# Gen func “The number of partitions of n where each part occurs 2, 3, 5 times = number of partitions of n…”

The number of partitions of n where each part occurs 2, 3, 5 times = number of partitions of n with parts modulo 2,3,6,9,10 modulo 12

This is from Subbarao 1971 but I don't quite understand the method, or I have possibly misunderstood what the answer is supposed to be?

So I have $$\prod^\infty_{n=1}\,\,(1 + q^{2n} + q^{3n} + q^{5n})$$

I believe this should eventually be $$\prod^\infty_{n=1} \frac{1}{(1-q^{12n+2})(1-q^{12n+3})(1-q^{12n+6})(1-q^{12n+9})(1-q^{12n+10})}$$ But I'm finding it impossible to get from A to B. I'm pretty new to this stuff.. Can anybody help?

The trick is the same as in the proof that the numbers of partitions with odd parts and with distinct parts are the same. Then $$\begin{split} \prod_{n=1}^{\infty}\left(1+q^{2n}+q^{3n}+q^{5n}\right) &=\prod_{n=1}^{\infty}\left(1+q^{2n}\right)\left(1+q^{3n}\right)=\prod_{n=1}^{\infty}\frac{1-q^{4n}}{1-q^{2n}}\frac{1-q^{6n}}{1-q^{3n}}\\ &=\prod_{n=1}^{\infty}\frac{1-q^{4n}}{\left(1-q^{4n}\right)\left(1-q^{4n+2}\right)}\frac{1-q^{6n}}{1-q^{3n}}\\ &=\prod_{n=1}^{\infty}\frac{1}{1-q^{4n+2}}\frac{1-q^{6n}}{1-q^{3n}}\\ &=\prod_{n=1}^{\infty}\frac{1}{\left(1-q^{12n+2}\right)\left(1-q^{12n+6}\right)\left(1-q^{12n+10}\right)}\frac{\left(1-q^{12n}\right)\left(1-q^{12n+6}\right)}{1-q^{3n}}\\ &=\prod_{n=1}^{\infty}\frac{1}{\left(1-q^{12n+2}\right)\left(1-q^{12n+10}\right)}\frac{1-q^{12n}}{1-q^{3n}}\\ &=\prod_{n=1}^{\infty}\frac{1}{\left(1-q^{12n+2}\right)\left(1-q^{12n+10}\right)}\frac{1}{\left(1-q^{12n+3}\right)\left(1-q^{12n+6}\right)\left(1-q^{12n+9}\right)}\\ &=\prod_{n=1}^{\infty}\frac{1}{\left(1-q^{12n+2}\right)\left(1-q^{12n+3}\right)\left(1-q^{12n+6}\right)\left(1-q^{12n+9}\right)\left(1-q^{12n+10}\right)}, \end{split}$$ as desired.