Timeline for Does the Apéry-like sequence $A_n=(n!)^2\sum_{k=0}^n { \rho \choose k}^2 { \rho+n-k \choose n-k}, \rho=e^{2 \pi i/3}$ change signs infinitely often?

6 events

when toggle format	what		by	license	comment
Dec 6 at 8:09	comment	added	CHUAKS		Should be "independent of $a$" in the last comment.
Nov 23 at 16:30	comment	added	CHUAKS		If one defines the initial condition $p_0=1,p_1=x-a$, the recurrence $p_{n+1}=(x-n-a)p_n-nap_{n-1}$ indeed defines the Charlier polynomial $p_n=(-1)^n {}_2F_0(-n,-x;;-1/a)$. We then have $\alpha_k=\infty, \beta_k=(k-1)$ independent. If $a<0$ roots are no longer all real but limit still holds for the real roots. The unexpected observation is when $a$ is any complex number, the kth smallest real part of the roots still converge to $k-1$. Maybe this is related to universality of eigenvalues.
Nov 12 at 15:14	comment	added	Hjalmar Rosengren		Yes, your other example seems to be very different. I am not sure exactly what those polynomials are. The recursion looks a little bit like for Charlier polynomials, which is $$xp_n=p_{n+1}+(n+a)p_n+nap_{n-1}.$$ Charlier polynomials have their orthogonality measure supported on $\mathbb N$.
Nov 12 at 14:10	comment	added	CHUAKS		Yes thanks for the detail reply. I am convinced of the $(-1)^nS_n(3/4-x,1/2,1/2,1/2)$ now. But this cannot be applied to the distinct roots $a_n=an,b_n=bn$ case. Since $\alpha_{k-1}<\alpha_k$, an $\epsilon$ left neighbour contains only the $r_{n,k}$ so cannot be dense. The coefficients grows linearly where as those of $S_n$ are quadratic.
Nov 12 at 9:24	history	edited	Hjalmar Rosengren	CC BY-SA 4.0	added 7 characters in body
Nov 12 at 8:31	history	answered	Hjalmar Rosengren	CC BY-SA 4.0