# The behaviour of $\operatorname{Im}(!n)$

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:

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?

• Just out of curiosity : how did you get the last result ? Interesting problem $\to +1$. – Claude Leibovici Apr 12 at 10:19
• @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. – Carolus Apr 12 at 11:02

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.