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## MaBloWriMo continues with a hint

Yesterday, as a puzzle, I asked what the generating function

$\displaystyle \log \sum_{n \ge 0} n! z^n = z + \frac{3}{2} z^2 + \frac{13}{3} z^3 + \frac{71}{4} z^4 + \dots$

counts. Today I’ll give some hints; unfortunately I did not have enough time to write up a satisfying solution. As commenter Zahlen points out, it’s better to think of $\sum_{n \ge 0} n! z^n$, not as an ordinary generating function, but as an exponential generating function. That makes it the exponential generating function of the sequence $(n!)^2$ of squared factorials.

The reason we’d like to do this, although Zahlen didn’t make this explicit, is that this maneuver opens up the possibility of appealing to the exponential formula. Loosely speaking, the exponential formula can be interpreted as saying that if some exponential generating function $f(z)$ counts objects that have a decomposition into “connected components,” then $\log f(z)$ counts connected objects. For example, when $f(z) = \frac{1}{1 - z}$ (the exponential generating function of the factorials), the logarithm is

$\displaystyle \log f(z) = \log \frac{1}{1 - z} = \sum_{n \ge 1} \frac{z^n}{n}$

and this can be interpreted as reflecting the cycle decomposition of a permutation.

So: what is the relevant “connected components” decomposition here?