# An exact formula for counting solutions of the Frobenius equation summed to 8

Let us consider a Frobenius equation with: $$x_1+\dots+x_n=8, \tag{1}$$ where $$(x_1, \dots, x_n)$$ must consist of non-negative integers, i.e. $$x_j \in \mathbb{N}$$ as Natural numbers.

Here is my question: Is there a general formula for Eq.(1) counting all the possible solutions $$(x_1, \dots, x_n)$$
for given the positive integer $$n \in \mathbb{Z}^+$$? This should be related to the Partition, but I am not sure the exact forms are known? Say, can we find the total number of possible solutions as a function $$f(n)$$, and what is $$f(n)=?$$

I am interested in finding $$f(20)=?$$ $$f(36)=?$$

p.s. Sorry if this question is too simple for number theorists. But please provide me answer and Refs if you already know the answer. Many thanks!

p.s.2. A more advanced generalized version of question is asked in https://mathoverflow.net/questions/352331/request-for-an-exact-formula-related-to-a-partition-in-number-theory

• Search this site for stars and bars. – Rob Pratt Feb 10 at 4:32
• – ccorn Feb 10 at 4:52
• @Rob Pratt, many thanks +1 – wonderich Feb 10 at 4:52

The number of solutions will be $$\begin{eqnarray*} [x^8]: (1+x+x^2+\cdots)^n = [x^8]: \frac{1}{(1-x)^n}. \end{eqnarray*}$$ Now use $$\begin{eqnarray*} [x^m]: \frac{1}{(1-x)^n}= \binom{m+n-1}{n-1}. \end{eqnarray*}$$