Ramanujan's Generating Function for Partitions Modulo 7

Ramanujan's Partition Congruences

Based on the empirical analysis of a table of partitions Ramanujan conjectured his famous partition congruences $$\boxed{\begin{align}p(5n + 4)&\equiv 0\pmod{5}\notag\\ p(7n + 5)&\equiv 0\pmod{7}\notag\\ p(11n + 6)&\equiv 0\pmod{11}\notag\end{align}}\tag{1}$$ and gave some of the most beautiful proofs for them (see here). In addition to these proofs he gave the following generating functions for $p(5n + 4), p(7n + 5)$: $$\sum_{n = 0}^{\infty}p(5n + 4)q^{n} = 5\frac{\{(1 - q^{5})(1 - q^{10})(1 - q^{15})\cdots\}^{5}}{\{(1 - q)(1 - q^{2})(1 - q^{3})\cdots\}^{6}}\tag{2}$$ and $$\begin{align}\sum_{n = 0}^{\infty}p(7n + 5)q^{n}&= 7\frac{\{(1 - q^{7})(1 - q^{14})(1 - q^{21})\cdots\}^{3}}{\{(1 - q)(1 - q^{2})(1 - q^{3})\cdots\}^{4}}\notag\\ &\,\,\,\,\,\,\,\,+ 49q\frac{\{(1 - q^{7})(1 - q^{14})(1 - q^{21})\cdots\}^{7}}{\{(1 - q)(1 - q^{2})(1 - q^{3})\cdots\}^{8}}\tag{3}\end{align}$$ We have already established $(2)$ in one of our posts and this post deals with the identity $(3)$ concerning generating function of partitions modulo $7$.

The identity $(3)$ is exceedingly difficult to prove using elementary techniques and most of the proofs involve heavy symbolic manipulation which is normally done using a software like MAPLE or MACSYMA. I tried to obtain some help from MSE to get a proof which used something possible via hand calculation but did not get any good answers there. I chanced to read an old paper titled "Some Identities involving the Partition Function" by Oddmund Kolberg which provided very nice and elementary proofs of the identities $(2)$ and $(3)$ in a systematic fashion. This post tries to elaborate on the concise proof available in that paper.

The Basic Technique

The main idea of the proof is to start with the following well known identities of Euler and Jacobi: $$f(-q) = (1 - q)(1 - q^{2})(1 - q^{3})\cdots = \sum_{n = -\infty}^{\infty}(-1)^{n}q^{n(3n + 1)/2}\tag{4}$$ and $$f^{3}(-q) = \{(1 - q)(1 - q^{2})(1 - q^{3})\cdots\}^{3} = \sum_{n = 0}^{\infty}(-1)^{n}(2n + 1)q^{n(n + 1)/2}\tag{5}$$ and split each of the series above based on the powers of $q$ modulo some prime number. Here we focus on the specific case where the prime number concerned is $7$. Thus we can write $$\begin{align}f(-q)&= \sum_{n = -\infty}^{\infty}(-1)^{n}q^{n(3n + 1)/2}\notag\\ &= g_{0} + g_{1} + g_{2} + g_{3} + g_{4} + g_{5} + g_{6}\notag\\ &= \sum_{s = 0}^{6}g_{s}\notag\\ &= \sum_{s = 0}^{6}\left(\sum_{n(3n + 1)/2\,\equiv\, s\pmod{7}}(-1)^{n}q^{n(3n + 1)/2}\right)\tag{6}\end{align}$$ and $$\begin{align}f^{3}(-q)&= \sum_{n = 0}^{\infty}(-1)^{n}q^{n(n + 1)/2}\notag\\ &= h_{0} + h_{1} + h_{2} + h_{3} + h_{4} + h_{5} + h_{6}\notag\\ &= \sum_{s = 0}^{6}h_{s}\notag\\ &= \sum_{s = 0}^{6}\left(\sum_{n \geq 0,\,n(n + 1)/2\,\equiv\,s\pmod{7}}(-1)^{n}(2n + 1)q^{n(n + 1)/2}\right)\tag{7}\end{align}$$ Clearly the expression $n(3n + 1)/2$ falls into one of the four class classes $0, 1, 2, 5$ modulo $7$ so that $$g_{3} = g_{4} = g_{6} = 0$$ Similarly we have $$h_{2} = h_{4} = h_{5} = 0$$ On the other hand we can evaluate $g_{2}$ by noting that $$\begin{align}n(3n + 1)/2 &\equiv 2\pmod{7}\notag\\ \Leftrightarrow 24\{n(3n + 1)/2\} + 1&\equiv 49 \equiv 0\pmod{7}\notag\\ \Leftrightarrow (6n + 1)^{2}&\equiv 0\pmod{7}\notag\\ \Leftrightarrow 6n + 1&\equiv 0\pmod{7}\notag\\ \Leftrightarrow n&\equiv 1\pmod{7}\notag\end{align}$$ We thus have $$\begin{align}g_{2}&= \sum_{n = -\infty}^{\infty}(-1)^{7n + 1}q^{(7n + 1)(21n + 4)/2}\notag\\ &= -\sum_{n = -\infty}^{\infty}(-1)^{7n}q^{2}q^{49n(3n + 1)/2}\notag\\ &= -q^{2}f(-q^{49})\notag\end{align}$$ In a similar fashion we can calculate $h_{6} = -7q^{6}f^{3}(-q^{49})$ so that $h_{6} = 7g_{2}^{3}$.

We thus have the following values of $g_{s}, h_{s}$: $$\begin{align}g_{2} = -q^{2}f(-q^{49}),\, g_{3} = g_{4} = g_{6} &= 0\notag\\ h_{6} = -7q^{6}f^{3}(-q^{49}) = 7g_{2}^{3},\,h_{2} = h_{4} = h_{5} &= 0\tag{8}\end{align}$$ Using the relation $\sum h_{s} = (\sum g_{s})^{3}$ we get $\sum h_{s} = (g_{0} + g_{1} + g_{2} + g_{5})^{3}$ and on equating the terms based on their modulo classes on each side (after cubing RHS) we get $$\boxed{\begin{align}3(g_{0}^{2}g_{2} + g_{1}^{2}g_{0} + g_{2}^{2}g_{5})&= h_{2} = 0\notag\\ 3(g_{1}^{2}g_{2} + g_{2}^{2}g_{0} + g_{5}^{2}g_{1})&= h_{4} = 0\notag\\ 3(g_{0}^{2}g_{5} + g_{2}^{2}g_{1} + g_{5}^{2}g_{2})&= h_{5} = 0\notag\\ g_{2}^{3} + 6g_{0}g_{1}g_{5}&= h_{6} = 7g_{2}^{3}\notag\end{align}}\tag{9}$$ Putting $\alpha = g_{0}/g_{2}, \beta = g_{1}/g_{2}, \gamma = g_{5}/g_{2}$ we can rewrite the above equations as $$\boxed{\begin{align}\alpha \beta^{2} + \alpha^{2} + \gamma &= 0\notag\\ \beta \gamma^{2} + \beta^{2} + \alpha &= 0\notag\\ \gamma \alpha^{2} + \gamma^{2} + \beta &= 0\notag\\ \alpha\beta\gamma &= 1\notag\end{align}}\tag{10}$$ We now relate these quantities with the partition function $p(n)$.

Relation with Partition Function $p(7n + 5)$

We have the generating function of the partition function $p(n)$ given by the formula $$\sum_{n = 0}^{\infty}p(n)q^{n} = \frac{1}{f(-q)}\tag{11}$$ We split the series on the left based on powers of $q$ modulo $7$ so that $\sum p(n)q^{n} = \sum_{s = 0}^{6}P_{s}$ with $P_{s}$ defined by $$P_{s} = \sum_{n = 0}^{\infty}p(7n + s)q^{7n + s}$$ We thus have the following equations $$\begin{align}P_{0} + P_{1} + P_{2} + P_{3} + P_{4} + P_{5} + P_{6} &= \frac{1}{f(-q)}\notag\\ g_{0} + g_{1} + g_{2} + g_{3} + g_{4} + g_{5} + g_{6} &= f(-q)\notag\end{align}$$ Multiplying the above equations and arranging terms based on their modulo classes we get $$\begin{align}g_{0}P_{0} + g_{6}P_{1} + g_{5}P_{2} + g_{4}P_{3} + g_{3}P_{4} + g_{2}P_{5} + g_{1}P_{6}&= 1\notag\\ g_{1}P_{0} + g_{0}P_{1} + g_{6}P_{2} + g_{5}P_{3} + g_{4}P_{4} + g_{3}P_{5} + g_{2}P_{6}&= 0\notag\\ g_{2}P_{0} + g_{1}P_{1} + g_{0}P_{2} + g_{6}P_{3} + g_{5}P_{4} + g_{4}P_{5} + g_{3}P_{6}&= 0\notag\\ g_{3}P_{0} + g_{2}P_{1} + g_{1}P_{2} + g_{0}P_{3} + g_{6}P_{4} + g_{5}P_{5} + g_{4}P_{6}&= 0\notag\\ g_{4}P_{0} + g_{3}P_{1} + g_{2}P_{2} + g_{1}P_{3} + g_{0}P_{4} + g_{6}P_{5} + g_{5}P_{6}&= 0\notag\\ g_{5}P_{0} + g_{4}P_{1} + g_{3}P_{2} + g_{2}P_{3} + g_{1}P_{4} + g_{0}P_{5} + g_{6}P_{6}&= 0\notag\\ g_{6}P_{0} + g_{5}P_{1} + g_{4}P_{2} + g_{3}P_{3} + g_{2}P_{4} + g_{1}P_{5} + g_{0}P_{6}&= 0\notag\end{align}$$ Using Cramer's rule we can calculate the desired sum $P_{5}$ corresponding to $p(7n + 5)$ as $P_{5} = D_{5}/D$ where $D$ and $D_{5}$ are the following determinants $$D = \begin{vmatrix}g_{0} & g_{6} & g_{5} & g_{4} & g_{3} & g_{2} & g_{1}\\ g_{1} & g_{0} & g_{6} & g_{5} & g_{4} & g_{3} & g_{2}\\ g_{2} & g_{1} & g_{0} & g_{6} & g_{5} & g_{4} & g_{3}\\ g_{3} & g_{2} & g_{1} & g_{0} & g_{6} & g_{5} & g_{4}\\ g_{4} & g_{3} & g_{2} & g_{1} & g_{0} & g_{6} & g_{5}\\ g_{5} & g_{4} & g_{3} & g_{2} & g_{1} & g_{0} & g_{6}\\ g_{6} & g_{5} & g_{4} & g_{3} & g_{2} & g_{1} & g_{0}\end{vmatrix}\tag{12}$$ $$D_{5} = -\begin{vmatrix}g_{1} & g_{0} & g_{6} & g_{5} & g_{4} & g_{2}\\ g_{2} & g_{1} & g_{0} & g_{6} & g_{5} & g_{3}\\ g_{3} & g_{2} & g_{1} & g_{0} & g_{6} & g_{4}\\ g_{4} & g_{3} & g_{2} & g_{1} & g_{0} & g_{5}\\ g_{5} & g_{4} & g_{3} & g_{2} & g_{1} & g_{6}\\ g_{6} & g_{5} & g_{4} & g_{3} & g_{2} & g_{0}\end{vmatrix}\tag{13}$$ Calculating these determinants above is bit of a challenge but is made possible by the vanishing of some of $g's$ and the relations $(9)$.

Evaluation of determinants $D, D_{5}$

If we take a closer look at $D$ we can see that it is the determinant of a circulant matrix $A$ and therefore we can easily find the eigenvalues of this matrix and thereby calculate the determinant $D$ as a product of eigenvalues. If $\omega$ is a $7^{\text{th}}$ root of unity (including $\omega = 1$ also) then we can see that the expression $$g_{0} + \omega g_{1} + \omega^{2}g_{2} + \omega^{3}g_{3}+ \omega^{4}g_{4}+ \omega^{5}g_{5}+ \omega^{6}g_{6}$$ is an eigenvalue of $A$ and the vector $$(1, \omega, \omega^{2}, \omega^{3}, \omega^{4}, \omega^{5}, \omega^{6})$$ is the corresponding eigenvector. If $\omega$ is a primitive $7^{\text{th}}$ root of unity then all the $7^{\text{th}}$ roots of unity are given by powers of $\omega$ and thus for each of $t = 0, 1, 2, 3, 4, 5, 6$ we obtain the eigenvalue $$\lambda_{t} = g_{0} + \omega^{t} g_{1} + \omega^{2t}g_{2} + \omega^{3t}g_{3}+ \omega^{4t}g_{4}+ \omega^{5t}g_{5}+ \omega^{6t}g_{6}$$ and therefore we have the determinant $D$ given by $$D = \prod_{t = 0}^{6}(g_{0} + \omega^{t} g_{1} + \omega^{2t}g_{2} + \omega^{3t}g_{3}+ \omega^{4t}g_{4}+ \omega^{5t}g_{5}+ \omega^{6t}g_{6}) = \prod_{t = 0}^{6}\sum_{s = 0}^{6}\omega^{st}g_{s}$$ Using the definition of $g_{s} = g_{s}(q)$ we can easily see that $\omega^{st}g_{s}(q) = g_{s}(\omega^{t}q)$ and hence we have $$\begin{align}D&= \prod_{t = 0}^{6}\sum_{s = 0}^{6}\omega^{st}g_{s} = \prod_{t = 0}^{6}\sum_{s = 0}^{6}g_{s}(\omega^{t}q) = \prod_{t = 0}^{6}f(-\omega^{t}q)\notag\\ &= \prod_{t = 0}^{6}\prod_{n = 1}^{\infty}(1 - \omega^{nt}q^{n}) = \prod_{n = 1}^{\infty}\prod_{t = 0}^{6}(1 - \omega^{tn}q^{n})\notag\\ &= \prod_{n \not\equiv 0\pmod{7}}(1 - q^{7n})\prod_{n \equiv 0\pmod{7}}(1 - q^{n})^{7} = \frac{f^{8}(-q^{7})}{f(-q^{49})}\tag{14}\end{align}$$ Before calculating $D_{5}$ we need to perform a direct calculation of the determinant $D$ using the usual definition of a determinant. To simplify the calculation we use the fact that $$g_{0} = \alpha g_{2}, g_{1} = \beta g_{2}, g_{5} = \gamma g_{2}, g_{3} = g_{4} = g_{6} = 0$$ and then we have $$D = g_{2}^{7}\begin{vmatrix}\alpha & 0 & \gamma & 0 & 0 & 1 & \beta\\ \beta & \alpha & 0 & \gamma & 0 & 0 & 1\\ 1 & \beta & \alpha & 0 & \gamma & 0 & 0\\ 0 & 1 & \beta & \alpha & 0 & \gamma & 0\\ 0 & 0 & 1 & \beta & \alpha & 0 & \gamma\\ \gamma & 0 & 0 & 1 & \beta & \alpha & 0\\ 0 & \gamma & 0 & 0 & 1 & \beta & \alpha\end{vmatrix} = g_{2}^{7}D'$$ Thanks to the presence of the zeroes in $D'$ and the fact that $\alpha\beta\gamma = 1$ the determinant $D'$ can be calculated by hand with not so unreasonable effort and we get $$\begin{align}D'&= (\alpha^{7} + \beta^{7} + \gamma^{7}) - 7(\alpha\beta^{5} + \beta\gamma^{5} + \gamma\alpha^{5})\notag\\ &\,\,\,\,\,\,\,\,+ 14(\alpha^{2}\beta^{3} + \beta^{2}\gamma^{3} + \gamma^{2}\alpha^{3}) + 8\tag{15}\end{align}$$ In the same manner we can calculate $D_{5} = -g_{2}^{6}D'_{5}$ where $$D'_{5} = \begin{vmatrix}\beta & \alpha & 0 & \gamma & 0 & 1\\ 1 & \beta & \alpha & 0 & \gamma & 0\\ 0 & 1 & \beta & \alpha & 0 & 0\\ 0 & 0 & 1 & \beta & \alpha & \gamma\\ \gamma & 0 & 0 & 1 & \beta & 0\\ 0 & \gamma & 0 & 0 & 1 & \alpha\end{vmatrix}$$ and $$\begin{align}D'_{5}&= (\alpha\beta^{5} + \beta\gamma^{5} + \gamma\alpha^{5}) - 4(\alpha^{2}\beta^{3} + \beta^{2}\gamma^{3} + \gamma^{2}\alpha^{3})\notag\\ &\,\,\,\,\,\,\,\,+ 3(\alpha^{3}\beta + \beta^{3}\gamma + \gamma^{3}\alpha) - 8\tag{16}\end{align}$$ Since we know the value of $D'$ in a closed form as $$D' = \frac{D}{g_{2}^{7}} = -\frac{f^{8}(-q^{7})}{q^{14}f^{8}(-q^{49})}\tag{17}$$ the real issue is to evaluate $D'_{5}$ in a closed form. For this we try to find relation between $D'$ and $D'_{5}$. We use a new set of variables $y_{1}, y_{2}, y_{3}$ defined by $$y_{1} = \alpha^{3}\beta, y_{2} = \beta^{3}\gamma, y_{3} = \gamma^{3}\alpha\tag{18}$$ and then apply equation $(10)$ to evaluate the following expressions $$\begin{align}\alpha^{2}\beta^{3}&= y_{1}y_{2} = - y_{1} - 1\tag{19a}\\ \beta^{2}\gamma^{3} &= y_{2}y_{3} = - y_{2} - 1\tag{19b}\\ \gamma^{2}\alpha^{3} &= y_{3}y_{1} = - y_{3} - 1\tag{19c}\end{align}$$ $$y_{1}y_{2}y_{3} = 1\tag{20}$$ $$\alpha\beta^{5} = y_{1} - y_{2} + 1, \beta\gamma^{5} = y_{2} - y_{3} + 1, \gamma\alpha^{5} = y_{3} - y_{1} + 1\tag{21}$$ $$\begin{align}\alpha^{7}&= - y_{1}^{2} + y_{1} - y_{3} - 1\tag{22a}\\ \beta^{7}&= - y_{2}^{2} + y_{2} - y_{1} - 1\tag{22b}\\ \gamma^{7}&= - y_{3}^{2} + y_{3} - y_{2} - 1\tag{22c}\end{align}$$ Using equation $(15)$ and the above relations between $y_{i}$ we get $$\begin{align}D'&= -(y_{1}^{2} + y_{2}^{2} + y_{3}^{2}) - 14(y_{1} + y_{2} + y_{3}) - 58\notag\\ &= -(y_{1} + y_{2} + y_{3})^{2} - 16(y_{1} + y_{2} + y_{3}) - 64\notag\\ &= -(y_{1} + y_{2} + y_{3} + 8)^{2}\notag\end{align}$$ Using equation $(17)$ we get $$y_{1} + y_{2} + y_{3} + 8 = \pm\frac{f^{4}(-q^{7})}{q^{7}f^{4}(-q^{49})}$$ and clearly we can see that the expression $y_{1}$ begins with $-q^{-7}$ so that we need to take the negative sign on RHS in above equation. We finally have $$y_{1} + y_{2} + y_{3} = -\frac{f^{4}(-q^{7})}{q^{7}f^{4}(-q^{49})} - 8\tag{23}$$ In the same manner we can see that $$D'_{5} = 7(y_{1} + y_{2} + y_{3}) + 7 = -7\frac{f^{4}(-q^{7})}{q^{7}f^{4}(-q^{49})} - 49$$ We now have $$\begin{align}\sum_{n = 0}^{\infty}p(7n + 5)q^{7n + 5}&= P_{5} = \frac{D_{5}}{D} = \frac{-g_{2}^{6}D'_{5}}{g_{2}^{7}D'} = -\frac{D'_{5}}{g_{2}D'}\notag\\ &= \frac{1}{q^{2}f(-q^{49})}\cdot\frac{q^{14}f^{8}(-q^{49})}{f^{8}(-q^{7})}\left(7\frac{f^{4}(-q^{7})}{q^{7}f^{4}(-q^{49})} + 49\right)\notag\\ &= q^{5}\left(7\frac{f^{3}(-q^{49})}{f^{4}(-q^{7})} + 49q^{7}\frac{f^{7}(-q^{49})}{f^{8}(-q^{7})}\right)\notag\end{align}$$ Cancelling $q^{5}$ from both sides and replacing $q^{7}$ by $q$ we get the Ramanujan's identity $$\sum_{n = 0}^{\infty}p(7n + 5)q^{n} = 7\frac{f^{3}(-q^{7})}{f^{4}(-q)} + 49q\frac{f^{7}(-q^{7})}{f^{8}(-q)}$$ The above proof is totally elementary and it is probable that Ramanujan did something similar to obtain his formula. The same technique of splitting series for $f(-q)$ and $1/f(-q)$ into multiple parts based on the modulo classes of the powers of $q$ can be used to prove the identity $(2)$ for partitions modulo $5$. In the above technique the only hard part is the calculation of determinants by brute force which is somewhat laborious. For a person like Ramanujan with unmatched powers of symbolic manipulation this would have been a child's play.

Partition Congruence $p(49n + 47) \equiv 0\pmod{49}$

Ramanujan's motivation for the identity $(3)$ was to prove the partition congruence $$p(49n + 47) \equiv 0\pmod{49}\tag{24}$$ and he gave a very short proof of this congruence in the following manner. Using binomial theorem it is easy to check that $$(1 - q)^{7} = 1 - q^{7} + 7J$$ where $J$ is a power series in $q$ with integral coefficients and hence we can write $$\frac{1 - q^{7}}{(1 - q)^{7}} = 1 + 7J$$ where $J$ again represents some power series in $q$ with integer coefficients (note that each instance of $J$ may represent a different power series in $q$). Replace $q$ with $q^{2}, q^{3}, \ldots$ and multiplying the results we get $$\frac{(1 - q^{7})(1 - q^{14})(1 - q^{21})\cdots}{\{(1 - q)(1 - q^{2})(1 - q^{3})\cdots\}^{7}} = \frac{f(-q^{7})}{f^{7}(-q)}= 1 + 7J\tag{25}$$ Using identity $(3)$ we can write $$\dfrac{{\displaystyle \sum_{n = 0}^{\infty}p(7n + 5)q^{n + 1}}}{7f^{2}(-q^{7})} = qf^{3}(-q)\frac{f(-q^{7})}{f^{7}(-q)} + 7q^{2}\frac{f^{5}(-q^{7})}{f^{8}(-q)}$$ and using $(25)$ we can now write $$\dfrac{{\displaystyle \sum_{n = 0}^{\infty}p(7n + 5)q^{n + 1}}}{7f^{2}(-q^{7})} = qf^{3}(-q) + 7J = \sum_{n = 0}^{\infty}(-1)^{n}(2n + 1)q^{n(n + 1)/2 + 1} + 7J$$ We next need to analyze the coefficient of $q^{7n}$ on both sides. On the RHS we can see that $$\begin{align}\frac{n(n + 1)}{2} + 1 &\equiv 0\pmod{7}\notag\\ \Leftrightarrow n(n + 1) + 2 &\equiv 0\pmod{7}\notag\\ \Leftrightarrow n(n + 1) &\equiv 5\pmod{7}\notag\\ \Leftrightarrow 4n(n + 1) + 1 &\equiv 0\pmod{7}\notag\\ \Leftrightarrow (2n + 1)^{2} &\equiv 0\pmod{7}\notag\\ \Leftrightarrow (2n + 1) &\equiv 0\pmod{7}\notag\end{align}$$ so that the coefficient of $q^{7n}$ on the RHS is a multiple of $7$. It follows that the coefficient of $q^{7n}$ in $\sum_{n = 0}^{\infty}p(7n + 5)q^{n + 1}$ is a multiple of $49$. Thus we have $p(49n + 47)\equiv 0\pmod{49}$.

Print/PDF Version