In the paper "Note on certain modular relations considered by Messrs Ramanujan, Darling and Rogers" (Proceedings of London Mathematical Society (1922) s2-20 (1): 408-416) Mordell gives proofs of the identities \begin{align} \sum_{n = 0}^{\infty}p(5n + 4)q^{n} &= 5\frac{\{(1 - q^{5})(1 - q^{10})(1 - q^{15})\cdots\}^{5}}{\{(1 - q)(1 - q^{2})(1 - q^{3})\cdots\}^{6}}\tag{1}\\ \sum_{n = 0}^{\infty}p(7n + 5)q^{n} &= 7\frac{\{(1 - q^{7})(1 - q^{14})(1 - q^{21})\cdots\}^{3}}{\{(1 - q)(1 - q^{2})(1 - q^{3})\cdots\}^{4}}\notag\\ &\,\,\,\,+ 49q\frac{\{(1 - q^{7})(1 - q^{14})(1 - q^{21})\cdots\}^{7}}{\{(1 - q)(1 - q^{2})(1 - q^{3})\cdots\}^{8}}\tag{2} \end{align} based on theory of elliptic modular functions. He uses a slightly old notation $$\Delta(\omega_{1}, \omega_{2}) = \left(\frac{2\pi}{\omega_{2}}\right)^{12}q\prod_{n = 1}^{\infty}(1 - q^{n})^{24}$$ where $q = e^{2\pi i\omega}, \omega = \omega_{1}/\omega_{2}$.

Clearly both the series given by Ramanujan lead to partition congruences modulo $5, 5^{2}, 7, 7^{2}$. Mordell shows that these results are equivalent to $$\sum_{j = 0}^{4}\left(\frac{\Delta(5\omega_{1}, \omega_{2})}{\Delta\{(\omega_{1} + j\omega_{2})/5, \omega_{2}\}}\right)^{1/24} = 5\left(\frac{\Delta(5\omega_{1}, \omega_{2})}{\Delta(\omega_{1}, \omega_{2})}\right)^{1/4}\tag{3}$$ and \begin{align} \sum_{j = 0}^{6}\left(\frac{\Delta(7\omega_{1}, \omega_{2})}{\Delta\{(\omega_{1} + j\omega_{2})/7, \omega_{2}\}}\right)^{1/24} &= 7\left(\frac{\Delta(7\omega_{1}, \omega_{2})}{\Delta(\omega_{1}, \omega_{2})}\right)^{1/6}\notag\\ &\,\,\,\,+ 49\left(\frac{\Delta(7\omega_{1}, \omega_{2})}{\Delta(\omega_{1}, \omega_{2})}\right)^{1/3}\tag{4} \end{align} Mordell goes on to mention that identities like $(3), (4)$ don't exist if we replace replace $5, 7$ by $11$. On the other hand he says that there is similar identity for modulo $13$.

This seems very weird because we do have a congruence identity $p(11n + 6) \equiv 0\pmod{11}$ and we don't have a similar partition congruence $\pmod{13}$. What Mordell probably means is that we do not have an expansion like $(1)$ or $(2)$ for $p(11n + 6)$ and at the same time we do have some formula like $(1), (2)$ for $p(13n + a)$ for some integer $a, 0 < a < 13$.

Is my understanding mentioned in last paragraph correct? If so, do we have an identity like $(1), (2)$ available in literature for modulo $13$? Any references regarding such an identity and its proof would be really helpful.


Herbert S. Zuckerman in his paper Identities analogous to Ramanujan's identities involving the partition function (published in Duke Mathematical Journal Vol 5,1939, pages 88-110) gives the following formula (which is rather cumbersome) $$\sum_{n=0}^{\infty} p(13n+6)q^n=11\prod_{n=1}^{\infty} \frac{(1-q^{13n})}{(1-q^n)^2}+468q \prod_{n=1}^{\infty} \frac{(1-q^{13n})^3}{(1-q^n)^4}+6422q^2 \prod_{n=1}^{\infty} \frac{(1-q^{13n})^5}{(1-q^n)^6}+43940q^3 \prod_{n=1}^{\infty} \frac{(1-q^{13n})^7}{(1-q^n)^8}+171366q^4 \prod_{n=1}^{\infty} \frac{(1-q^{13n})^9}{(1-q^n)^{10}}+371293q^5 \prod_{n=1}^{\infty} \frac{(1-q^{13n})^{11}}{(1-q^n)^{12}}+371293q^6 \prod_{n=1}^{\infty} \frac{(1-q^{13n})^{13}}{(1-q^n)^{14}}$$ All the coefficients on the right hand side except the first one ($11$) are divisible by $13$. Thus even though there is (not-so) nice generating function for $\sum p(13n+6)q^n$ there is no corresponding simple congruence.

Oddmund Kolberg in his paper Some identities involving the partition function (published in Mathematica Scandinavica, Vol 5, 1957, pages 77-92) gave a much simpler and systematic procedure to prove Ramanujan's generating functions described in this question. The same technique of Kolberg has been successfully applied by Göksal Bilgici and A. Bülent Ekin in their paper Some congruences for modulus 13 related to partition generating function (published in Ramanujan Journal, Vol 33, 2014, pages 197-218) to prove the above mentioned formula.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.