Stirling transform

In combinatorial mathematics, the Stirling transform of a sequence { a_n : n = 1, 2, 3, ... } of numbers is the sequence { b_n : n = 1, 2, 3, ... } given by

${\displaystyle b_{n}=\sum _{k=1}^{n}\left\{{\begin{matrix}n\\k\end{matrix}}\right\}a_{k}}$,

where ${\displaystyle \left\{{\begin{matrix}n\\k\end{matrix}}\right\}}$ is the Stirling number of the second kind, which is the number of partitions of a set of size ${\displaystyle n}$ into ${\displaystyle k}$ parts. This is a linear sequence transformation.

The inverse transform is

${\displaystyle a_{n}=\sum _{k=1}^{n}(-1)^{n-k}\left[{n \atop k}\right]b_{k}}$,

where ${\textstyle (-1)^{n-k}\left[{n \atop k}\right]}$ is a signed Stirling number of the first kind, where the unsigned ${\displaystyle \left[{n \atop k}\right]}$ can be defined as the number of permutations on ${\displaystyle n}$ elements with ${\displaystyle k}$ cycles.

Berstein and Sloane (cited below) state "If a_n is the number of objects in some class with points labeled 1, 2, ..., n (with all labels distinct, i.e. ordinary labeled structures), then b_n is the number of objects with points labeled 1, 2, ..., n (with repetitions allowed)."

If

${\displaystyle f(x)=\sum _{n=1}^{\infty }{a_{n} \over n!}x^{n}}$

is a formal power series, and

${\displaystyle g(x)=\sum _{n=1}^{\infty }{b_{n} \over n!}x^{n}}$

with a_n and b_n as above, then

${\displaystyle g(x)=f(e^{x}-1)}$.

Likewise, the inverse transform leads to the generating function identity

${\displaystyle f(x)=g(\log(1+x))}$.