Timeline for Number of walks on integer lattice with self-edge at zero

9 events

when toggle format	what		by	license	comment
Jul 2, 2016 at 1:49	vote	accept	Raisin Bread
Jun 27, 2016 at 13:25	comment	added	Johann Cigler		@MaxAlekseyev: Thank you. This is a nice proof.
Jun 26, 2016 at 14:08	comment	added	Max Alekseyev		@JohannCigler: First, the g.f. given in A098615. Second, directly this can be proved as follows. The sum equals the coefficient of $x^{n+1}$ in $(1-x^2)(1+x^2)^n\frac{x}{1-x-x^2}$. From here, one can get its g.f. using Lagrange inversion -- e.g., see mathoverflow.net/q/32188
Jun 25, 2016 at 13:14	comment	added	Johann Cigler		This is a very nice argument. But I do not see how one gets $\sum_{k=0}^{\lfloor(n+1)/2\rfloor}\left({n\choose k}-{n\choose k-1}\right)F(n-2k+1)$ from the generating function $1/(1-x-2x^2 C(x^2))$.
Jun 24, 2016 at 16:14	history	edited	Max Alekseyev	CC BY-SA 3.0	added 169 characters in body
Jun 24, 2016 at 15:57	comment	added	Max Alekseyev		@FedorPetrov: Indeed, this is a nice argument that simplifies computation.
Jun 24, 2016 at 15:53	history	edited	Max Alekseyev	CC BY-SA 3.0	added 483 characters in body
Jun 24, 2016 at 15:46	comment	added	Fedor Petrov		Paths of first type are generalized Catalan numbers: the number of them equals $0$ if $ab\leq 0$ and equals (number of all usual paths from $a$ to $b$)-(number of all usual paths from $a$ to $-b$) otherwise.
Jun 24, 2016 at 15:32	history	answered	Max Alekseyev	CC BY-SA 3.0