Lattice path combinatorics Continued fraction representation by polynomials Chebyshev polynomials of the second kind a b s t r a c tWe determine the distribution of duration in the gambler's ruin problem given that one specific player wins. In this version we allow ties in the single games. We present a unified approach which uses generating functions to prove and extend some results that were obtained in [Frederick Stern, Conditional expectation of the duration in the classical ruin problem, Math.

An interesting 2-adic property of the Stirling numbers of the second kind S.n; k/ was conjectured by the author in 1994 and proved by De Wannemacker in 2005: 2 .S.2 n ; k// D d 2 .k/ 1, 1 Ä k Ä 2 n . It was later generalized to 2 .S.c2 n ; k// D d 2 .k/ 1, 1 Ä k Ä 2 n , c 1 by the author in 2009. Here we provide full and two partial alternative proofs of the generalized version. The proofs are based on non-standard recurrence relations for S.n; k/ in the second parameter and congruential identities.

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International audience Let $n$ and $k$ be positive integers, $d(k)$ and $\nu_2(k)$ denote the number of ones in the binary representation of $k$ and the highest power of two dividing $k$, respectively. De Wannemacker recently proved for the Stirling numbers of the second kind that $\nu_2(S(2^n,k))=d(k)-1, 1\leq k \leq 2^n$. Here we prove that $\nu_2(S(c2^n,k))=d(k)-1, 1\leq k \leq 2^n$, for any positive integer $c$. We improve and extend this statement in some special cases. For the difference, we obtain lower bounds on $\nu_2(S(c2^{n+1}+u,k)-S(c2^n+u,k))$ for any nonnegative integer $u$, make a conjecture on the exact order and, for $u=0$, prove part of it when $k \leq 6$, or $k \geq 5$ and $d(k) \leq 2$. The proofs rely on congruential identities for power series and polynomials related to the Stirling numbers and Bell polynomials, and some divisibility properties.