Maximizing Signless Stirling Numbers Let $c(n, k)$ denote the number of permutations in $S_n$ whose cycle decomposition has $k$ cycles. For a fixed $n$, I want to find $k$ such that $c(n, k)$ is maximized.
I know that the $k$ I seek is either the floor or ceiling of $1+\frac{1}{2}+\cdots+\frac{1}{n}$, but I'm not quite sure how to do this. This maybe suggests using an exponential generating function, but I don't know what the EGF of the sequence $c(n, k)$, for a fixed $n$ is. Could I have some hints? 
 A: I’m not sure how useful this is for you, but in this paper [PDF] Erdős showed that there is a $k_n$ maximizing $c(n,k)$ and that it is
$$k_n=\left\lfloor\ln(n+1)+\gamma-1+\frac{\zeta(2)-\zeta(3)}{\ln(n+1)+\gamma-\frac32}+\frac{h}{\left(\ln(n+1)+\gamma-\frac32\right)^2}\right\rfloor\;,$$
where $\gamma$ is the Euler-Mascheroni constant, $\zeta$ is the Riemann zeta function, and $-1.1<h<1.5$. He also noted that for $n>188$ this can be simplified to
$$\left\lfloor\ln n-\frac12\right\rfloor\le k_n\le\lfloor\ln n\rfloor\;.$$
A: I have  not looked  at the Erdös  paper yet  but I would  like to
thank  Brian  M. Scott  for  the quick  reply  giving  such a  precise
reference.

What follows are some informal  observations where the reader is asked
to be patient with  the lack of rigor. Some time ago  I used the Polya
Enumeration  Theorem to  prove the  following asymptotic  for Stirling
numbers of the first kind
(this is the MSE link)
$$\left[ n\atop k\right]
\sim \frac{(n-1)!}{(k-1)!} \log^{k-1} (n-1).$$
We ask for what $k$ this is maximized. Consider the function
$$\frac{Q^x}{\Gamma(x+1)}$$
with $Q>1$  a constant .  We  have by inspection that  the growth from
the  exponential   term  dominates  until  the   Gamma  function  term
takes over, for an ultimate limit of zero.

To locate the point where this happens differentiate to get
$$\log Q \times \frac{Q^x}{\Gamma(x+1)}
- \frac{Q^x}{\Gamma(x+1)^2} \Gamma'(x+1) = 0.$$
This gives
$$\log Q - \frac{\Gamma'(x+1)}{\Gamma(x+1)} = 0$$
or in terms of the digamma function
$$\psi(x+1) = \log Q.$$
But on the real line we have $\psi(x) \sim \log x$ so that the
conclusion is that
$$x\sim Q.$$

Returning to the Stirling numbers we see that here
$Q = \log(n-1)$
giving the approximation
$$k\sim\log n$$
for the $k$ that maximizes $\left[n\atop k\right]$
with $n$ fixed.
Addendum. Where generating functions are concerned we have the species
$$\mathfrak{P}(\mathcal{U}\mathfrak{C}(\mathcal{Z}))$$
which gives the bivariate generating function
$$\exp\left(u\log\frac{1}{1-z}\right)$$
so that with $n$ fixed
$$\left[n\atop k\right]
= n! [z^n] \frac{1}{k!} \left(\log\frac{1}{1-z}\right)^k.$$
The mechanics of extracting coefficient asymptotics from this are
discussed in the text Analytic Combinatorics by Flajolet and Sedgewick
and in the slides from that text which refer to the so-called
standard function scale.
Remark. The fomulae from the PET computation are exact, 
even though we have used only the first term here.
