I wonder if there is a classical way of resolution of the following sequence : $$P(m,1):=\frac{2}{m!}, \qquad P(m,n+1) := \frac{2}{m!} + mP(m+1,n). \quad (1)$$ The results I obtained with the first values of $n$ are quite complicated and doesn't give a clear induction pattern...
Motivation : I am studying the generalized function $$T_{\frac{\mathrm{sgn}(x)}{x^n}}:= \frac{(-1)^{n-1}}{(n-1)!}(\mathrm{sgn}(x)\ln(|x|))^{(n)} \quad (2)$$ where the derivation is taken in the sense of the distributions. I don't know if this definition is canonical (it is made up by myself for a specific use), but I was wondering what would be $x^n T_{\frac{\mathrm{sgn(x)}}{x^{n+m}}}$ equal to. I obtained $$ x^n T_{\frac{\mathrm{sgn(x)}}{x^{n+m}}} = T_{\frac{\mathrm{sgn(x)}}{x^{m}}}-P(m,n) \delta_0^{(m-1)} \quad (3)$$ where $\delta_0$ is the Dirac distribution and $P$ is defined as in (1).
Nota: It is not an homework...Moreover, I can survive with only (3), but if there exists an explicit form, it would be much more satisfying, isn't it?
You can recursively use the definition to reach
\begin{align} P(m,n)&=\frac{2}{m!}+\frac{2m}{(m+1)!}+\dots+\frac{2m(m+1)\dots(m+k-1)}{(m+k)!}\\ &\ +m(m+1)\cdots(m+k)P(m+k+1,n-k-1) \end{align} and by plugging in $k=n-2$ to be able to use the $P(m,1)$ definition, this eventually leads to $$ P(m,n)=\sum_{k=0}^{n-1}\frac{2(m+k-1)!}{(m-1)!(m+k)!}=\frac{2}{(m-1)!}\sum_{k=0}^{n-1}\frac{1}{m+k} = \frac{2(H_{m+n-1}-H_{m-1})}{(m-1)!} $$
where $H_n$ is $n$-th Harmonic number: $$ H_n = 1+\frac{1}{2}+\dots+\frac{1}{n}. $$