=1

1=)(0, 12222

1

0=

2

1=

(2)

mmm

n tttm

tm

ta

which gives the even case.


If 12= mn , then the weighted sum of tilings s , where n has zero 2 value, is given by mtt )(1 22 , 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 1n , then

1.2= if),(

;2= if),()(=)1,(

21

221(2)

mnttG

mntGtGta

m

mmn (21)

Proof: We provide both algebraic and combinatorial proofs of this result. Taking 1= q in (17) and

replacing x with 2x and t with 2t in

210 1

1=)(

xtxxtG m

mm

implies

,))()(()(=

)()(1=1

1=)1,;(

2221

0

1221

0

221

0

2422

2

mmm

m

mm

m

mm

m

xtGtGxttG

xtGxtxxxt

xtxtxa

which gives the result. We provide a bijective proof of (21) in the case when 1=t , the general case being similar, and show

1.2= if,

;2= if,=1,1)(

1

1(2)

mnF

mnFa

m

mn (22)

To do so, define the sign of n by )(21)(=)(

sgn , and let en and o

n denote the subsets

of n whose members have positive and negative sign, respectively. Then

|||=|1,1)((2) on

enna and it suffices to identify a subset *

n of en having cardinality 1mF or

1mF , along with a sign-changing involution of *nn .

Let mn 2= and nn consist of those coverings 21= such that ii 212 = for all i . If


nn , then let oi denote the smallest index i such that ii 212 , i.e., dsii =212 or sd

. Let )(f denote the covering that is obtained from by exchanging the positions of the

1)(2 oi -st and )(2 oi -th pieces of , leaving all other pieces undisturbed. Then the mapping f

is seen to be a sign-changing involution of nn .

We now define an involution of n . Let nn * consist of those members containing an even

number of pieces and ending in a domino. Note that enn * and that 1

* |=| mn F since

members of *n are synonymous with members of m ending in a domino, upon halving.

Observe further that if *nn 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 *nn ,

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

*nn . Combining the two mappings f and g yields a sign-changing involution of *

nn ,

as desired. If 12= mn , then apply the mapping f defined above to n . Note that the set of survivors has

cardinality 1mF , upon halving, since they are of the form 12221= for some , with

ii 212 = for each ][i and s=12 . This completes the proof of (22).

Let )( 2nt denote the sum of the 2 values taken over all of the members of n .

Proposition 2.10 . If 1n , then

odd. is if,10

21)(

;even is if,10

4

=)(

1

2

nFLn

nFnL

t

nn

nn

n (23)

Proof: To find )( 2nt , we consider the contribution of the dominos that cover the numbers 12 i and i2

for some i fixed within all of the members of n . Let 12= mn .

Note that there are imi FF 22212 dominos that cover the numbers 12 i and i2 within all of the

members of n .

Summing over all i , we have


.==)( 2212

1

0=22212

1=2 imi

m

iimi

m

in FFFFt

To simplify this sum, we recall the Binet formulas 5

=nn

nF

and 0,,= nL nnn where

and denote the positive and negative roots, respectively, of the equation 0=12 xx . Then for m even, we have

,=

==

)(==

=

)(

)()(=

))((=5

212

121112

342142

12

0=12242

12

0=12

342

12

0=142

12

0=12

1

0=

1241241

2=

142142

12

0=

12121

0=

222212121

0=2212

1

0=

mm

mmmmmmmm

imim

m

imim

m

im

im

m

iim

m

im

m

i

mimim

mi

imim

m

i

mmm

i

imimiim

iimi

m

i

FmL

LFmLFFFFmL

FFmLLmL

LLL

FF

by Identities 28, 26 and 33 in Benjamin and Quinn (2003) and since 11= mmm FFL . Substituting

2

1=

nm gives the second formula when 4) mod( 1n . A similar calculation gives the same

formula when 4) mod( 3n . If mn 2= and m is odd, then similar reasoning shows that


,2=22=

22=

22=

)()()(=

5=)(5

22112

342

2

3

0=142

2

1

0=2

242

2

3

0=2

2242241

2

1=

2422422

1

0=

221

0=

12212

1

0=2

mmmmmmm

im

m

iim

m

im

im

m

im

mimim

mi

imim

m

i

mmm

i

imi

m

in

FmLFFFFmL

FFmL

LmL

FFt

and the first formula in (23) follows when 4) mod( 2n , upon replacing m with 2

n. A similar

calculation gives the same formula when 4) mod( 0n . We close this section with a general formula for ),((2) tqan .

Theorem 2.11. If 0n , then

1.2= if,

;2= if,1

=),(

122

0=0=

22

0=

1

0=(2)

mntqji

jm

j

jim

mntqji

jm

j

jimq

tqa

imji

j

m

i

imji

j

m

i

m

n (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 mn 2= is even. If n contains no squares, then it

consists of m odd-positioned dominos, whence the mq term. So suppose that contains i dominos, where 10 mi , and that j of the dominos are odd-positioned. There are im 22 squares and 1 im possible positions to insert each of the j odd-positioned dominos relative

to the squares, whence there are

j

jim choices concerning their placement. There are im

possible positions to insert each of the ji even-positioned dominos, whence there are


ji

jm 1 choices concerning their placement. Thus, there are

ji

jm

j

jim 1 members

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 0=q 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 :

0.,1)(1)(= 22

0=0=0=

mtqj

jmq

ji

jm

j

jimtq jmjj

m

j

imji

j

m

i

(25)

A similar identity can be obtained by comparing the even cases of (24) and (13). Setting 1= q

in (24), comparing with (21), and replacing t with t gives a pair of formulas for )(tGm .

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 1ik that are covered by the left

half of a domino. For example, if 24=n , 4=k , and 24= ddssddssdsddsdss (see

Figure 3 below), then 8=)(s and 52=2315113=)(4 . If q and t are indeterminates,

then define the distribution polynomial ),()( tqb kn by

1,,:=),( )()()(

ntqtqb sk

n

kn

with 1:=),()(

0 tqb k . For example, if 6=n and 3=k , then

.1)(1)(2)1)((=),( 27225222222(3)

6 tqtqttqttttqb

Note that 1

)( =)(1, nk

n Gtb for all k and n .

Figure 3. The tiling 24= ddssddssdsddsdss has 52=)(4 . In what follows, we will often suppress arguments and write nb for ),()( tqb k

n . Considering

whether the last piece within a member of n is a square or a domino yields the recurrence

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24


2,,= 21

1

nbqtbb nn

nn

if n is divisible by k , and the recurrence

2,,= 21 nbtbb nnn

if n is not, with initial conditions 1=0b and tb =1 . In [4], Carlitz studied the polynomials

),((1)1 tqbn 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 ),()( tqb k

n and their

generating function, with specific consideration of the case 2=k . 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 nb .

Theorem 3.1. We have

),()(

)()(=

11

2

2

3

0=1

3

0=1

11

3

0=0

kk

kkk

k

rkr

k

r

kk

rkr

k

r

kkk

krr

k

r

nn

n

xcxxcx

xGxcxGxqcqxGxb

(26)

where

)(1

)1)((=)(

10=

11

0=

2

1

01

kik

j

i

ikkk

j

i

jj

kj

jk

Gxq

xqGqxxc

and

.

)(1

)1)((=)(

10=

11

1=

2

1

0

kik

j

i

ikkk

j

i

jj

kj

jkk

Gxq

xqGqxGxc

Proof: It is more convenient to first consider the generating function for the numbers ),(:= )(

1 tqbb knn .


Then the sequence nb has initial values 0=0b and 1=1b and satisfies the recurrences

0,and2,= 21 mkrbbtb rmkrmkrmk (27)

with

1.,= 11

1

mbqbtb mkmk

mkmk (28)

Let

,=)(0

mrmk

mr xbxc

where ][kr . Then multiplying the recurrences (27) and (28) by mx , and summing the first over

0m and the second over 1m , gives

).()(1=)(

),()(=)(

,,3,4,=),()(=)(

11

1

12

21

xqcxqxxtcxc

xxcxtcxc

krxcxtcxc

kk

kk

k

rrr

By induction on r , we obtain

.1),()(=)( 11

1 krxqcxqGxxcGGxc kk

krkrrr

(29)

Taking kr = and 1= kr in (29) gives

)(11

=)( 11

1

1

xqcxG

Gxq

xG

Gxc k

kk

kk

k

kk

and

),(1

)1)((

1

1)(=

)(1

)(

1

)(=

)()(11

=

)()(=)(

11

11

1

1

11

11

11

11212

1

112

1

11

111

1

11

11

111

xqcxG

xGxq

xG

xG

xqcxG

GxqGGGqx

xG

GGGxG

xqcxqGxqcxG

Gxq

xG

GxGG

xqcxqGxxcGGxc

kk

k

kk

k

k

kk

kk

k

kk

kkkk

k

kkkk

kk

kk

kk

k

kk

k

kkk

kk

kkkkkk

where we have used the identity 1

112 1)(=

kkkk GGG [see, e.g., (Benjamin and Quinn (2003),


Identity 246]. Iterating the last recurrence gives

,

)(1

)1)((=

1

)1)((

1

1)(=)(

10=

11

0=

2

1

0

11)(

1)(11

1

1=1

11

01

kik

j

i

ikkk

j

i

jj

kj

j

kki

kikk

ikj

ikkj

kjkk

jk

Gxq

xqGqx

Gxq

xqGxq

Gxq

xqGxc

which implies

.

)(1

)1)((=

)(1

)1)((

1=)(

10=

11

1=

2

1

0

10=

1)(11

1

0=

2

11

kik

j

i

ikkk

j

i

jj

kj

jk

kik

j

i

kikk

j

i

jkjj

kj

jk

k

kk

Gxq

xqGqxG

Gxq

xqGqxG

xG

Gxc

Then, by (29), we have

),()(

)()(=

)()())()((=

)(==

11

2

1=1

11

2

1=

2

1=

11

11

1

2

1=

1=01=0

kk

kkk

k

rkr

k

r

kkk

krkr

k

r

kk

rr

k

r

kk

kkk

kkkk

kkr

kk

krr

rk

r

kr

rk

r

rmkrmk

m

k

r

nn

n

xcxxcx

xGxqcqxGxcxG

xcxxcxxqcqxGxcxGGx

xcxxbxb

where )(1 xck and )(xck are as given. Formula (26) now follows upon noting

.1

==),(0

10

)(

0

nn

n

nn

n

nkn

n

xbx

xbxtqb

One can find explicit expressions for the nb using Theorem 3.1 and the following formulas that


involve the q -binomial coefficient [see, e.g., (Andrews (1976) or Stanley (1997)]:

a

qjaij

i

j

xj

a

xq

x

=

)(10=

(30)

and

,=)( 2

1

0=1=

aja

q

aj

a

ij

i

yxa

jqxqy

(31)

where j is a non-negative integer. Theorem 3.2. The following formulas hold for nb . If 0m , then

.1)(= 112

1

1)(

0=

2

1

0=1

kqkq

jamk

ajk

ak

akm

a

jj

km

jkkmk j

am

a

jGGqqGb

(32)

If 0m , then

.1

1)(= 11

121)(

0=

2

1

0=2

kqkq

jamk

ajk

ak

akm

a

jj

km

jkmk j

am

a

jGGqqb

(33)

If 1m and 30 kr , then

,11

1)(

11)(=

11

11

21)(1

0=

2

11

0=1

1

111

2

1

1)(1

0=

2

11

0=2

kqkq

jamk

ajk

ak

akm

a

jj

km

jr

km

kqkq

jamk

ajk

ak

akm

a

jj

km

jrkrmk

j

am

a

jGGqqGq

j

am

a

jGGqqGGb

(34)

with 1= rr Gb . Proof: Let 1= kmkn , where 0m . Then the coefficient of nx on the right-hand side of (26) is given by


.

)(1

)1)((][ =))(]([

10=

11

1=

2

1

0

kik

j

i

ikkk

j

i

jj

kj

m

jkk

m

Gxq

xqGqxxGxcx

By (30) and (31), we have for each 0j ,

ak

a

kqjak

kij

i

jk Gx

j

a

Gxq

xG1

10=

1 =

)(1

)(

and

,1)(=)1)(( 12

1

1)(

0=

11

1=

ajk

a

kq

ak

akj

a

ikkk

j

i

Gxa

jqxqG

so that coefficient of nx is given by

,1)( 112

1

1)(

0=1

2

1

0=

amk

kqkq

ajk

ak

akm

aj

k

jj

km

jk G

j

am

a

jGq

G

qG

which yields (32). Similar proofs apply to formulas (33) and (34), in the latter case, upon extracting the coefficient of nx from two separate terms in (26). When 1=k in Theorem 3.2, the inner sum in (32) reduces to a single term since 0=0G and

gives

0,,=),( 22

0=

(1)

ntj

jnqtqb jn

q

jn

jn (35)

which is well-known [see, e.g., Shattuck and Wagner (2005)] When 2=k in Theorem 3.2, we get for all 0m the formulas

22

22

0=

2

0=

(2)2

11)(1)(=),(

qq

jamaaam

a

jm

jm j

am

a

jtqqtqb

(36)

and


.1)(1)(=),(22

22

0=

2

0=

(2)12

qq

jamaaam

a

jm

jm j

am

a

jtqqttqb

(37)

The polynomials ),((2) tqbn are considered in more detail below.

3.2. The Case Let us write (2)

nb for ),((2) tqbn . The even and odd terms of the sequence (2)nb satisfy the following

two-term recurrences. Proposition 3.3. If 2m , then

,1)(= )2(42

32(2)22

212(2)2

mm

mm

m bqbtqb (38)

with 1=(2)

0b and qtb 2(2)2 = , and

,1)(= (2)

3212(2)

12212(2)

12

mm

mm

m bqbtqb (39)

with tb =(2)

1 and tqtb )(1= 3(2)3 .

Proof: To show (39), first note that the total weight of all the members of 12 m ending in d or ss is

(2)12 mb and (2)

122

mbt , respectively. The weight of all members of 12 m ending in ds is

)( (2)32

(2)12

12

mmm bbq . To see this, we insert a d just before the final s in any 12 m ending in

s . By subtraction, the total weight of all tilings that end in s is (2)32

(2)12 mm bb , and the inserted

d contributes 12 m towards the 2 value since it covers the numbers 12 m and m2 . For (38), note that by similar reasoning, the total weight of all members of m2 ending in d , ss , and ds

is (2)22

12

m

m bq , (2)22

2mbt , and (2)

4232(2)

22

mm

m bqb , respectively.

We were unable to find, in general, two-term recurrences comparable to (38) and (39) for the sequences ),()( tqb k

rmk , where k and r are fixed and 0m . Let nnn

xtqbtqxb ),(=),;( (2)

0 .

Using (38) and (39), it is possible to determine explicit formulas for the generating functions m

mmxb(2)

20 and m

mmxb(2)

120 and thus for ),;( tqxb , 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 2=k in Theorem 3.1.


Proposition 3.4. We have

.

)1)((1

)(1)(1=),;(

222

0=

22

1=

222

0 xqt

xqxtxqxtqxb

ij

i

ij

i

jj

j

(40)

Taking 0=q and 1= q in (40) shows that )(0,(2) tbn and )1,((2) tbn are the same as )(0,(2) tan

and )1,((2) tan 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 1= q , and replacing t with t , then gives a pair of formulas for the Fibonacci polynomials. Corollary 3.5. If 0m , then

j

am

a

jttGtG jamja

m

a

m

jmm

11)(1)(=)()(

0=0=1 (41)

and

.1)(1)(=)(0=0=

1

j

am

a

jttG jamja

m

a

m

jm (42)

Taking 1=q in (36) and (37), and noting )(=)(1, 1

(2) tGtb nn , gives another pair of formulas.

Corollary 3.6. If 0m , then

j

am

a

jttG jama

m

a

m

jm

11)(1)(=)( 2

0=0=12 (43)

and

.1)(1)(=)( 2

0=0=22

j

am

a

jtttG jama

m

a

m

jm (44)


Let )( 2nt denote the sum of the 2 values taken over all of the members of n . We conclude

with the following explicit formula for )( 2nt .

Proposition 3.7. If 0n , then

.40

5)2(215)2(6

8

3)(21)(21)(=)( 2

21

212

2

nnnnnn

FnnFnnFnFnt (45)

Proof:

To find )( 2nt , first note that 1=20|,1);(=)( q

nnn

qxbdq

dxt

. By Proposition 3.4 and partial

fractions, we have

.)2(1

74

)(1

33

)8(1

79

)4(1

43

)8(1

71=

)2(1

)(11)(2

21

)(1

)2(1

)(11)(

1

)(1

)2(1

)(1)(1=|,1);(

32222222

12

22

02

22

12

22

02

22

12

222

0

21=

xx

x

xx

x

xx

x

xx

x

xx

x

x

x

xxnn

x

xxx

x

xxnn

x

xxx

x

xxnxxqxb

dq

d

n

nn

n

n

nn

nn

nn

nq

Note that

;50

2)17)((51)16)((5=

)(1

1][

,5

2)2(1)(=

)(1

1][

,=1

1][

1232

1222

12

nnn

nnn

nn

FnnFnn

xxx

FnFn

xxx

Fxx

x

see sequences A000045, A001629, and A001628, respectively, in Sloane (2010). Thus, the

coefficient of nx in 1=|,1);( qqxbdq

d is given by

,40

5)2(215)2(6

8

3)(21)(21)( 2

21

212

nnnnn FnnFnnFnFn

which completes the proof.


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 1=k , 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, k mod , 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.

Generalizations of Two Statistics on Linear Tilings

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