What distribution do the rows of the Stirling numbers of the second kind approach? In wikipedia about the Pascal triangle:
Relation to binomial distribution
"When divided by 2n, the nth row of Pascal's triangle becomes the binomial distribution in the symmetric case where p = 1/2. By the central limit theorem, this distribution approaches the normal distribution as n increases."

How can I find the distribution approached by the rows of the Stirling numbers of the second kind?

$$\begin{array}{llllllll}
 1 & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} \\
 1 & 1 & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} \\
 1 & 3 & 1 & \text{} & \text{} & \text{} & \text{} & \text{} \\
 1 & 7 & 6 & 1 & \text{} & \text{} & \text{} & \text{} \\
 1 & 15 & 25 & 10 & 1 & \text{} & \text{} & \text{} \\
 1 & 31 & 90 & 65 & 15 & 1 & \text{} & \text{} \\
 1 & 63 & 301 & 350 & 140 & 21 & 1 & \text{} \\
 1 & 127 & 966 & 1701 & 1050 & 266 & 28 & 1
\end{array}$$
After all the Stirling numbers of the second kind satisfy the recurrence:
$$\left\{{n+1\atop k}\right\} = k \left\{{ n \atop k }\right\} + \left\{{n\atop k-1}\right\}$$
Where as the Pascal triangle satisfies the recurrence:
$${n \choose k} = {n-1 \choose k-1} + {n-1 \choose k}$$
If such a distribution does exist I would like to find the x value for the maximum of the distribution for each row and see if it says anything about x/LambertW(x).

I think this is what you are looking for:
Stirling Behavior is Asymptotically Normal,
L. H. Harper,
The Annals of Mathematical Statistics
Vol. 38, No. 2 (Apr., 1967), pp. 410-414
Link
