Let $p(n)$ denote the number of partitions of a positive integer $n$. It seems to me that we have for all $n>25$
$$
p(n)^2>p(n-1)p(n+1).
$$
In other words, the sequence $(p(n))\_{n\in \mathbb{N}}$ is log-concave, or satisfies $PF\_2$, with
$$
\det \begin{pmatrix} p(n) & p(n+1) \cr p(n-1) & p(n) \end{pmatrix}>0
$$
for $n>25$. Is this true ? I could not find a reference in the literature so far. On the other hand, the partition function is really studied a lot.
So it seems likely that this is known.   

Similarly, property $PF\_3$, with the corresponding $3\times 3$ determinant, seems to hold for all $n>221$, too, and also
$PF\_4$ for all $n>657$.   

The question is also motivated from the study of Betti numbers for nilpotent Lie algebras, in particular filiform nilpotent Lie algebras.



Is the sequence of partition numbers log-concave?
The first two terms of the Hardy-Ramanujan formula give
$$p(n) = \frac{1}{4 \sqrt{3} n} \exp(\pi \sqrt{2n/3}) + O \left(\exp(\pi \sqrt{n/6} ) \right)$$
so
$$\log p(n) = \pi \sqrt{2/3} \sqrt{n} - \log n - \log (4 \sqrt{3}) + O(\exp(-\pi \sqrt{n/6} ) ).$$
So
$$\log p(n+2) - 2 \log p(n+1) + \log p(n) = $$
$$ \pi \sqrt{2/3} \left( \sqrt{n+2} - 2\sqrt{n+1} + \sqrt{n} \right) - \left( \log(n+2) - 2 \log(n+1) + \log n \right) + O(\exp(-\pi \sqrt{n/6} ) )$$
$$= \left[ \left( \frac{- \pi \sqrt{2/3}}{4} \right) n^{-3/2} + O(n^{-5/2}) \right] + O(n^{-2}) + O(\exp(-\pi \sqrt{n/6} ) ).$$
So this quantity is negative for $n$ sufficiently large.


The larger determinants seem harder; there is probably a smarter way to do this.




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With the help of Mathematica, I set $q(n) = a \exp(c \sqrt{n})/n$ and computed that
$$\det \begin{pmatrix} q(n) & q(n+1) & q(n+2) \\ q(n-1) & q(n) & q(n+1) \\ q(n-2) & q(n-1) & q(n) \end{pmatrix} = q(n)^3 \left( \frac{c^3}{32 n^{9/2}} + O(n^{-10/2}) \right).$$
The error in approximating $p(n)$ by $q(n)$ (for $a = 1/(4 \sqrt{3})$ and $c = \pi \sqrt{2/3}$) will be exponentially smaller than $n^{-9/2}$, so the $3 \times 3$ determinant is positive for $n$ large. 


The $4 \times 4$ determinant vanishes to order at least $n^{-12/2}$, and I gave up waiting for the computation to finish when I asked for more terms.


