How many $s,t,u$ satisfy: $s +2t+3u +\ldots = n$? 
Given $n\in \mathbb{N}^+$, what is the possible number of combinations
  $s,t,u,\ldots\in\mathbb{N}$, such that: $$s +2t+3u +\ldots = n\quad?$$
  Additionally, is there an efficient way to find these combinations
  other than an elimination process?

This problem comes from the formula for series reversion, and gives the number of terms in each inverse coefficient.

The  number of solutions of the equation equals the coefficient of $z^n$ in the expression 
$$
\frac{1}{(1-z)(1-z^2)(1-z^3)\ldots }=\sum_{n}p(n)z^n,
$$
where $p(n)$  is the Euler partition function, so that the number of combinations is $p(n)$. 
(Christoph): There is a bijection between the partition function and the desired function, since from a partition $a_1+a_2+\ldots+a_k=n$ you set $s$ to the number of $1$s appearing, $t$ to the number of $2$s appearing, etc. 
