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I am suppose to prove that the number of partitions of $n$ in which each part appears $2$, $3$, or $5$ times equals the number partitions of $n$ in to parts which are congruent to $2$, $3$, $6$, $9$, or $10$ modulo $12$.

I tried using Remmels Thm in order to prove this but I ran into a problem classifying the pairwise disjoint multisets. I now think the approach that should be used is find the generating function for each and show equality using algebra. I think the generating function for

$A$: partitions of $n$ in which each part appears $2$, $3$, or $5$ times

$$ G(A) = (1+x^2+x^4+...)(1+x^3+x^6+..)(1+x^5+x^10+...) = \frac{1}{1-x^2}\frac{1}{1-x^3}\frac{1}{1-x^5} $$

$B$: partitions of $n$ into parts which are congruent to $2$,$3$,$6$,$9$, or $10$ modulo $12$

$$ G(B) = \frac{1}{(1-x^{12k-2})(1-x^{12k-3})(1-x^{12k-6})(1-x^{12k-9})(1-x^{12k-10})} $$

Is this correct? Not sure how to manipulate the functions into each other

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    $\begingroup$ A: No. It should be $\prod_{i \geq 1} \left(1 + x^{2i} + x^{3i} + x^{5i}\right)$. What you computed is instead the number of partitions of $n$ in which each part equals $2$, $3$ or $5$. $\endgroup$ – darij grinberg Nov 14 '18 at 3:20
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    $\begingroup$ And your second GF is $$\prod_{k=1}^\infty\frac1{(1-x^{2k-2})(1-x^{2k-3})(1-x^{2k-6})(1-x^{2k-9})(1-x^{2k-10})}.$$ $\endgroup$ – Lord Shark the Unknown Nov 14 '18 at 4:54
  • $\begingroup$ Okay thanks! I have been working on it but I don't seem to get the algebra $\endgroup$ – fireshock Nov 14 '18 at 5:27

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