# Log concavity of the power partition function

Recently I proved that the power partition function is log concave. The power partition function (usually denoted $P_k(n)$) counts the number of ways to partition an integer into perfect kth powers. Log concavity means for a sequence $a(n)$, we have that $a(n)^2>a(n-1)*a(n+1)$.

I know that for every k there is *some* $N_k$ such that for all $n>N_k$ the power partition function is log concave. What I want is to know the smallest such $N_k$.

Right now, I only know the smallest $N_k$ for the first three powers:

$k=1 \Rightarrow N_1=25$

$k=2 \Rightarrow N_1=1024$

$k=3 \Rightarrow N_1=15656$

I brute forced this in mathematica, but I cannot go any further. When looking, I moved all the terms of the inequality to the left and waited for the graph to always be positive.

$\frac{P_k(n)}{P_k(n-1)P_k(n+1)}>0$

The function toggles between positive and negative values until it is eventually always positive. Capturing the last negative term in the sequence gives the smallest $N_k$. I waited for about 50 terms in a row to be positive and then looked backwards and the last negative term. But my method takes too long now.

Can anyone think of a faster way to return the smallest $N_k$ for $k=4$ or $5$?

This is a theory question irrelevant to Sage. https://math.stackexchange.com/ would be better place for such question.

I agree that the question is not Sage specific. However it is not irrelevant to Sage as it is something that could be programmed.

Note that your second equation makes no sense: all of

`P_k(n)`

,`P_k(n-1)`

and`P_k(n+1)`

are positive numbers so that the ratio`P_k(n) / (P_k(n-1) P_k(n+1))`

isalwayspositive.Also for

`k=2`

I found`N_k = 1042`

rather than the`1024`

that you wrote. Possibly you inverted the last two digits?