Request for an exact formula related to a partition in number theory The Frobenius equation is the Diophantine equation $$
a_1 x_1+\dots+a_n x_n=b,$$ 
where the $a_j$ are positive integers, $b$ is an integer, and a solution $$(x_1, \dots, x_n)$$ 
must consist of non-negative integers, i.e. 
$$
x_j \in \mathbb{N}
$$
as Natural numbers. For negative $b$, there are no solutions.


*

*My question: Is there any known formula that counts the number of solutions, by giving $a_1, \dots, a_n$, $b$, and $n$? Let us call this function as $F (a_1, \dots, a_n; b, n)$, what is known for this: 


$$F (a_1, \dots, a_n; b, n)=?$$

For all the $a_j=1$, we can simplify the above Frobenius equation to:
$$
x_1+\dots+x_n=b, \tag{1}$$
where $b \in \mathbb{Z}^+$ is a positive integer. 


*

*Here is another simpler question:  Is there a general formula for Eq.(1) counting all the possible solutions $$(x_1, \dots, x_n)$$
for given the positive integer $n \in \mathbb{Z}^+$ and $b \in \mathbb{Z}^+$? 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 soultions as a function $f(n,b)$, and what is
$$
f(n,b)=?
$$
It seems the answer is known:

$$
f(n,b)= \binom{b+n-1}{n-1}.
$$

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!
 A: As usual, it depends on what you call a "formula". There are several approaches used to study the general equation: numerical semigroups, Fourier analysis, partial fraction decomposition, generating funtions, counting lattice point in polytopes, multiple zeta functions, etc.
For $n=2$ (and $a_1$, $a_2$ relatively prime) there is a simple formula, namely
$$F(a_1,a_2;b,2)=\frac{b}{a_1a_2}
-\left\{\frac{{a_2}^{-1}b}{a_1}\right\}
-\left\{\frac{{a_1}^{-1}b}{a_2}\right\}+1,$$
where ${a_1}^{-1}$ is an inverse modulo $a_2$, ${a_2}^{-1}$ is an inverse module $a_1$,
and $\{x\}$ denotes the fractional part of $x$. There is no known similar formula for $n\ge3$ (and there is probably no hope in finding a simple one). In the simplest case $a_i=1$ for all $i$, your formula is correct.
There is a beautiful book that discusses this topic in Chapter 1: "Computing the continuous discretely" by Beck and Robins. Also, as a reference, you may look at any book about "Numerical semigroups".
(There are several people in MO that are far better qualified at answering this, and I hope they will see your question and answer it.)
A: $F(a_1,\dots,a_n;b)$ equals the coefficient of $z^b$ in the generating function
$$f(z):=\frac{1}{1-z^{a_1}}\frac{1}{1-z^{a_2}}\cdots \frac{1}{1-z^{a_n}}.$$
For a fixed choice of $a_1,\dots,a_n$, explicit formula for $F(a_1,\dots,a_n;b)$ as a function of $b$ can obtained via partial fraction decomposition.
For the second question, see Stars and bars.
A: You can take generating function $$f(z):=\frac{1}{1-z^{a_1}}\frac{1}{1-z^{a_2}}\cdots \frac{1}{1-z^{a_n}}$$ as in Max Alekseyev's answer and calculate $F (a_1, \dots, a_n; b, n)$ as $$
\frac{1}{2 \pi i} \int_{|s|=\rho} f(s) \frac{d s}{s^{b+1}} \quad (0<\rho<1).
$$
It gives the answer
$$
F (a_1, \dots, a_n; b)=\frac{b^{n-1}}{(n-1) !  a_{1} \ldots a_{n}}+\sum_{k=0}^{n-2} c_{k} b^{k}.
$$
It is a classical applications of  contour integration taken from the book "Residues and their applications" by  A.O. Gelfond (1966, pp. 98-99, Russian). If $(a_j,a_k)=1$ ($j\ne k$) then all poles (excepting $s=1$) are simple and formula can be simplified:
$$
F (a_1, \dots, a_n; b)=\frac{(-1)^{n-1}}{(n-1) !} \frac{d^{n-1}}{d s^{n-1}}\left[s^{-b-1} \prod_{k=1}^{n} \frac{1-s}{1-s^{a_{k}}}\right]_{s=1}+R 
$$
where $|R|<C$ for some constant $C$. 
A: Some number theory terminology for your second question is the number of integer compositions of $b$ with $n$ parts, where the parts are required to be positive integers.  There are $\binom{b-1}{n-1}$ of these: think of having $b$ 1s in a row and, among the $b-1$ spaces between them, placing $n-1$ plus signs.  Combine the adjacent 1s and separate parts by +, e.g., $11+1+1+11 \sim 2+1+1+2$ is one of the $\binom{5}{3}=10$ 4-part compositions of 6.
By the way, integer partitions are equivalent to solutions where the order of the summands does not matter.  Equivalently, if the summands are placed in a specified order, typically nonincreasing.  For example, $2+1+1+2$ and $1+2+1+2$, etc., would all correspond to $2+2+1+1$.  There are generally fewer partitions than compositions and there is not such a simple formula for the number of them.
Your formula, $\binom{b+n-1}{n-1}$, is for the number of compositions with nonnegative integer parts.  The "stars & bars" argument in combinatorics verifies the formula: Any arrangement of $b$ 1s and $n-1$ plus signs gives one of these compositions, e.g., $1111+11++ \sim 4+2+0+0$ is one of the $\binom{9}{3} = 84$ 4-part weak compositions of 6.
As @EFinat-S wrote, there's a nice formula for two variables, but for arbitrary linear Diophantine equations, there is nothing like that.  The end of Beck & Robbins chapter 1 touches on this (e.g., the "chicken McNugget problem") and goes on to more advanced approaches---it's a great book.
