What is the closed form solution to the sum of inverse products of parts in all compositions of n?

Question

My question is exactly as in the title:

What is the generating function or closed form solution to the sum of inverse products of parts in all compositions of $n$?

This question was inspired by just browsing around OEIS sequences. Here is the OEIS sequence for the numerators and denominators of the sum of inverse products of parts in all compositions of $n$:

https://oeis.org/A323339

https://oeis.org/A323340

The progress that I have made is finding the solution to the sum of products of parts in all compositions of $n$, the solution was inspired by this thread:

Number of parts equal to $~1~$ in all compositions of $~n~$

But I am still unclear how to solve this problem.

Please help.

Answer 1

The generating function you requested is already given in OEIS at A323339:
"FORMULA $\ \ $ G.f. for fractions: 1 / (1 + \log(1 - x)). - Ilya Gutkovskiy, Nov 12 2019
EXAMPLE $\ \ \ \ $ 1/1, 1/1, 3/2, 7/3, 11/3, 347/60, 3289/360, 1011/70, 38371/1680, 136553/3780, 4320019/75600, 12528587/138600, 40771123/285120, ... = A323339/A323340"

Numbers and types of compositions without zero are enumerated by the Ordinary Bell polynomials

$$\hat{B}_n(x_1,...,x_n)=\sum_{k=0}^n\sum_{\pi_{n,k}(j_1,...,j_n)}\frac{k!}{j_1!...j_n!}x_1^{j_1}...x_n^{j_n}.$$

$\pi_{n,k}(j_1,...,j_{n-k+1})$ are the partitions of the integer $n$ into exactly $k$ parts, the $x_i$ are symbols for the parts $i\ $ ($i\in\{1,...,n\}$).

For your problem of reciprocals, we have to use the reciprocals of the parts:

$$a(n)=\sum_{k=0}^n\sum_{\pi_{n,k}(j_1,...,j_n)}\frac{k!}{j_1!1^{j_1}...j_n!n^{j_n}}$$

Let $s$ denote the Stirling numbers of the first kind.

$$s(n,k)=(-1)^{n-k}\sum_{\pi_{n,k}(j_1,...,j_n)}\frac{n!}{j_1!1^{j_1}...j_n!n^{j_n}}$$

$$a(n)=\sum_{k=0}^n\frac{k!}{n!}|s(n,k)|$$

The generating function for the numbers you requested is therefore:

$$\sum_{n=0}^\infty\sum_{k=0}^n\frac{k!}{n!}|s(n,k)|\ x^n.$$

Answer 2

The generating function, as stated on the OEIS page, is $$\frac1{1+\log(1-x)}.$$ Here is a quick proof of this result. Each composition of some positive integer is simply a finite sequence $a_1,a_2,\dots,a_m$ of positive integers. In the generating function, such a sequence contributes a term of $$\frac{1}{a_1\cdots a_m}x^{a_1+\dots+a_m}=\left(\frac{x^{a_1}}{a_1}\right)\cdots\left(\frac{x^{a_m}}{a_m}\right).$$ As a result, the sum of terms corresponding to compositions with exactly $m$ terms is $$\sum_{a_1=1}^\infty \cdots\sum_{a_m=1}^\infty \left(\frac{x^{a_1}}{a_1}\right)\cdots\left(\frac{x^{a_m}}{a_m}\right)=\left(\sum_{a=1}^\infty \frac{x^a}a\right)^m=(-\log(1-x))^m.$$ Summing this over all $m\geq 0$ gives the desired generating function.