What is the asymptote for the positions of the largest Stirling numbers of the second kind?

Question

The infinite lower triangular array of Stirling numbers of the second kind starts:

$$\begin{array}{llllllll} 1 & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} \\ 1 & 1 & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} \\ 1 & 3 & 1 & \text{} & \text{} & \text{} & \text{} & \text{} \\ 1 & 7 & 6 & 1 & \text{} & \text{} & \text{} & \text{} \\ 1 & 15 & 25 & 10 & 1 & \text{} & \text{} & \text{} \\ 1 & 31 & 90 & 65 & 15 & 1 & \text{} & \text{} \\ 1 & 63 & 301 & 350 & 140 & 21 & 1 & \text{} \\ 1 & 127 & 966 & 1701 & 1050 & 266 & 28 & 1 \end{array}$$

https://oeis.org/A008277

Leaving only the largest elements of the table, we have:

$$\begin{array}{llllllll} 1 & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} \\ 1 & 0 & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} \\ 0 & 3 & 0 & \text{} & \text{} & \text{} & \text{} & \text{} \\ 0 & 7 & 0 & 0 & \text{} & \text{} & \text{} & \text{} \\ 0 & 0 & 25 & 0 & 0 & \text{} & \text{} & \text{} \\ 0 & 0 & 90 & 0 & 0 & 0 & \text{} & \text{} \\ 0 & 0 & 0 & 350 & 0 & 0 & 0 & \text{} \\ 0 & 0 & 0 & 1701 & 0 & 0 & 0 & 0 \end{array}$$

The column index of those elements is a sequence starting: $$a(n) = 1, 1, 2, 2, 3, 3, 4, 4, 4, 5, 5, 5, 6, 6, 6, 7, 7, 7, 8, 8, 8, 9, 9, \ 9, 10, 10, 10, 10, 11, 11, 11, 12, 12, 12, 12, 13, 13, 13, 14, 14, \ 14, 14, 15, 15, 15, 15, 16, 16, 16, 16, 17, 17, 17, 17, 18, 18, 18, \ 18, 19, 19,...$$ http://oeis.org/A024417

For the first few terms this agrees with the nearest integer of: $$\frac{n}{W(n)}-1$$ but starts to differ at $n=54$ and $n=58$.

The plot up to $n = 160$:

Where the sequence $a(n)$ is the blue dots and the asymptotic is the red curve.

So is: $$a(n) \sim \frac{n}{W(n)}-1$$ where $W(n)$ is the LambertW function, a good asymptotic?

Mathematica code for the plot:

Clear[a, n, t, m, aa, bb, x]
a[n_] := (m = Max[t = Table[StirlingS2[n, k], {k, 1, n}]];
   Position[t, m][[1, 1]]);
aa = Table[a[n], {n, 1, 160}]
(*From Jean-François Alcover, Nov 15 2011*)
bb = N[Table[n/ProductLog[n] - 1, {n, 1, 160}]]
Show[ListPlot[aa], ListLinePlot[bb, PlotStyle -> Red]]

Answer 1

Wikipedia has a different bound, namely $\frac{n}{\log n}$, and gives a paper that proves this bound.