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What's going on with the behaviour of the subfactorial's imaginary part? Background: Out of curiosity I tried to construct some recurrence relations using the Pochhammer symbol and out of those came some subfactorials. For example:

$$a_{n+1}=a_n+(3)_n=a_n+3(3+1)(3+2)...(3+n-1).$$

Mathematica gave me:

$a_n$ = $1/2 (-1)^n$ Gamma[$n+3$] Subfactorial[$-n-3$]$-$Subfactorial[$-3$].

Not having seen seen negative subfactorials before I googled "negative subfactorial" or "subfactorial of negative numbers" and some similar phrases, which gave 0 hits. Here's a plot:

plot

I also plotted the Gamma function (just to have something to relate to). So, I don't understand the behaviour of the imaginary part. Looking closer at the values of $\operatorname{Im}(!n)$ it appears that

$$\sum _{n=-\infty}^0 \operatorname{Im}(!n)=-\frac{\pi}{e^2}.$$

Anyone who can shed some light on this? Some intuition? Better methods of visualization?

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  • $\begingroup$ Just out of curiosity : how did you get the last result ? Interesting problem $\to +1$. $\endgroup$ – Claude Leibovici Apr 12 at 10:19
  • $\begingroup$ @ClaudeLeibovici I found it through numerical calculations in Mathematica. It was clear that it was converging really fast so I thought it might be something interesting. $\endgroup$ – Carolus Apr 12 at 11:02
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Mathematica obviously defines subfactorial (cf. Eq. $(4)$ here) of an arbitrary complex number $s$ as $$ !s=\frac{\Gamma(s+1,-1)}{e},\tag1 $$ where $\Gamma(s,z)$ is the upper incomplete gamma function. The expression for the function for integer negative $s$ can be found here and reads for $z=-1$: $$ \Gamma(-n,-1) =\frac{(-1)^{n+1}}{n!}\left[i\pi+\operatorname{Ei}(1)-e\sum_{k=0}^{n-1}k!\right],\tag2 $$ where $\operatorname{Ei}(z)$ is the exponential integral.

Thus, $$ \operatorname{Im}\; !(-n)=\frac{(-1)^{n}}{(n-1)!}\frac\pi e $$ and the summation formula: $$ \sum_{n=1}^\infty \operatorname{Im}\; !(-n)=-\frac{\pi}{e^2} $$ immediately follows.

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