Maximum of $\sum_{k=0}^{P} (-1)^{k+P}\binom{P}{k} k^N $ with $P\leq N$ For a combinatorics problem, I would need to identify the value of $P$ that maximizes the function
$$\Omega(P) = \sum_{k=0}^{P} (-1)^{k+P}\binom{P}{k} k^N$$
with $P\leq N$. Numerically, I see that the maximum is roughly located at $P=3N/4$, but I don't know how to prove it.
I am wondering whether an extrapolation of $\Omega$ to real values of $P$ could be found, which would simplify the search for a maximum. Any idea?
Thanks a lot!
 A: I'm going to use lower case letters instead of upper. The sum can be expressed as  $p! \, S(n,p),$ where $S(n,p)$ is the Stirling number of the second kind.  We can use asymptotic expansions of Stirling numbers to solve your problem for $n \to \infty.$  I want to reproduce analytically your observed zeroth order approximation, $p^*=\frac{3}{4}n.$  To find a maximum, take a derivative and set to zero, and an approximation to a derivative in $p$ is the difference
$$\Delta_p \;  p!S(n,p) = (p-1)!\big( p S(n,p) - S(n,p-1) \big) = 0$$
That is, to find $p^*$ we want $p$ such that
$$p=\frac{S(n,p-1)}{S(n,p)} = \frac{S(n+1,p)}{S(n,p)} - p $$
where the recursion relation has been used.  Now most approximations to the Stirling numbers of second kind depend on solving a transcendental equation (perhaps the Lambert W is involved.)  However, Guy Louchard in 'Asymptotics of the Stirling numbers of the second kind revisited' records a formula (among many) that is good enough in the region of interest, and is due to Moser and Wyman.  It can be written as
$$ S(n,p) \sim \binom{n}{p}\Big(\frac{2}{p}\Big)^{p-n} .$$
Insert, simplify, and we have the simple equation
$$p=\frac{1}{2} \frac{(n+1)p}{2(1+n-p)} $$ which has two solutions, one of which is $p^*=\frac{3}{4}(n+1).$
I've made plots of the approximation using the binomial and simple factor vs. the ratio of Stirling numbers, and the approximate curve results in a root that is a bit too large, but not terribly so. To go beyond the zeroth order would seem to be a tedious analytic task, and it depends on whether the Lambert W is in your toolbox.
Added 11/29/2021
With an expansion found in arXiv 1506.00699v1 that implies a more accurate $S(n,p)$ in the vicinity where the line $p$ intersects it, a more complicated equation for p can be derived:
$$p=\frac{\sqrt{(n+1)n}}{2\log{2}}\exp\Big(\frac{n(n+1)-4p^2\log^2{2}}{n(n+1)+2(\log{2} -1)} \Big)$$
An exact solution can be found in terms of the Lambda W.  Instead, take the solution and expand it asymptotically as $n\to\infty.$ Using some facts about the Lambert W and some simplification, then it can be shown
$$ p^* \sim \frac{1}{2\log{2}}\big(n+\frac{1}{2}-\frac{1}{8n} +...\big) $$
This solution shows the true zeroth order factor is $ 1/(2\log{2}) \approx 0.72135$, which is about 4% less than the empirically derived factor of 0.75 by the OP.
A: Hint: Consider the differences $\Omega(P+1)-\Omega(P)$, and use Pascal's identity.
