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3 added 494 characters in body

I found this paper of Temme (available here) that gives an explicit but somewhat complicated asymptotic for the Stirling number S(n,m) of the second kind, by the methods alluded to in previous answers (generating functions -> contour integral -> steepest descent)

Here's the asymptotic (as copied from that paper). One first sets

$t_0 := \frac{n-m}{m}$

and finds the positive real number $x_0$ solving the transcendental equation

$\frac{1-e^{-x_0}}{x_0} = \frac{m}{n}$

(one has the asymptotics $x_0 \approx 2(1-m/n)$ when $m/n$ is close to 1, and $x_0 \approx n/m$ when $m/n$ is close to zero.) One then defines

$A := \phi(x_0) - m t_0 + (n-m) t_0$

where

$\phi(x) := - n \ln x + m \ln(e^x - 1).$

(Note: $x_0$ is the stationary point of $\phi(x)$.) One has an integral representation

$S(n,m) = \frac{n!}{m!} \frac{1}{2\pi i} \int e^{\phi(x)} \frac{dx}{x}$

where the integral is a small contour around the origin. The saddle point method then gives

$S(n,m) = (1+o(1)) e^A m^{n-m} f(t_0) \binom{n}{m}$

where

$f(t_0) := \sqrt{\frac{t_0}{(1+t_0)(x_0-t_0)}}$

and o(1) goes to zero as $n \to \infty$ (uniformly in m, I believe).

In principle, one can now approximate $m! S(n,m)$ to within o(1) and compute its maximum in finite time, but this seems somewhat tedious. It does seem though that the maximum is attained when $m/n = c+o(1)$ for some explicit constant $0 < c < 1$.

EDIT: Actually, it's clear that the maximum is going to be obtained in the range $n/e \leq m \leq n$ asymptotically, because $m! S(n,m)$ equals $n! \approx (n/e)^n$ when $m=n$, and on the other hand we have the trivial upper bound $m! S(n,m) \leq m^n$. Among other things, this makes $x_0$ and $t_0$ bounded, and so the f(t_0) term is also bounded and not of major importance to the asymptotics. The other terms however are still exponential in n...

EDIT: There is also the identity

$\sum_{k=1}^n (k-1)! S(n,k) = (-1)^n Li_{1-n}(2)$

where $Li_s$ is the polylogarithm function. So, up to a factor of n, the question is the same as that of obtaining an asymptotic for $Li_{1-n}(2)$ as $n \to -\infty$. This seems quite doable (presumably from yet another contour integration and steepest descent method) but a quick search of the extant asymptotics didn't give this immediately.

2 added 455 characters in body

I found this paper of Temme (available here) that gives an explicit but somewhat complicated asymptotic for the Stirling number S(n,m) of the second kind, by the methods alluded to in previous answers (generating functions -> contour integral -> steepest descent)

Here's the asymptotic (as copied from that paper). One first sets

$t_0 := \frac{n-m}{m}$

and finds the positive real number $x_0$ solving the transcendental equation

$\frac{1-e^{-x_0}}{x_0} = \frac{m}{n}$

(one has the asymptotics $x_0 \approx 2(1-m/n)$ when $m/n$ is close to 1, and $x_0 \approx n/m$ when $m/n$ is close to zero.) One then defines

$A := \phi(x_0) - m t_0 + (n-m) t_0$

where

$\phi(x) := - n \ln x + m \ln(e^x - 1).$

(Note: $x_0$ is the stationary point of $\phi(x)$.) One has an integral representation

$S(n,m) = \frac{n!}{m!} \frac{1}{2\pi i} \int e^{\phi(x)} \frac{dx}{x}$

where the integral is a small contour around the origin. The saddle point method then gives

$S(n,m) = (1+o(1)) e^A m^{n-m} f(t_0) \binom{n}{m}$

where

$f(t_0) := \sqrt{\frac{t_0}{(1+t_0)(x_0-t_0)}}$

and o(1) goes to zero as $n \to \infty$ (uniformly in m, I believe).

In principle, one can now approximate $m! S(n,m)$ to within o(1) and compute its maximum in finite time, but this seems somewhat tedious. It does seem though that the maximum is attained when $m/n = c+o(1)$ for some explicit constant $0 < c < 1$.

EDIT: Actually, it's clear that the maximum is going to be obtained in the range $n/e \leq m \leq n$ asymptotically, because $m! S(n,m)$ equals $n! \approx (n/e)^n$ when $m=n$, and on the other hand we have the trivial upper bound $m! S(n,m) \leq m^n$. Among other things, this makes $x_0$ and $t_0$ bounded, and so the f(t_0) term is also bounded and not of major importance to the asymptotics. The other terms however are still exponential in n...

1

I found this paper of Temme (available here) that gives an explicit but somewhat complicated asymptotic for the Stirling number S(n,m) of the second kind, by the methods alluded to in previous answers (generating functions -> contour integral -> steepest descent)

Here's the asymptotic (as copied from that paper). One first sets

$t_0 := \frac{n-m}{m}$

and finds the positive real number $x_0$ solving the transcendental equation

$\frac{1-e^{-x_0}}{x_0} = \frac{m}{n}$

(one has the asymptotics $x_0 \approx 2(1-m/n)$ when $m/n$ is close to 1, and $x_0 \approx n/m$ when $m/n$ is close to zero.) One then defines

$A := \phi(x_0) - m t_0 + (n-m) t_0$

where

$\phi(x) := - n \ln x + m \ln(e^x - 1).$

(Note: $x_0$ is the stationary point of $\phi(x)$.) One has an integral representation

$S(n,m) = \frac{n!}{m!} \frac{1}{2\pi i} \int e^{\phi(x)} \frac{dx}{x}$

where the integral is a small contour around the origin. The saddle point method then gives

$S(n,m) = (1+o(1)) e^A m^{n-m} f(t_0) \binom{n}{m}$

where

$f(t_0) := \sqrt{\frac{t_0}{(1+t_0)(x_0-t_0)}}$

and o(1) goes to zero as $n \to \infty$ (uniformly in m, I believe).

In principle, one can now approximate $m! S(n,m)$ to within o(1) and compute its maximum in finite time, but this seems somewhat tedious. It does seem though that the maximum is attained when $m/n = c+o(1)$ for some explicit constant $0 < c < 1$.