# Advanced Discrete Math Generating function Problem

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

• 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 $5$. – darij grinberg Nov 14 '18 at 3:20
• And your second GF is $$\prod_{k=1}^\infty\frac1{(1-x^{2k-2})(1-x^{2k-3})(1-x^{2k-6})(1-x^{2k-9})(1-x^{2k-10})}.$$ – Lord Shark the Unknown Nov 14 '18 at 4:54
• Okay thanks! I have been working on it but I don't seem to get the algebra – fireshock Nov 14 '18 at 5:27