Combinatorial Proofs of Various q-Pell Identities via Tilings

Abstract

Recently, Benjamin, Plott, and Sellers proved a variety of identities involving sums of Pell numbers combinatorially by interpreting both sides of a given identity as enumerators of certain sets of tilings using white squares, black squares, and gray dominoes. In this article, we state and prove q-analogues of several Pell identities via weighted tilings.

Tiling Proofs of Some Formulas for the Pell Numbers of Odd Index

“…Description of the set C n : Let C n denote the set of tilings of [n] in which each square is marked with a member of [6]. Then half the right-hand side gives the signed sum over all the members of C n according to the number of dominos k.…”

“…Cardinality of the set C n : To complete the proof, we show a n+1 = 2|C n |, n 0. Given a square painted black or white and c = c 1 c 2 · · · c n ∈ C n , where c i denotes the member of [6] assigned to the i th square, we'll construct a member of P 2n+1 , denoted c , in n steps according to the iterative procedure below. …”

“…Description of the set C 2n : Let C k,j consist of the (2n−2j)-tilings containing exactly k − j dominos in which each square is marked with a member of [6]. One fourth the right-hand side then gives the signed sum over all members of…”

“…Members of Y n ending in at least two dominos are clearly synonymous with members of Y n−2 , and are counted twice in the preceding, hence the second term on the right side of (6). This accounts for all members of Y n , except for those whose first black or white square (which must be white) covers cell 2n − 2, and there are exactly two such tilings (a green square, followed by d n−2 w 2 or d n−2 wb).…”

“…For example, p 0 = 1 counts the empty tiling and p 2 = 5 since a board of length 2 may be covered (exactly) by either a domino or by two squares, each painted black or white. Benjamin, Plott, and Sellers [5] have provided combinatorial proofs of several recent Pell number identities which were q-generalized by Briggs, Little, and Sellers [6]. Let a n := p 2n−1 , n 1, denote the n th Pell number of odd index with a 0 := 0.…”