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I am trying to understand the basic result in this paper:

http://www.aimath.org/news/partition/folsom-kent-ono.pdf

My problem is with the example at the end of page 2. I understand it's supposed to be a particular instance of theorem 1.1 with $l=13, b_1=1,b_2=3,m=1$, but I can't see how they pass from $p(\frac{l^{b_2}n+1}{24})$ (which might be a fraction for all I know, since it seems like it's equal to $\frac{13^3n+1}{24}$) to $p(13^3n+1007)$.

Where did the 1007 come from? Where did the division by 24 disappear?

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up vote 3 down vote accepted

They define $p(\alpha)=0$ if $\alpha$ is not an integer. So you only have to care about those $n$ for which $13^3n+1$ is divisble by 24.

Reducing mod 24 this means that we need $13n +1 \equiv 0 {\pmod {24}}$ or equivalently $13n \equiv 23 \pmod {24}$ or equivalently that $n \equiv 11 \pmod {24}$. Now if you parametrize these numbers i.e $n= 24N+11$ and plug that in you get $$(13^3(24N+11)+1)/24=13^3N+1007$$.

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