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0

I realize there is already an answer here, giving the generating function. But I don't understand how this person got the generating function. It might help to add some variables to the generating function: $((z_1x)^0 + (z_1x)^2 + (z_1x)^3 + \cdots) ((z_2x^2)^0 + (z_2x^2)^2 + (z_2x^2)^3 + \cdots) ((z_3x^3)^0 + (z_3x^3)^2 + (z_3x^3)^3 + \cdots) \cdots \\ =...


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As mentioned by Lord Shark the Unknown, it is easy to prove this using the fact that uniformly converging analytic functions converge to analytic functions. We have uniform convergence of an infinite product easily here on any disk of radius $r<1$ since $$\left|\log\left[\frac1{1-z^m}\right]\right|=|\log(1-z^m)|\le|z|^m+\frac{|z|^{2m}}{|1-z^m|^2}\le r^m+...


1

The generating function for partitions where there are at most $r-1$ parts of each size is $$ (1+x+x^2+\dots+x^{r-1})(1+x^2+x^4+\dots+x^{2(r-1)})(1+x^3+\dots+x^{3(r-1)})\cdots $$ When expanding this out, the choice of the summand $x^{kj}$ in the factor $1+x^k+\dots+x^{k(r-1)}$ corresponds to having $j$ parts of size $k$ in the partition. We can write this as ...


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This is a stars and bars problem. The way you have done it, you want the number of integer solutions to $$r + y + b_k + v = 102$$ subject to $$\begin{align}r&\geq0\\y&\geq0\\b_k&\geq1\\v&\geq3\end{align}$$ So, put $1$ bean in the black jar, $3$ beans in the violet jar, and distribute the remaining $98$ beans in the $4$ jars, This gives $${...


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