For a given integer $N$, we want to maximize $(\frac{N}{k})^k$. The solution in real numbers is obtained when $k = \frac{N}{e}$. The solution in integers is then either $\lfloor\frac{N}{e}\rfloor$ or $\lceil\frac{N}{e}\rceil$. However, it seems that the correct solution is always rounding to the nearest integer.
Is this assumption correct and if so, what's the proof?
What about writing the taylor development to the 3rd derivative ? That's a good place to start in the maximum.
Let's imagine that N/e is closer to the upper integer. You have n < n+1/2 < N/e < n+1.
f' = 0
f'' = something < 0 but it acts symetrically, and so f(n) will be pushed farther from the maximum than f(n+1) which is closer to it N/E
f''' > 0 so this will push f(n) negatively and f(n+1) positively
So in this case it seems true that if N/e is closer to n+1 than to n, then f(n+1) is bigger than f(n)
Second case:
f'' will push f(n) negatively less than f(n+1) but f''' will push f(n) negatively and f(n+1) positively so you are trying to find a place where this effect is bigger than the previous one to find a counter example if any. I haven't done the calculation here but you can come back with questions if any
Note : 1/e is irrational so you can find N such that N/e is arbitrarily close to m+0.5 Note 2: I say "seem" in the first case at the end because a Taylor development is cool but we have to analyze the rest too and minor it.
This is not a full proof but it gives you a good step forward
