TITLE: Log-concavity of repeated convolution
QUESTION [9 upvotes]: Let $A = (a_0,a_1,\ldots,a_k)$ be a sequence of strictly positive numbers, and let $A^{\ast k}$ denote the $k$-fold repeated convolution (defined by $A^{\ast 1} = A$ and $A^{\ast k+1} = A^{\ast k} \ast A$).
Is it true that for any such sequence $A$, there exists $n$ such that $A^{\ast n}$ is log-concave?
As an example, any sequence of the form $A = (1,x,1)$ is log concave if $x\geq 1$, and $A\ast A$ is log concave if $x\geq \sqrt{2/3}$.  In general, I have checked numerically that $A^{\ast k}$ is log concave if $x\geq \sqrt{2/(k+1)}$.  I have also checked this statement numerically for many other sequences of longer length, but I have no idea how to go about proving anything.

REPLY [5 votes]: Odlyzko and Richmond proved your conjecture:
http://www.dtc.umn.edu/~odlyzko/doc/arch/unimodal.convolut.pdf
Odlyzko, A. M.(1-BELL); Richmond, L. B.(3-WTRL)
On the unimodality of high convolutions of discrete distributions. 
Ann. Probab. 13 (1985), no. 1, 299–306.