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Nov 14, 2013 at 18:31 comment added David E Speyer @StephenDeSalvo Thanks! You are, of course, correct.
Nov 14, 2013 at 16:30 comment added Stephen DeSalvo And very important! In the comment immediately above, "This is easily fixed by ... " replace the ... with "replacing all $n$'s with $n-1/24$ and multiplying the first term in the expansion by $(1-1/(c \sqrt{n-1/24}))$." This $n-1/24$ in place of $n$ is essential otherwise the relative error is still $O(n^{-1/2})$.
Nov 2, 2013 at 3:09 comment added Stephen DeSalvo The asymptotic expansion for $p(n)$ in the first line is wrong. The error term is actually $O(\exp(\pi \sqrt{2n/3})/\sqrt{n})$. This fact completely negates even this asymptotic argument, since the Big-Oh error term is larger than the $n^{-3/2}$, which is the rate of decay of log-concavity. This is easily fixed by multiplying the first term in the expansion by $(1-1/c\sqrt{n}))$, where $c = \pi \sqrt{2/3}$. This does not affect the final answer (in this case!), but without it the argument as written is invalid. See for example Section 6.2 of arxiv.org/pdf/1310.7982v1.pdf
Aug 2, 2013 at 21:34 comment added Dietrich Burde @Stephen DeSalvo: I have accepted the answer, although it does not completely answer the question. I do not think that accepting is only possible if everything is answered - often this is not possible. The reference of Janoski's thesis is almost an answer. Unfortunately I am not yet convinced.
Aug 2, 2013 at 18:46 comment added Stephen DeSalvo Since this answer was accepted, I feel obligated to point out that this does not answer the original question, which was about all $n > 25$. This answer addresses log-concavity of the partition function for all $n>n_0$ for some unspecified $n_0$. If this was the intended question, I suggest editing the original question to reflect this, otherwise a casual reader might confuse the asymptotic analysis as a rigorous proof for all $n>25$.
Aug 2, 2013 at 15:56 history edited David E Speyer CC BY-SA 3.0
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Aug 2, 2013 at 15:47 history edited David E Speyer CC BY-SA 3.0
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Aug 2, 2013 at 8:39 vote accept Dietrich Burde
Aug 1, 2013 at 23:28 comment added Igor Rivin @DavidSpeyer so is this Polya frequency related to monotonicity as in the moment problem?
Aug 1, 2013 at 22:54 comment added David E Speyer @IgorRivin In answer to your question, in principle, we should be able to determine the sign of $PF_k$ for $k$ fixed and $n$ large in the same way. Discard all the terms except for $\exp(\pi \sqrt{2(n+i-j)/3})/(4 \sqrt{3} (n+i-j))$. Divide all entries in the matrix by $\exp(\pi \sqrt{2n/3})/(4 \sqrt{3} n)$. The resulting matrix entries have asymptotic expansions in powers of $1/\sqrt{n}$; keep expanding to see whether the leading term is positive or negative. But, when I started doing this, I got a lot of cancellation, and I don't want to work that hard.
Aug 1, 2013 at 22:47 comment added David E Speyer @IgorRivin $PF$ stands for "Polya frequency". A sequence $a_n$, $n \geq 0$ which obeys $\det (a_{n+i-j})_{1 \leq i,j \leq r} \geq 0$ for all $r$ and $n$ such that the determinant makes sense is called a Polya frequency sequence. There are a lot of beautiful theorems about analytic functions whose coefficients form a Polya frequency sequence. I don't know any results about functions whose coefficients obey this condition for $n$ large, but I'm not an expert.
Aug 1, 2013 at 19:44 comment added Igor Rivin Presumably the higher determinants reduce (asymptotically) to the same question for the main term of the asymptotic expansion. How would one go to check that a function satisfies $PF_k$ (in the OP's notation)?
Aug 1, 2013 at 18:15 comment added Suvrit +1: so we have yet another candidate for the phenomenon of eventual counterexamples :-)
Aug 1, 2013 at 15:59 history answered David E Speyer CC BY-SA 3.0