Gen func "The number of partitions of n where each part occurs 2, 3, 5 times = number of partitions of n..."

Question

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?

Answer 1

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 $$ \begin{split} \prod_{n=1}^{\infty}\left(1+q^{2n}+q^{3n}+q^{5n}\right) &=\prod_{n=1}^{\infty}\left(1+q^{2n}\right)\left(1+q^{3n}\right)=\prod_{n=1}^{\infty}\frac{1-q^{4n}}{1-q^{2n}}\frac{1-q^{6n}}{1-q^{3n}}\\ &=\prod_{n=1}^{\infty}\frac{1-q^{4n}}{\left(1-q^{4n}\right)\left(1-q^{4n-2}\right)}\frac{1-q^{6n}}{1-q^{3n}}\\ &=\prod_{n=1}^{\infty}\frac{1}{1-q^{4n-2}}\frac{1-q^{6n}}{1-q^{3n}}\\ &=\prod_{n=1}^{\infty}\frac{1}{\left(1-q^{12n-10}\right)\left(1-q^{12n-6}\right)\left(1-q^{12n-10}\right)}\frac{\left(1-q^{12n}\right)\left(1-q^{12n-6}\right)}{1-q^{3n}}\\ &=\prod_{n=1}^{\infty}\frac{1}{\left(1-q^{12n-10}\right)\left(1-q^{12n-2}\right)}\frac{1-q^{12n}}{1-q^{3n}}\\ &=\prod_{n=1}^{\infty}\frac{1}{\left(1-q^{12n-10}\right)\left(1-q^{12n-2}\right)}\frac{1}{\left(1-q^{12n-9}\right)\left(1-q^{12n-6}\right)\left(1-q^{12n-3}\right)}\\ &=\prod_{n=1}^{\infty}\frac{1}{\left(1-q^{12n-10}\right)\left(1-q^{12n-9}\right)\left(1-q^{12n-6}\right)\left(1-q^{12n-3}\right)\left(1-q^{12n-2}\right)},\\ &=\prod_{n=0}^{\infty}\frac{1}{\left(1-q^{12n+2}\right)\left(1-q^{12n+3}\right)\left(1-q^{12n+6}\right)\left(1-q^{12n+9}\right)\left(1-q^{12n+10}\right)}, \end{split} $$ as desired.

Note: There was an index error earlier, now corrected thanks to the suggestion by PSWadder.

Answer 2

I believe the proof used by Alexander needs re-indexing. This is because, it will miss the $$ q^{2i} $$ term when all is multiplied out on the RHS for example. But clearly will have this term when LHS is multiplied out.

I provide an alternative proof below:

$$ \prod_{i=1}^\infty (1+q^{2i}+q^{3i}+q^{5i}) = \prod_{i=1}^\infty (1+q^{2i})(1+q^{3i}) \\ = \prod_{i=1}^\infty (1+q^{2i})\frac{1-q^{2i}}{1-q^{2i}}(1+q^{3i})\frac{1-q^{3i}}{1-q^{3i}} \\ = \prod_{i=1}^\infty\frac{1-q^{4i}}{1-q^{2i}}\frac{1-q^{6i}}{1-q^{3i}} \\ = \prod_{i=1}^\infty\frac{1}{(1-q^{12i-10})(1-q^{12i-9})(1-q^{12i-6})(1-q^{12i-3})(1-q^{12i-2})}$$

Which is generating function for number of partitions of n with parts 2, 3, 6, 9, or 10 mod 12.

I can go from line 3 to line 4 by realizing this is the infinite product of $$\frac{1}{1-q^k}$$ where k is the even positive numbers without multiples of 4 union multiples of three without multiples of 6. This is the same as the list of all positive integers congruent to 2, 3, 6, 9, or 10 mod 12.