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 $5$?
