Number of integral soultions to linear equations without unit coefficients

Question

To determine the number of integral solutions for the linear equation

$$ x_1+x_2+x_3+\cdots+x_k = N$$

we have an expression $$ ^{N+k-1}C_{k-1}$$

But I want to know if the coefficients of $x_{1}+x_{2}+x_{3}$ were not unity, i.e. it were of type

$$ a_1 x_1+a_2 x_2+a_3 x_3+\cdots+a_k x_k = N$$

then how can we determine the number of integral solutions $>0 $ to this equation? How do we work towards the solution for this?

P.S. I am really not aware if any other question exists as the duplicate of this. Please pardon me if a question exactly like this exists.

Robert Israel · Accepted Answer · 2014-09-05 15:26:29Z

4

The number $s(N)$ of solutions of your equation has generating function $$ \sum_{N=0}^\infty s(N) z^N = \prod_{i=1}^k \dfrac{1}{1 - z^{a_i}}$$

answered Sep 5 '14 at 15:26

Robert Israel

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$\begingroup$ So, if I have a linear equation like $2*(x_{1}+x_{2}+\cdots+x_{6})+x_{7} = C$ , (C being some integer) then the right hand side of the above equation becomes : $\prod_{k=1}^7 \frac{1}{1-a_{i}}$ = $(1-x^{2})^{-6}(1-x)^{-1} = (1-x)^{-7} (1+x)^{-6} $. Suppose I have to extract the answer for S(10), then should I find the coefficient of $x^{10}$ in the right hand side for my answer? $\endgroup$ – nerdier.js Sep 5 '14 at 18:43
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$\begingroup$ Yes, that's right. $\endgroup$ – Robert Israel Sep 5 '14 at 18:51

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