Generalization of a Statistic on Linear Domino Arrangements

GENERALIZATION OF A STATISTIC ON LINEAR DOMINO ARRANGEMENTS

TOUFIK MANSOUR AND MARK SHATTUCK

ABSTRACT. In this paper, we generalize an earlier statistic on square-and-domino tilings by considering only those squares covering a multiple of k, where k is a fixed positive integer. We consider the distribution of this statistic jointly with the one that records the number of dominos in a tiling. We derive both finite and infinite sum expressions for the corresponding joint distribution polynomials, the first of which reduces when k 1to a = prior result. The cases q 0 and q 1 are noted for general k. Finally, the case k 2 = =− = is considered specifically, where further results may be given, including a combinatorial proof when q 1. =−

1. INTRODUCTION

Let Fn be the Fibonacci number defined by the recurrence Fn Fn 1 Fn 2 if n 2, = − + − ≥ with initial conditions F 0 and F 1. See, for example, sequence A000045 in [12]. Let 0 = 1 = Gn Gn(t) be the Fibonacci polynomial defined by Gn Gn 1 tGn 2 if n 2, with G0 0 = = − + − ≥ = and G1 1; note that Gn (1) Fn for all n. See, for example, [10]. Finally, the q-binomial = x = coefficient k q is defined by

x i 1 ¡ ¢ k 1 q − + x − , if k 0; i 1 1 qi = − ≥ Ãk! = (0, if k 0. q Q < Polynomial generalizations of Fn have arisen in connection with statistics on binary words [3], Morse code sequences [4], lattice paths [5], and linear domino arrangements [10, 11]. Let us recall now a statistic related to domino arrangements. If n 1, then let F ≥ 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 F are also called (linear) tilings or domino arrangements. (If n 0, then n = F0 consists of the empty tiling having length zero.) Note that members of Fn correspond uniquely to words in the alphabet {d, s} compris- ing i d’s and n 2i s’s for some i,0 i n . In what follows, we will frequently identify − ≤ ≤ ⌊ 2 ⌋ tilings c by such words c c . For example, if n 4, then F {dd,dss, sds, ssd, ssss}. 1 2 ··· = 4 = Note that Fn Fn 1 for all n. Given π Fn , let ρ(π) denote the sum of the numbers | |= + ∈ covered by squares in π. For example, if n 15 and π sds2d 2sd 2s F (see Figure 1 = = ∈ 15 below), then ρ(π) 1 4 5 10 15 35. = + + + + =

1 2 3 4 5 6 7 8 9 101112131415

FIGURE 1. The tiling π sds2d 2sd 2s F has ρ(π) 35. = ∈ 15 = Date: 2013-06-14. 2000 Mathematics Subject Classiﬁcation. 11B39, 05A15. Key words and phrases. tilings, Fibonacci numbers, polynomial generalization. 1 2 TOUFIK MANSOUR AND MARK SHATTUCK

The statistic ρ was introduced in [11], where its distribution was studied on r -mino arrangements. Let ν(π) denote the number of dominos in the tiling π. Then the joint distribution for the ρ and ν statistics on Fn is given by

n n 2j 1 ρ(π) ν(π) − + n j j (1) q t q( 2 ) − t , n 0, = j ≥ π Fn j 0 Ã ! 2 X∈ X= q where q and t are indeterminates. Equation (1) is the r 2 case (corresponding to square- = and-domino tilings) of [11, Theorem 2.1], which is a result on more general r -mino arrangements. Here, we will provide a different generalization of (1). Note that (1) reduces n n j to the well-known formula Fn 1 − when q t 1. + = j 0 j = = Recently, generalizations of the Fibonacci= sequence have been studied which specify P ¡ ¢ the recurrence for each value of the index mod k, where k is a ﬁxed positive integer. For example, the recurrence

(2) Qm a j Qm 1 b j Qm 2, m j (mod k), = − + − ≡ with Q 0 and Q 1, was considered in [8], where a Binet-like formula is derived. See 0 = 1 = also [6] for the case when b 1 for all j and [13] for the case k 2. These generalizations j = = so far have been studied primarily from an algebraic standpoint such as through the use of generating functions [6] or orthogonal polynomials [8]. In [7], a special case of (2) and a closely related sequence are studied from a more combinatorial viewpoint in terms of statistics on linear tilings and new generalizations of Fn are obtained which extend prior ones. In this paper, we continue this study by considering a generalization of the ρ statistic defined above, where one looks only at squares that cover multiples of k. More precisely, let ρk record the sum divided by k of all the multiples of k which are covered by squares within a member of F . Note that ρ reduces to ρ when k 1. n k = In the next section, we obtain an explicit formula for all k (see Theorem 2.2 below) for the joint distribution a(k)(q, t) : qρk (π)t ν(π). n = π Fn X∈ This yields an infinite family of q-generalizations for the numbers Gn(t) defined above, and setting q 1 yields seemingly new expressions for G (t). When k 1 in our formula, = n = we obtain the explicit expression (1) above, but with a different proof than that given in (k) [11]. We also note some special cases of q and provide an infinite expansion for an (q, t) (see Theorem 2.7 below). In the third section, we consider specifically the case k 2, = where further combinatorial results may be given. In particular, we provide a combinatorial proof explaining the values of a(2)( 1,1) as well as an explicit expression for the sum n − of the ρ2 values taken over all the members of Fn . Note that ρ2 records half the sum of the even numbers covered by squares within a tiling.

2. GENERAL FORMULAS Suppose k is a ﬁxed positive integer. Given π F , let ν(π) denote the number of ∈ n dominos of π and let ρk (π) denote the sum divided by k of all the multiples of k covered by squares of π. For example, if π s2d 3sdsdsd 2s2d 2 F (see Figure 2 below), then = ∈ 25 ν(π) 9 and = 9 12 15 21 ρ3(π) + + + 19. = 3 = GENERALIZATION OF A STATISTIC ON LINEAR DOMINO ARRANGEMENTS 3

(k) If q and t are indeterminates, then deﬁne the distribution polynomial an (q, t) by a(k)(q, t) : qρk (π)t ν(π), n 1, n = ≥ π Fn X∈ with a(0)(q, t) : 1. For example, if n 6 and k 3, then n = = = a(3)(q, t) 2t 2 t 3 q(1 t)(t 2qt q2 q2t). 6 = + + + + + + (k) Note that an (1, t) Gn 1 for all k and n. = + 1 2 3 4 5 6 7 8 9 10111213141516171819202122232425

FIGURE 2. The tiling π s2d 3sdsdsd 2s2d 2 F has ρ (π) 19. = ∈ 25 3 = (k) In what follows, we will often suppress arguments and write an for an (q, t). Consid- ering whether the last piece within a member of Fn is a square or a domino yields the recurrence n (3) an q k an 1 tan 2, n 2, = − + − ≥ if n is divisible by k, and the recurrence

(4) an an 1 tan 2, n 2, = − + − ≥ if n is not, with the initial conditions a 1 and 0 = q, if k 1; a1 = = (1, if k 1. > To solve recurrences (3) and (4), we first ascertain an explicit formula for the generating function of the numbers an. Theorem 2.1. We have (5) j 1 j + jk k 1 k 1 G q( 2 )x a xn − xr G txk − ( tx)r G k . n r 1 k 1 r j n 0 = Ãr 0 + − r 0 − − − ! j 0 (1 2tqi xkG ( t)k q2i x2k ) ≥ = = i 0 k 1 X X X X≥ = − − + − Proof. It is more convenient to first consider the generatingQ function for the numbers an′ : (k) = a (q, t). Then the sequence a′ has initial values a′ 0 and a′ 1 and satisfies the n 1 n 0 = 1 = recurrences−

(6) amk′ r amk′ r 1 tamk′ r 2, 2 r k and m 0, + = + − + + − ≤ ≤ ≥ with m (7) amk′ 1 q amk′ tamk′ 1, m 1. + = + − ≥ Let m ar (x) amk′ r x , = m 0 + ≥ where r [k]. Then multiplying the recurrencesX (6) and (7) by xm, and summing the ﬁrst ∈ over m 0 and the second over m 1, gives ≥ ≥ ar (x) ar 1(x) tar 2(x), 3 r k, = − + − ≤ ≤ a (x) a (x) txa (x), 2 = 1 + k a1(x) 1 qxak (qx) txak 1(x). = + + − 4 TOUFIK MANSOUR AND MARK SHATTUCK

By induction on r , we obtain

ar (x) Gr 1a2(x) tGr 2a1(x), 2 r k. = − + − ≤ ≤ Therefore, ar (x) Gr 1(a1(x) txak (x)) tGr 2a1(x), = − + + − which implies

(8) ar (x) Gr a1(x) txGr 1ak (x), 2 r k. = + − ≤ ≤ Taking r k in (8) gives = 1 txGk 1 a1(x) − − ak (x). = Gk By induction on r , we obtain r Gr ( t) xGk r ar (x) + − − ak (x), 1 r k. = Gk ≤ ≤ Since a1(x) 1 qxak (qx) txak 1(x), the last relation may be rewritten as = + + − G qxG (9) a (x) k k a (qx). k k 2 k 2 k = 1 2txGk 1 ( t) x + 1 2txGk 1 ( t) x − − + − − − + − Iterating (9) yields j 1 j 1 + j + ( 2 ) Gk q x ak (x) . = j i k 2i 2 j 0 i 0(1 2tq xGk 1 ( t) q x ) X≥ = − − + − Thus, we have Q j 1 j + j G q( 2 )x r k ar (x) (Gr ( t) xGk r ) , 1 r k, = + − − j i k 2i 2 ≤ ≤ j 0 i 0(1 2tq xGk 1 ( t) q x ) X≥ = − − + − which implies Q k k n mk r r k an′ x amk′ r x + x ar (x ) n 0 = r 1 m 0 + = r 1 ≥ = ≥ = X X X X j 1 j + jk k k G q( 2 )x xr G xk ( tx)r G k . r k r j = Ãr 1 + r 1 − − ! j 0 (1 2tqi xkG ( t)k q2i x2k ) = = i 0 k 1 X X X≥ = − − + − The result now follows since Q n n 1 n an x an′ 1x an′ x . n 0 = n 0 + = x n 0 X≥ X≥ X≥

(k) We now derive an explicit formula for the polynomials an (q, t). Theorem 2.2. If n km r, where m 0 and 0 r k 1, then = + ≥ ≤ ≤ − r 1 (10) an Gr 1S(m) ( t) + Gk 1 r S(m 1), = + + − − − − where m m j j 1 a m j a k j ( +2 ) m (k 1)j a m j a S(m) ( 1) Gk q t − − d d + − + − , m 0, = j 0 − a j + − Ã j ! Ã j ! ≥ X= X= q q with S( 1) 0 and − = d Gk 1 GkGk 2. ± = − ± − p GENERALIZATION OF A STATISTIC ON LINEAR DOMINO ARRANGEMENTS 5

Proof. Note ﬁrst that

2 k 2 d Gk 1 G ( t) − , ± = − ± k 1 − − q − 2 m 1 by the identity Gm Gm 1Gm 1 ( t) − , which can be shown by induction (see, e.g., [2, − + − = − Identity 8] for the t 1 case). Then =

i k 2i 2 i i 1 2tq xGk 1 ( t) q x (1 d tq x)(1 d tq x). − − + − = − + − − Let n mk r , where m 0and0 r k 1. By Theorem 2.1, we have = + ≥ ≤ ≤ −

m r 1 m 1 an Gr 1[x ](a(x)) ( t) + Gk 1 r [x − ](a(x)), = + + − − − where

j 1 j + j G q( 2 )x a(x) k . = j i k 2i 2 j 0 i 0(1 2tq xGk 1 ( t) q x ) X≥ = − − + − Using the expansion [1] Q

y j a y a j i = Ã j ! i 0(1 q y) a j q = − X≥ k 2 Q and the fact d d ( t) − , we have + − = −

j 1 j + j G q( 2 )x [xm ](a(x)) [xm ] k =  j i i  j 0 i 0(1 d tq x)(1 d tq x) X≥ = − + − − j j 1  G q( +Q2 ) j j k m j (d tx) (d tx) [x + ] + − = j j 2j Ã j i · j i ! j 0 d d t i 0(1 d tq x) i 0(1 d tq x) X≥ + − = − + = − − j j 1 m G q( +2 ) m Q Q k a a m j a m j a (d t) + − (d t) + − k j k j = j 0 ( 1) t a j Ã j ! + · Ã j ! − X= − X= q q m m j j 1 a m j a k j ( +2 ) m (k 1)j a m j a ( 1) Gk q t − − d d + − + − , = j 0 − a j + − Ã j ! Ã j ! X= X= q q which completes the proof.

Letting k 1 in Theorem 2.2 gives the following expression for a(1)(q, t). = n Corollary 2.3. If n 0, then ≥

n n 2j 1 n j (1) ( − 2 + ) j (11) an (q, t) q − t . = j 0 Ã j ! 2 X= q 6 TOUFIK MANSOUR AND MARK SHATTUCK

1 1 Proof. When k 1, we have d since G0 0 and G 1 . Taking k 1 in (10) then = ± =± pt = − = t = gives

n n a n j a j 1 1 a 1 + − n j a (1) j ( +2 ) n n j a an (q, t) S(n) ( 1) q t ( 1) + − + − = = − pt j · − pt j j 0 a j µ ¶ Ã !q µ ¶ Ã !q X= X= n n j 1 n j a n j a ( + ) − n a q 2 t 2 ( 1) − + − = j 0 a j − Ã j ! Ã j ! X= X= q q n n n j 1 j a 2n j a ( − + ) n a q 2 t 2 ( 1) − − − = j 0 a n j − Ãn j ! Ã n j ! X= =X− − q − q n j j j n j 1 a a n j n a ( 1) q( −2+ )t 2 ( 1) + − − = j 0 − a 0 − Ã n j ! Ãn j ! X= X= − q − q n n j 1 j n n 2j 1 j − + n j /2 − + n j j ( 1) q( 2 ) − t 2 q( 2 ) − t , = − n j = j j 0 Ã − !q2 j 0 Ã !q2 j Xeven= X= where we have used the identity

n m n m a m n a +2 , if n m (mod 2); (12) − ( 1)a + − m q2 ≡ (0 m n). a 0 − Ã m ! Ã m ! = (¡0,¢ otherwise, ≤ ≤ X= q q

Note that (12) may be obtained by writing

a m a 1 xm xm ( 1)a + xa xa − m · m = m (1 qi x) · m (1 qi x) = m (1 q2i x2) a 0 Ã !q a 0 Ã !q i 0 i 0 i 0 X≥ X≥ = + = − = − Q a m 2Qa m Q + x + , = a 0 Ã m ! 2 X≥ q and extracting the coefﬁcient of xn from both sides.

Remark: Formula (11) corresponds to the r 2 case of [11, Theorem 2.1], which is a result = on more general r -mino arrangements where no restriction is placed on the positions of r -minos or squares. The proof there was combinatorial, though it does not seem that it can be extended to prove Theorem 2.2 above.

(k) Taking q 1 and r k 1 in (10), and noting an (1, t) Gn 1, yields the following iden- = = − = + tity. GENERALIZATION OF A STATISTIC ON LINEAR DOMINO ARRANGEMENTS 7

Corollary 2.4. If m 0 and k 1, then ≥ ≥ k m k j j m (k 1)j a m j a a m j a (13) G(m 1)k Gk ( 1) Gk t − − d d + − + − , + = j 0 − a j + − Ã j !Ã j ! X= X= where d Gk 1 pGkGk 2. ± = − ± − We have the following explicit formula for the number of members of Fn (weighted according to the value of ν) in which no square covers a multiple of k.

Corollary 2.5. If n km r, wherem 0 and 0 r k 1, then = + ≥ ≤ ≤ − (k) m r 1 m r (14) an (0, t) t Gr 1T (m) ( 1) + t + Gk 1 r T (m 1), = + + − − − − where m m 1 m 2i i T (m) + Gk −1 (GkGk 2) . = i 0 Ã2i 1! − − X= + Proof. Setting q 0 in (10) implies = m m 1 (k) m a m a r 1 m r − a m 1 a amk r (0, t) t Gr 1 d d − ( 1) + t + Gk 1 r d d − − , + = + a 0 + − + − − − a 0 + − X= X= with m m 1 m 1 m a m a d + d + 1 m 1 m 2i 2i 1 d d − + − − 2 + Gk −1 ( GkGk 2) + . a 0 + − = d d = 2pGkGk 2 i 0 Ã2i 1! − − X= + − − − X= + p

(1) n For example, when k 1 in(14), we see that a (0, t) equals t 2 for n even and zero for n = n odd. Taking k 2 in (14) gives a(2) (0, t) t m and a(2) (0, t) (m 1)t m for m 0. These = 2m = 2m 1 = + ≥ formulas are readily seen directly. + We next consider the case q 1. Recall that for any generating function in q, the eval- =− uation at q 1 gives the difference in cardinalities between those membersof a structure =− having an even value for the statisticcounted by q with those having an odd value. Letting (i) n q 1 and t 1 in (5) gives the following formulas, where fi : n 0 an ( 1,1)x : =− = = ≥ − P (1 x x3 x4)(1 x6) f1 − − − − , = 1 x12 − (1 x x3 x4 2x5 x6 x7 x9 x10)(1 x12) f2 + + + + − + + − − , = 1 x24 − 1 x 2x2 x3 x4 f3 + + − + , = 1 x6 − (1 x 2x2 3x3 2x4 x5 x6)(1 x4 x8) f4 + + + − + − + + . = 1 5x8 x16 − + The ﬁrst three generating functions show that the sequences a(i)( 1,1), i 1,2,3, are n − = periodic with periods 12, 24, and 6, respectively. The sequences a(1)( 1,1) and a(2)( 1,1) n − n − are seen to satisfy the stronger conditions pn 6 pn and pn 12 pn for all n 0. From + =− + =− ≥ the appearance of the generating function f , it seems that the sequence a(4)( 1,1) would 4 n − not be periodic, which is indeed the case. It turns out that there are no other values of k for which the sequence a(k)( 1,1) is periodic. n − 8 TOUFIK MANSOUR AND MARK SHATTUCK

Proposition 2.6. The sequence a(k)( 1,1) is never periodic (or eventually periodic) when n − k 4. ≥ Proof. Substituting q 1 and t 1 into the inﬁnite part of (5) gives =− = j 1 j + jk F ( 1)( 2 )x k − j i k k 2k j 0 i 0(1 2Fk 1( 1) x ( 1) x ) X≥ = − − − + − F 2m( 1)m x2mk Q k − = k k 2k k 2k 2 2 2k m m 0 (1 2Fk 1x ( 1) x )((1 ( 1) x ) 4Fk 1x ) ≥ − − + − + − − − X 2m 1 m 1 (2m 1)k F + ( 1) + x + k − + k 2k 2 2 2k m 1 m 0 ((1 ( 1) x ) 4Fk 1x ) + ≥ + − − − X k k 2k 1 Fk 3x ( 1) x + − + − , = 2 2 k 2k 4k 1 (Fk 4Fk 1 2( 1) )x x + − − + − + and thus k 1 r k k 1 r r k k 2k (k) n ( r −0 Fr 1x x r −0 ( 1) Fk 1 r x )(1 Fk 3x ( 1) x ) a ( 1,1)x = + − = − − − + − + − . n − = 2 2 k 2k 4k n 0 P 1 (PFk 4Fk 1 2( 1) )x x X≥ + − − + − + Let a(x) and b(x) denote the numerator and the denominator in the (unsimpliﬁed) ex- (k) n pression above for n 0 an ( 1,1)x . Suppose now that ≥ − P a(x) d(x) c(x) , b(x) = + 1 xℓ − where ℓ is a positive integer, c(x) is any polynomial (possibly zero), and d(x) isoftheform m 1 d(x) x + e(x), with m denoting the degree of c(x) (we take m to be 1 if c(x) is the = − zero polynomial) and e(x) being a polynomial of degree at most ℓ 1. Then (1 xℓ)(a(x) − − − b(x)c(x)) b(x)d(x) implies that the equation b(x) 0 musthave at least one root of unity = d(x) = among its roots since e(x) m 1 is of degree at most ℓ 1, with e(x) not identically zero. x + = 1 − Then the equation b(u) 0, where u x 2k , mustalsohaveatleast onerootof unityamong = = its roots, since r a root of unity implies r 2k is as well. The equation b(u) 0 is given by = 2 2 k 2 (15) 1 (Fk 4Fk 1 2( 1) )u u 0. + − − + − + = If k 4, then ≥ 2 2 k Fk 4Fk 1 2( 1) 5, − − + − ≤− since 2 2 Fk 4Fk 1 (Fk 2Fk 1)(Fk 2Fk 1) Fk 3(Fk 2Fk 1) 7. − − = − − + − =− − + − ≤− Note that an equation of the form

1 au u2 0, a 5, − + = ≥ a pa2 4 has (real) roots − . So the only possible roots of unity that are also roots to such an 2 ± 2 a pa2 4 a pa2 4 equation are 1. However, the equations − 1 and − 1 have solutions ± 2 + 2 =± 2 − 2 =± a 2 in each case, but a 5. Thus no roots of unity satisfy equation (15) when k 4, =± ≥ ≥ which implies the result. GENERALIZATION OF A STATISTIC ON LINEAR DOMINO ARRANGEMENTS 9

Remark: When k 1,2,3, the equation (15) is satisﬁed by roots of unity and it works out = that the sequences a(k)( 1,1) are periodic in these cases. n − s 1 i Let (x : q)s i −0(1 q x). We conclude this section with the following inﬁnite expan- = = − sion for the numbers a(k)(q, t) for all k 1. Q n ≥ Theorem 2.7. If n km r, where m 1 and 0 r k 1, then = + ≥ ≤ ≤ − m sm m m (16) an t Gr 1 q d bs d cs = + s 0 + + − ≥ r 1 Xm r ¡ s(m 1)¢ m 1 m 1 ( 1) + t + Gk 1 r q − d − bs d − cs , + − − − s 0 + + − X≥ ¡ ¢ where j j 1 s 1 s ( +2 ) ( +2 ) s ( 1) Gk q + + d bs − + , = j j s i j s t (q : q)s (q : q)j s i 0(q d q d ) X≥ − = + − − Q j j 1 s 1 s ( +2 ) ( +2 ) s ( 1) Gk q + + d cs − − , = j j s i j s t (q : q)s (q : q)j s i 0(q d q d ) X≥ − = − − + and Q

d Gk 1 GkGk 2. ± = − ± − p 2 k 2 Proof. Note ﬁrst that d Gk 1 Gk 1 ( t) − , as in the proof of Theorem 2.2, and ± = − ± − − − thus q s k 2s 2 1 2tq xGk 1 ( t) q x (1 ρs x)(1 θs x), − − + − = − − s s where ρs d tq and θs d tq . = + = − Let n mk r , where m 1and0 r k 1. By partial fractions, let us write = + ≥ ≤ ≤ − j j 1 ( +2 ) j G q x bs cs k , j j 0 (1 2tqi xG ( t)k q2i x2) = s 0 1 ρs x + s 0 1 θs x i 0 k 1 ≥ − ≥ − X≥ = − − + − X X where bs and cs areQ constants to be determined. By Theorem 2.1,

j 1 j + j G q( 2 )x m k an Gr 1[x ] = +  j i k 2i 2  j 0 i 0(1 2tq xGk 1 ( t) q x ) X≥ = − − + − j 1  j + j Q G q( 2 )x r 1 m 1 k ( t) + Gk 1 r [x − ] + − − −  j i k 2i 2  j 0 (1 2tq xGk 1 ( t) q x ) X≥ i 0 − − + −  =  m bs Q cs Gr 1[x ] = + 1 ρ x + 1 θ x µs 0 s s 0 s ¶ X≥ − X≥ − r 1 m 1 bs cs ( t) + Gk 1 r [x − ] + − − − s 0 1 ρs x + s 0 1 θs x µ ≥ − ≥ − ¶ m m X r 1 X m 1 m 1 Gr 1 (bs ρs cs θs ) ( t) + Gk 1 r (bs ρs − cs θs − ). = + s 0 + + − − − s 0 + X≥ X≥ 10 TOUFIK MANSOUR AND MARK SHATTUCK

We also have j j 1 ( +2 ) Gk q bs = j s 1 j j j s ρ − (1 ρi /ρs ) (1 ρi /ρs ) (1 θi /ρs ) X≥ s i 0 − i s 1 − i 0 − = = + j 1 = s j + j 1 Q Q( 1) G q( 2 )ρ + Q − k s = j j s 1 i s j s i j s i j s d t i −0(q q ) i s 1(q q ) i 0(d tq d tq ) X≥ + = − − + − − j 1= +s 1 = s j ( + ) ( + ) Q ( 1) G qQ2 + 2 ρ Q − k s = j 1 j s i j s t + (q : q)s (q : q)j s (d q d q ) X≥ − i 0 + − − j 1 s=1 s j ( + ) ( + ) s ( 1) G q 2 +Q 2 + d − k + = j j s i j s t (q : q)s (q : q)j s (q d q d ) ≥ i 0 X − = + − − and, similarly, Q j j 1 s 1 s ( +2 ) ( +2 ) s ( 1) Gk q + + d cs − − , = j j s i j 0 t (q : q)s (q : q)j s (q d q d ) X≥ − i 0 − − + = which gives (16). Q

3. THE CASE k 2 = (2) In this section, we consider further results concerning the polynomial sequence an (2) = an (q, t). Taking k 2 in (10), and noting d d 1 in this case, gives the explicit formu- = + = − = las m m j 1 a m j a (2) ( +2 ) m j a2m q t − + − = j 0 a j Ã j ! Ã j ! X= X= q q m 1 m 1 j 1 a m j 1 a − ( + ) m 1 j − (17) t q 2 t − − + − − , m 0, − j 0 a j Ã j ! Ã j ! ≥ X= X= q q and m m j 1 a m j a (2) ( +2 ) m j (18) a2m 1 q t − + − , m 0. + = j 0 a j Ã j ! Ã j ! ≥ X= X= q q Though we are unable to give simpler expressions for the polynomials (17) and (18), they are seen to be solutions to the following relatively simple recurrences. Proposition 3.1. If m 2, then ≥ (2) m (2) 2 (2) (19) a2m (q qt t)a2m 2 qt a2m 4, = + + − − − with a(2) 1 and a(2) q t, and 0 = 2 = + (2) m (2) 2 (2) (20) a2m 1 (q 2t)a2m 1 t a2m 3, + = + − − − with a(2) 1 and a(2) q 2t. 1 = 3 = + Proof. We provide a combinatorial argument, the initial values being clear. To show (19), first note that if m 2, then the total weight of all the members of F2m ending in ss is m (2) ≥ (2) q a2m 2, while the weight of those ending in d is ta2m 2. To determine the weight of − − the members of F2m ending in ds, first insert a domino before the final square within any member of F2m 2 ending in s. By subtraction, the total weight of all the members of − GENERALIZATION OF A STATISTIC ON LINEAR DOMINO ARRANGEMENTS 11

F (2) (2) 2m 2 ending in s is a2m 2 ta2m 4, and the inserted domino increases both the ν and − − − − ρ2 values by 1 (note that the final square moves from position 2m 2to2m). Thus, the F (2) (2)− total weight of all members of 2m ending in ds is qt(a2m 2 ta2m 4), which gives (19). − − − By similar reasoning, the total weight of all members of F2m 1 ending in ss, d and ds is m (2) (2) (2) (2) + q a2m 1, ta2m 1 and t(a2m 1 ta2m 3), respectively, which gives (20). − − − − − We were unable to find, in general, two-term recurrences comparable to (19) and (20) (k) for the sequences amk r (q, t), where k and r are fixed and m 0. Let + ≥ (2) n f (x; q, t) an (q, t)x , = n 0 X≥ which we’ll also denote by f (x). Proposition 3.2. We have j 1 + 2j q( 2 )x (21) f (x; q, t) (1 x tx2) . = + − j i 2 2 j 0 i 0(1 tq x ) X≥ = − Proof. This follows from setting k 2 in (5) above,Q but we give an alternative derivation = (2) m m using Proposition 3.1 as follows. Let b(x) m 0 a2m x . Multiplying (19) by x , and = ≥ summing over m 2, implies ≥ P b(x) 1 (t q)x qx(b(qx) 1) tx(1 q)(b(x) 1) qt 2 x2b(x), − − + = − + + − − or 1 qx b(x) b(qx). = 1 tx + (1 tx)(1 qtx) − − − Iterating the last equation gives

j 1 + j q( 2 )x b(x) j = j 0 (1 tx) (1 tqi x)2 X≥ − i 1 − = j 1 + j Q q( 2 )x (1 tx) . = − j i 2 j 0 i 0(1 tq x) X≥ = − (2) m Similarly, if c(x) m 0 a2m 1 x , then we haveQ = ≥ + j 1 + j P q( 2 )x c(x) , = j i 2 j 0 i 0(1 tq x) X≥ = − and thus Q f (x) b(x2) xc(x2) = + j 1 j 1 + 2j + 2j q( 2 )x q( 2 )x (1 tx2) x , = − j i 2 2 + j i 2 2 j 0 i 0(1 tq x ) j 0 i 0(1 tq x ) X≥ = − X≥ = − as desired. Q Q Substituting q 1 in (21) yields the following result. =− Corollary 3.3. We have (1 x tx2)(1 x tx2)(1 x tx2) (22) a(2)( 1, t)xn + + + − − + . n 2 4 4 8 n 0 − = 1 (2t 1)x t x X≥ − − + 12 TOUFIK MANSOUR AND MARK SHATTUCK

Corollary 3.4. The sequence a(2)( 1,1) is determined by the condition n − f (n 12) f (n), n 0, + =− ≥ with the values of a(2)( 1,1) for 0 n 11 given by 1,1,0,1,1,2, 1,1,0,1, 1,0. n − ≤ ≤ − − Proof. Letting t 1 in (22), we have = (1 x x2)(1 x x2)(1 x x2) (2) n + + + − − + an ( 1,1)x 4 8 n 0 − = 1 x x X≥ − + (1 x x3 x5 x6)(1 x4)(1 x12) + + + − + − = (1 x4 x8)(1 x4)(1 x12) − + + − (1 x x3 x4 2x5 x6 x7 x9 x10)(1 x12) + + + + − + + − − , = 1 x24 − which implies the result.

Combinatorial proof of Corollary 3.4. F e F o F Let n and n denote the subsets of n having even and odd ρ2 values, respectively. F F F e F o We first define an involution of n off ofaset n′ which pairs members of n and n . Let F ′ F consist of those tilings of the form n ⊆ n (23) π d i (sd 2i1 s)(sd 2i2 s) (sd 2iℓ s), = ··· if n is even, and of the form (24) π d i (sd 2i1 s)(sd 2i2 s) (sd 2iℓ s)sd j , = ··· if n is odd, for some ℓ where i, j,i ,i ,...,i 0. We define an involution of F F ′ as 1 2 ℓ ≥ n − n follows. Given λ F F ′ , let j denote the smallest index j 1 such that either ∈ n − n o ≥ (i) an odd number of dominos occurs between the (2j 1)-st and (2j )-th squares, or − (ii) an even number of dominos occurs between the (2j 1)-st and (2j )-th squares − with at least one domino between the (2j )-th and (2j 1)-st squares (or between + the (2j )-th square and the end of the tiling, if the (2j )-th square is right-most).

Now exchange positions of the (2jo)-th square and the domino that precedes it if (i) occurs, or exchange the positions of the (2jo)-th square and the domino that directly follows F it if (ii) occurs. Let λ′ denote the resulting member of n′ . Then λ and λ′ have opposite ρ2-parity (since their ρ2 values differ by one), and the mapping λ λ′ is an involution F F 2 2 4 3 2 F7→ of n n′ . For example, if n 28 and λ d sd s d sdsd s 28, then jo 3 and −2 2 4 2 2 2 = = ∈ = λ′ d sd s d sd sd s. See Figure 3 below, where the (2j 1)-st and (2j )-th squares = o − o are shaded in each tiling.

1 2 3 4 5 6 7 8 9 10111213141516171819202122232425262728 λ

λ′

FIGURE 3. The tiling λ has ρ (λ) 35, while ρ (λ′) 34. 2 = 2 =

ρ2(π) We now consider the signed sum of members of F ′ , i.e., ( 1) . First observe n π Fn′ − that if i is even in (23) and (24) above, then one may verify that ∈ P ℓ 1 ρ2(π) + (mod 2), ≡ Ã 2 ! GENERALIZATION OF A STATISTIC ON LINEAR DOMINO ARRANGEMENTS 13 whereas if i is odd, then

ℓ ρ2(π) (mod 2). ≡ Ã2!

For the remainder of the proof, we will assume that n is even, the proof in the odd case being similar. Assume further that n 2m, where m is odd, as the argument for the case = of even m is basically the same. First suppose that π F ′ is of the form in (23) above, with i even. Note that m odd ∈ n implies ℓ is odd. Let π¯ be the tiling of length m given by

i i i i π¯ d 2 sd 1 sd 2 sd ℓ ; = ··· F F note that all members of m arise uniquely as π ranges over all members of n′ for which i is even. Let s(σ) denote the number of squares in a tiling σ. Then we have

ℓ 1 ℓ 1 s(π¯) 1 ρ2(π) + + + (mod 2). ≡ Ã 2 ! ≡ 2 = 2

If π F ′ is of theform in (23) with i odd, then m odd implies ℓ is even. Let π∗ be the tiling ∈ n of length m 1 given by − i 1 − i1 i2 i π∗ d 2 sd sd sd ℓ ; = ··· note that all members of Fm 1 arise uniquely in this manner. Observe that in this case −

ℓ ℓ s(π∗) ρ2(π) (mod 2). ≡ Ã2! ≡ 2 = 2

Therefore, we have

( 1)ρ2(π) ( 1)ρ2(π) ( 1)ρ2(π) F − = F − + F − π n′ π n′ π n′ X∈ i Xeven∈ i X∈odd (s(σ) 1)/2 s(σ)/2 (25) ( 1) + ( 1) . = − + − σ Fm σ Fm 1 ∈X ∈X − To evaluate the last two sums, we consider the statistic s(σ)/2 on F where r 1 and ⌈ ⌉ r ≥ pair members of F of opposite parity with respect to this statistic. Given σ σ σ r = 1 2 ···∈ F , let a denote the smallest index a 1 such that either r o ≥

(i) σ2a 1 d, or − = (ii) σ2a 1σ2a ss. − = F Deﬁne an involution of r by replacing σ2ao 1 d with ss if (i) occurs or by replacing − = σ2ao 1σ2ao ss with d if (ii) occurs. Note that this mapping changes the value of s(σ)/2 − = ⌈ ⌉ by one, whence it changes its parity. If r 0 (mod 3), then there is a single unpaired tiling r /3 ≡ r /6 in Fr , namely, (sd) , which has sign ( 1)⌈ ⌉. If r 1 (mod 3), then the single unpaired (r 1)/3 (r 2)/6 − ≡ tiling (sd) − s has sign ( 1)⌈ + ⌉. If r 2 (mod 3), then each member of F is paired − ≡ r with another of opposite parity, whence the resulting sum is zero. 14 TOUFIK MANSOUR AND MARK SHATTUCK

Applying the preceding to (25) shows that if m 0 (mod 3), i.e., if m 6p 3 for some p ≡ = + (since m was assumed odd) and n 12p 6, then = +

(2) ρ2(π) an ( 1,1) ( 1) − = π F − X∈ n′ s(σ)/2 s(σ)/2 ( 1)⌈ ⌉ ( 1)⌈ ⌉ = − + − σ F6p 3 σ F6p 2 ∈X + ∈X + (6p 3)/6 p 1 ( 1)⌈ + ⌉ 0 ( 1) + . = − + = − (2) p 1 p Similarly, if n 12p 2, then an ( 1,1) ( 1) + ( 1) 0, and if n 12p 10, then (2) = p+ 1 p 1 − = − + − = (2) = + a ( 1,1) 0 ( 1) + ( 1) + . This yields the values of a ( 1,1) given in Corollary n − = + − = − n − 3.4 above in the case when n 2m, where m is odd. The other cases are obtained similarly. = Remark: Comparable proofs may be given to explain the periodic nature of the a(1)( 1,1) and a(3)( 1,1) values witnessed above. n − n −

LetUn(t) denote the n-th Chebyshev polynomial of the second kind deﬁned byUn 1(t) + = 2tUn(t) Un 1(t), with U0(t) 1 and U1(t) 2t (see, e.g., [9]). − − = = n d Theorem 3.5. The coefﬁcient of x for n 0 in f (x; q, t) q 1 is given by ≥ dq | = n 1 n (ipt) + (2n 1)(4t 1)( 1) 2n(n 1) 4t 1 + + − + + − − Un (y) 8(4t 1) 2ipt + ³ n ((4t 1)( 1) 4t 1 2n(n 2))Un 1(y) , + + − + − − + − ´ where y 1 and i p 1. = 2ipt = − Proof. Differentiating the generating function f (x; q, t) in (21) with respect to q, and substituting q 1, yields = d x2(1 tx2)(1 tx2) g (x; t) : f (x, q) q 1 − + . = dq | = = (1 x tx2)3(1 x tx2)2 − − + − By partial fractions, we may rewrite this as

3 2tx 2 x 1 2tx g (x; t) − + + =−16(1 x tx2) + 8(1 x tx2)2 − 16(1 x tx2) 1 +tx− 1 +2tx− x − − − − − . + 4t(1 x tx2)2 − 4t(1 x tx2)3 − − − − n 1 By the fact that n 0 Un(t)x 1 2tx x2 , we obtain ≥ = − + P 2t 2x nU (t)xn 1 − n − 2 2 n 1 = (1 2tx x ) X≥ − + and 8t 2 2 12tx 6x2 n 2 − − + n(n 1)Un(t)x − 2 3 . n 2 − = (1 2tx x ) X≥ − + GENERALIZATION OF A STATISTIC ON LINEAR DOMINO ARRANGEMENTS 15

Let y 1 , where i p 1. Extracting the coefﬁcient of xn from each summand then = 2ipt = − gives n n 3 2tx ( ipt) [x ] − − 3Un(y) 2iptUn 1(y) , −16(1 x tx2) =− 16 − − µ + − ¶ ¡ n ¢ n 2 x (2 n)( ipt) [x ] + + − Un(y), 8(1 x tx2)2 = 8 µ ¶ + − n n 1 2tx (ipt) [x ] + Un(y) 2iptUn 1(y) , −16(1 x tx2) =− 16 − − µ ¶ − − ¡ n ¢ n 1 tx (1 4t (t 1)n)(ipt) [x ] − + + + Un(y) 4t(1 x tx2)2 = 4t(1 4t) µ ¶ − − + n 1 (1 n)(2t 1)(ipt) − + − Un 1(y), − 4(1 4t) − 2 + n n 1 2tx x (tn (t 2)n 2(1 4t))(ipt) [x ] − − − + − + Un(y) −4t(1 x tx2)3 = 8t(1 4t) µ − − ¶ + (tn2 (4t 1)n 1 3t)(ipt)n + − − + Un 1(y). + 4(1 4t) − + Adding all of these expressions yields the desired result.

Let tn (ρ2) denote the sum of the ρ2 values of all the members of Fn . Letting t 1 in the n = prior theorem, and noting i Un ( i/2) Fn 1, gives the following expression for tn(ρ2). − = + Corollary 3.6. If n 0, then ≥ 2 2 n (2n 1)Fn 1 2Fn (2n 2n 5)Fn 1 (4n 8n 6)Fn (26) tn (ρ2) ( 1) + + − + − + + + − . = − 16 + 80

REFERENCES

[1] G. E. Andrews. The Theory of Partitions. Encyclopedia of Mathematics and Applications, (No. 2), Addison-Wesley, 1976. [2] A. T. Benjamin and J. J. Quinn. Proofs that Really Count: The Art of Combinatorial Proof. Mathematical Association of America, 2003. [3] L. Carlitz. Fibonacci notes 3: q-Fibonacci numbers. Fibonacci Quart. 12 (1974), 317–322. [4] J. Cigler. Some algebraic aspects of Morse code sequences. Discrete Math. Theor. Comput. Sci. 6 (2003), 55–68. [5] J. Cigler. q-Fibonacci polynomials and the Rogers-Ramanujan identities. Ann. Comb. 8 (2004), 269–285. [6] M. Edson, S. Lewis, and O. Yayenie. The k-periodic Fibonacci sequence and an extended Binet’s formula. Integers 11 (2011), #A32. [7] T. Mansour and M. Shattuck. Generalizations of two statistics on linear tilings. Appl. Appl. Math. 07:2 (2012), 508–533. [8] J. Petronilho. Generalized Fibonacci sequences via orthogonal polynomials. Appl. Math. Comput. 218 (2012), 9819–9824. [9] T. Rivlin. Chebyshev Polynomials, From Approximation Theory to Algebra and Number Theory. John Wiley, New York, 1990. [10] M. Shattuck and C. Wagner. Parity theorems for statistics on domino arrangements. Electron. J. Combin. 12 (2005), #N10. [11] M. Shattuck and C. Wagner. A new statistic on linear and circular r -mino arrangements. Electron. J. Combin. 13 (2006), #R42. [12] N. J. Sloane. The On-Line Encyclopedia of Integer Sequences. Published electronically at http://oeis.org, 2010. [13] O. Yayenie. A note on generalized Fibonacci sequences. Appl. Math. Comput. 217 (2011), 5603–5611. 16 TOUFIK MANSOUR AND MARK SHATTUCK

DEPARTMENT OF MATHEMATICS, UNIVERSITY OF HAIFA, 31905 HAIFA,ISRAEL E-mail address: [email protected]

DEPARTMENT OF MATHEMATICS, UNIVERSITY OF TENNESSEE, KNOXVILLE, TN 37996 E-mail address: [email protected]