generalizations of two statistics on linear tilings
TRANSCRIPT
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
!][!][
!][
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
AAM: Intern. J., Vol. 7, Issue 2 (December 2012) 511
,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
512 T. Mansour & M. Shattuck
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
AAM: Intern. J., Vol. 7, Issue 2 (December 2012) 513
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
514 T. Mansour & M. Shattuck
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
AAM: Intern. J., Vol. 7, Issue 2 (December 2012) 515
,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
516 T. Mansour & M. Shattuck
.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
AAM: Intern. J., Vol. 7, Issue 2 (December 2012) 517
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
518 T. Mansour & M. Shattuck
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.
AAM: Intern. J., Vol. 7, Issue 2 (December 2012) 519
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.
520 T. Mansour & M. Shattuck
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
AAM: Intern. J., Vol. 7, Issue 2 (December 2012) 521
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
522 T. Mansour & M. Shattuck
.==)( 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
AAM: Intern. J., Vol. 7, Issue 2 (December 2012) 523
,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
524 T. Mansour & M. Shattuck
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
AAM: Intern. J., Vol. 7, Issue 2 (December 2012) 525
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 .
526 T. Mansour & M. Shattuck
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),
AAM: Intern. J., Vol. 7, Issue 2 (December 2012) 527
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
528 T. Mansour & M. Shattuck
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
AAM: Intern. J., Vol. 7, Issue 2 (December 2012) 529
.
)(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)(=),(
jamaaam
a
jm
jm j
am
a
jtqqtqb
(36)
and
530 T. Mansour & M. Shattuck
.1)(1)(=),(22
22
0=
2
0=
(2)12
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.
AAM: Intern. J., Vol. 7, Issue 2 (December 2012) 531
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)
532 T. Mansour & M. Shattuck
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.
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 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.
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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.
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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,
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