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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 $5$?