Ramanujan’s Partition Identities, Congruences

We want to prove Ramanujan’s congruence identities for the partition function:

$\displaystyle \begin{aligned} p(5 n+4) & \equiv 0 \quad(\bmod 5) \\ p(7 n+5) & \equiv 0 \quad(\bmod 7) \\ p(11 n+6) & \equiv 0 \quad(\bmod 11) \end{aligned}$

We prove them using identities involving $q$ series of the generating function:

$\displaystyle P(q)=\sum_{n>0} p(n) q^{n}=1+1 q+2 q^{2}+3 q^{3}+5 q^{4}+\cdots = \displaystyle \prod_{n=1}^{\infty} \frac{1}{\left(1-q^n\right)} =(q;q)_{\infty}$

$\displaystyle (a;q)_n = \prod_{k-0}^{n}(1-aq^k)$

$\displaystyle \sum_{n \geq 0} p(5 n+4) q^{n}=5+30 q+135 q^{2}+490 q^{3}+1575 q^{4}+4565 q^{5}+\cdots$

Pentagonal identity:

$\displaystyle (q ; q)_{\infty} = \prod_{n=1}^{\infty}\left( 1-q^n\right) = \sum_{-\infty}^{+\infty}(-1)^{k} q^{k(3 k+1) / 2}=1-q-q^{2}+q^{5}+q^{7}-q^{12}-q^{15}++--\cdots$

(You can directly prove this or obtain it as consequence of Jacobi -Triple Product identity)

Let ${\omega}$ be a fifth root of unity.

$\displaystyle P(q)=\sum_{n>0} p(n) q^{n}=\frac{1}{(q ; q)_{\infty}}=\frac{(\omega q ; \omega q)_{\infty}\left(\omega^{2} q ; \omega^{2} q\right)_{\infty}\left(\omega^{3} q ; \omega^{3} q\right)_{\infty}\left(\omega^{4} q ; \omega^{4} q\right)_{\infty}}{(q ; q)_{\infty}(\omega q ; \omega q)_{\infty}\left(\omega^{2} q ; \omega^{2} q\right)_{\infty}\left(\omega^{3} q ; \omega^{3} q\right)_{\infty}\left(\omega^{4} q ; \omega^{4} q\right)_{\infty}}$

$\displaystyle \begin{aligned} &\prod_{n \geq 1}\left(1-q^{n}\right)\left(1-\omega^{n} q^{n}\right)\left(1-\omega^{2 n} q^{2 n}\right)\left(1-\omega^{3 n} q^{3 n}\right)\left(1-\omega^{4 n} q^{4 n}\right) \\ &=\prod_{5 | n}\left(1-q^{n}\right)^{5} \cdot \prod_{5 \uparrow n}\left(1-q^{n}\right)\left(1-\omega q^{n}\right)\left(1-\omega^{2} q^{n}\right)\left(1-\omega^{3} q^{n}\right)\left(1-\omega^{4} q^{n}\right) \\ &=\prod_{n \geq 1}\left(1-q^{5 n}\right)^{5} \cdot \prod_{5 \uparrow n}\left(1-q^{5 n}\right) \\ &=\prod_{n \geq 1}\left(1-q^{5 n}\right)^{5} \cdot \prod_{n \geq 1}\left(1-q^{5 n}\right) / \prod_{5 | n}\left(1-q^{5 n}\right) \\ &=\frac{\left(q^{5} ; q^{5}\right)_{\infty}^{6} }{\left(q^{25} ; q^{25}\right)_{\infty}} \end{aligned}$

$\displaystyle \implies P(q)= \frac{1}{(q ; q)_{\infty}}=\frac{(\omega q ; \omega q)_{\infty}\left(\omega^{2} q ; \omega^{2} q\right)_{\infty}\left(\omega^{3} q ; \omega^{3} q\right)_{\infty}\left(\omega^{4} q ; \omega^{4} q\right)_{\infty}}{\left(q^{5} ; q^{5}\right)_{\infty}^{6} /\left(q^{25} ; q^{25}\right)_{\infty}}$

Using the following Ramanujan’s identity (obtained from specializations of Jacobi’s Triple Product identity)

$\displaystyle (q ; q)_{\infty}=\frac{\left(q^{10} ; q^{25}\right)_{\infty}\left(q^{15} ; q^{25}\right)_{\infty}\left(q^{25} ; q^{25}\right)_{\infty}}{\left(q^{5} ; q^{25}\right)_{\infty}\left(q^{20} ; q^{25}\right)_{\infty}}-q\left(q^{25} ; q^{25}\right)_{\infty} -q^{2} \frac{\left(q^{5} ; q^{25}\right)_{\infty}\left(q^{20} ; q^{25}\right)_{\infty}\left(q^{25} ; q^{25}\right)_{\infty}}{\left(q^{10} ; q^{25}\right)_{\infty}\left(q^{15} ; q^{25}\right)_{\infty}}$

we get

$\displaystyle (\omega q ; \omega q)_{\infty}\left(\omega^{2} q ; \omega^{2} q\right)_{\infty}\left(\omega^{3} q ; \omega^{3} q\right)_{\infty}\left(\omega^{4} q ; \omega^{4} q\right)_{\infty}$

$\displaystyle =\left(q^{25} ; q^{25}\right)_{\infty}^{4} \left(\left(a^{4}-3 q^{5} b\right)+q\left(a^{3}+2 q^{5} b^{2}\right)+q^{2}\left(2 a^{2}-q^{5} b^{3}\right)+q^{3}\left(3 a+q^{5} b^{4}\right)+5 q^{4}\right)$

where

$\displaystyle a=\frac{\left(q^{10} ; q^{25}\right)_{\infty}\left(q^{15} ; q^{25}\right)_{\infty}}{\left(q^{5} ; q^{25}\right)_{\infty}\left(q^{20} ; q^{25}\right)_{\infty}} \text { and } b=\frac{\left(q^{5} ; q^{25}\right)_{\infty}\left(q^{20} ; q^{25}\right)_{\infty}}{\left(q^{10} ; q^{25}\right)_{\infty}\left(q^{15} ; q^{25}\right)_{\infty}}$

$\displaystyle \sum_{n \geq 0} p(n) q^{n}=\frac{\left(q^{25} ; q^{25}\right)_{\infty}^{5}}{\left(q^{5} ; q^{5}\right)_{\infty}^{6}} \left(\left(a^{4}-3 q^{5} b\right)+q\left(a^{3}+2 q^{5} b^{2}\right)+q^{2}\left(2 a^{2}-q^{5} b^{3}\right)+q^{3}\left(3 a+q^{5} b^{4}\right)+5 q^{4}\right)$

Collecting terms ${ n\equiv 4 \mod 5}$ , we get

$\displaystyle \sum_{n \geq 0} p(5 n+4) q^{5 n+4}=5 q^{4} \frac{\left(q^{25} ; q^{25}\right)_{\infty}^{5}}{\left(q^{5} ; q^{5}\right)_{\infty}^{6}}$

$\displaystyle \sum_{n \geq 0} p(5 n+4) q^{n}=5 \frac{\left(q^{5} ; q^{5}\right)_{\infty}^{5}}{(q ; q)_{\infty}^{6}}$

Another Proof: A simpler proof of the the congruence is given below.

We get the following from Jacobi Identity

$\displaystyle (q;q)_{\infty} = \prod_{n=1}^{\infty}\left( 1-q^n\right) = \sum_{-\infty}^{\infty}(-1)^{n} q^{\left(3 n^{2}+n\right) / 2}$

$\displaystyle (q;q)_{\infty}^3 = \prod_{n=1}^{\infty}\left( 1-q^n\right)^3= \sum_{n>0}(-1)^{n}(2 n+1) q^{\left(n^{2}+n\right) / 2} =1-3 q+5 q^{3}-7 q^{6}+9 q^{10}-11 q^{15}+13 q^{21} -15 q^{28}+17 q^{36}-19 q^{45}+-\cdots$

$\displaystyle \sum_{-\infty}^{\infty}(-1)^{n} q^{\left(3 n^{2}+n\right) / 2} = A_0+A_1+A_2$

where ${A_i}$ has terms with exponents ${i \mod 5}$ -all the exponents are ${0,1}$ or ${2}$ modulo ${5.}$

$\displaystyle \sum_{n>0}(-1)^{n}(2 n+1) q^{\left(n^{2}+n\right) / 2} = B_0+B_1+B_3$

where ${B_i}$ has terms with exponents ${i \mod 5}$ – the exponents that appear are ${0,1}$ or ${3}$ modulo ${5}$ and the terms ${B_3}$ with ${3 \mod 5}$ exponents are ${0}$ modulo ${5.}$ That is ${B_3 =0 \mod 5.}$

$\displaystyle \sum_{n>0}(-1)^{n}(2 n+1) q^{\left(n^{2}+n\right) / 2} \equiv B_0+B_1 \mod 5$

$\displaystyle (q;q)_{\infty}^5 =(1-q)^{5}\left(1-q^{2}\right)^{5}\left(1-q^{3}\right)^{5} \cdots\equiv \left(1-q^{5}\right)\left(1-q^{10}\right)\left(1-q^{15}\right) \cdots \mod 5 = (q^5;q^5)_{\infty}$

$\displaystyle P(q) =\sum_{n \geq 0} p(n) q^{n}=\frac{1}{(q;q)_{\infty}}=\frac{(q;q)_{\infty}^{4}}{(q;q)_{\infty}^{5}} \equiv \frac{(q;q)_{\infty}^{4}}{(q^5;q^5)_{\infty}} \equiv \frac{\left(A_0+A_1+A_2\right)\left(B_0+B_1\right) }{(q^5;q^5)_{\infty}} \mod 5$

$\displaystyle \implies \sum_{n \geq 0} p(5 n+4) q^{5 n+4} \equiv 0 \mod 5$

Proof of ${\mod 7}$ congruence:

$\displaystyle \sum_{-\infty}^{\infty}(-1)^{n} q^{\left(3 n^{2}+n\right) / 2} = A_0+A_1+A_2+A_5$

where ${A_i}$ has terms with exponents ${i \mod 7}$ -all the exponents are ${0,1,2}$ or ${5}$ modulo ${7.}$

$\displaystyle \sum_{n>0}(-1)^{n}(2 n+1) q^{\left(n^{2}+n\right) / 2} = B_0+B_1+B_3+B_6$

where ${B_i}$ has terms with exponents ${i \mod 7}$ – the exponents that appear are ${0,1,3}$ or ${6}$ modulo ${5}$ and the terms ${B_6}$ with ${6 \mod 7}$ exponents are ${0}$ modulo ${7.}$ That is ${B_6 =0 \mod 7.}$

$\displaystyle \sum_{n>0}(-1)^{n}(2 n+1) q^{\left(n^{2}+n\right) / 2} \equiv B_0+B_1+B_3 \mod 7$

$\displaystyle (q;q)_{\infty}^7 = (q^7;q^7)_{\infty} \mod 7$

$\displaystyle P(q) =\sum_{n \geq 0} p(n) q^{n}=\frac{1}{(q;q)_{\infty}}=\frac{(q;q)_{\infty}^{6}}{(q;q)_{\infty}^{7}} \equiv \frac{(q;q)_{\infty}^{6}}{(q^7;q^7)_{\infty}} \equiv \frac{\left(B_0+B_1+B_3\right)^2 }{(q^5;q^5)_{\infty}} \mod 7$

$\displaystyle \implies \sum_{n \geq 0} p(7 n+5) q^{7 n+5} \equiv 0 \mod 7$

Proof of ${\mod 11}$ congruence:

$\displaystyle \sum_{-\infty}^{\infty}(-1)^{n} q^{\left(3 n^{2}+n\right) / 2} = A_0+A_1+A_2+A_5+A_7+A_{15}$

$\displaystyle \sum_{n>0}(-1)^{n}(2 n+1) q^{\left(n^{2}+n\right) / 2} \equiv B_0+B_1+B_3+B_6+B_{10} \mod 7$

$\displaystyle P(q) =\sum_{n \geq 0} p(n) q^{n}=\frac{1}{(q;q)_{\infty}}=\frac{(q;q)_{\infty}^{21}}{(q;q)_{\infty}^{22}} \equiv \frac{(q;q)_{\infty}^{21}}{(q^{11};q^{11})_{\infty}^2} \equiv \frac{\left(B_0+B_1+B_3+B_6+B_{10} \right)^7 }{(q^{11};q^{11})_{\infty}^2} \mod 11$

$\displaystyle \implies \sum_{n \geq 0} p(11 n+6) q^{11 n+6} \equiv \frac{F} {(q^{11};q^{11})_{\infty}^2} \mod 11$

${F \equiv 0 \mod 11}$ can be obtained using the following relations:

$\displaystyle \left(\sum_{n>0}(-1)^{n}(2 n+1) q^{\left(n^{2}+n\right) / 2} \right)^4 = (q;q)_{\infty}^{12}= (q;q)_{\infty}^{11} \sum_{-\infty}^{\infty}(-1)^{n} q^{\left(3 n^{2}+n\right) / 2}$

$\displaystyle \left(B_{0}+B_{1}+B_{3}+B_{6}+B_{10}\right)^{4} \equiv (q^{11}; q^{11})_{\infty}\left(A_{0}+A_{1}+A_{2}+A_{5}+A_{7}+A_{15}\right) \mod 11$

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