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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.

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  • $\begingroup$ To make sure, how do you define the convolution of finite sequences? By extending them (by zeroes) to functions on $\mathbb Z$? $\endgroup$ – Iosif Pinelis Mar 6 at 18:03
  • $\begingroup$ If so, your conjecture seems somewhat plausible, in view of the central limit theorem of probability theory. Perhaps, the Fourier transform can be used here. $\endgroup$ – Iosif Pinelis Mar 6 at 18:06
  • $\begingroup$ That's right, one can imagine extending the sequences by zeros to functions on $\mathbb{Z}$. I agree that the central limit theorem supports this conjecture, although I don't know of a way to recover log-concavity from the Fourier transform. $\endgroup$ – felipeh Mar 6 at 18:19
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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.

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