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.



Log-concavity of repeated convolution
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. 


