Algebraic and Combinatorial Properties of Simple, Coloured Walks

Proof. The rst identity derives from Theorem 2.1, with f(t) = (t 2 + t + 1) ?1. The second identity derives from Theorem 3.6, for which we have: p 1;2] n;k = (?1) n?k n + k ? 1 2k + 1 : The other identities are proved in an analogous way, if we consider that the generating function of the sequence g k is g(t) = t=(t 2 + t + 1) 2 : References 1. Proof. The algebraic proof can be obtained by applying Theorem 2.1, if we consider that the generating function of the sequence fzk + 1g k2N is f(t) = (1+(z ?1)t)=((1?t) 2): As far as the combinatorial proof of Shapiro's identity is concerned, let us consider a line parallel to the main diagonal and at distance 1 from it. We deene the weight of a walk as the distance of its ending point from ; clearly, all the walks counted by s a;b;b 0 ] n;k have a weight of zk + 1. The total weight of all of Shapiro's symmetric underdiagonal walks is therefore: W n = n X k=0 s a;b;b 0 ] n;k (zk + 1); and the theorem identity is proved if we prove that W n = (b + 2a) n. Let us proceed by induction on n. The initial cases W 0 = 1 and W 1 = b + 2a are trivial and can be checked by inspection. Now, let us assume W n = (b + 2a) n and take a walk not ending on the main diagonal. Let w be its weight; the walk generates the following walks of length n + 1: a walks of weight (w + z), b walks of weight w and a walks of weight (w ? z). Clearly, the total balance for these walks is (b+2a)w. When we have a walk ending on the main diagonal, its weight is w = 1 and it can only produce b 0 walks of weight w and a walks of weight (w + z). The total balance is again b 0 w + a(w + z) = (b + 2a)w because b 0 = b ? a(z ? 1). Hence, every walk of length n and weight w produces some walks of length n+1 and a total weight of (b + 2a)w. Since all the walks are obtained in this way, …