El análisis de la función $$f(-q) = (1 - q)(1 - q^{2})(1 - q^{3})\cdots$$ by replacing $q$ with $q^{1/5}$, Ramanujan is able to calculate the sum $$\sum_{n = 0}^{\infty}p(n)q^{n/5} = \frac{1}{f(-q^{1/5})}$$ where $p(n)$ is the number of partitions of $n$. Further, on equating coefficients of $q^{4/5}$ on both sides, he gets the beautiful result $$\sum_{n = 0}^{\infty}p(5n + 4)q^{n} = 5\frac{\{(1 - q^{5})(1 - q^{10})(1 - q^{15})\cdots\}^{5}}{\{(1 - q)(1 - q^{2})(1 - q^{3})\cdots\}^{6}}$$ which can be used to show $$\begin{aligned}p(5n + 4) &\equiv 0\pmod{5}\\p(25n + 24) &\equiv 0\pmod{25}\end{aligned}$$ (esto se muestra en detalle en mi blog).
El próximo Ramanujan menciona que la misma técnica se puede aplicar mediante el análisis de $$\sum_{n = 0}^{\infty}p(n)q^{n/7} = \frac{1}{f(-q^{1/7})}$$ and equating the coefficient of $q^{5/7},$ to get another beautiful identity $$\sum_{n = 0}^{\infty}p(7n + 5)q^{n} = 7\frac{\{(1 - q^{7})(1 - q^{14})(1 - q^{21})\cdots\}^{3}}{\{(1 - q)(1 - q^{2})(1 - q^{3})\cdots\}^{4}} + 49q\frac{\{(1 - q^{7})(1 - q^{14})(1 - q^{21})\cdots\}^{7}}{\{(1 - q)(1 - q^{2})(1 - q^{3})\cdots\}^{8}}$$ This can be used to prove the congruences $$\begin{aligned}p(7n + 5) &\equiv 0\pmod{7}\\p(49n + 47) &\equiv 0\pmod{49}\end{aligned}$$ Unfortunately, in his characteristic style, Ramanujan omits the proof for the case of $f(-p^{1/7})$. I tried to use the approach mentioned by Ramanujan for $f(-q^{1/7})$, y es llevado a engorroso expresiones (multiplicando tres casi similar polinomios de 6 grados cada simbólica de los coeficientes). Huelga decir que ninguno de los en línea simbólica paquetes como wolfram alpha o sympy de manejar complejas de manipulación simbólica. Supongo que Ramanujan de alguna manera simplificada de estos cálculos y de ese modo se obtiene una muy simple resultado de apariencia.
¿Alguien tiene alguna referencia de un enfoque más sencillo para el establecimiento de la identidad relativa $\sum p(7n + 5)q^{n}$?
Actualización: Para agregar más detalles, el uso de Euler pentagonal fórmula $$f(-q^{1/7}) = \sum_{n = -\infty}^{\infty}(-1)^{n}(q^{1/7})^{n(3n + 1)/2}$$ The numbers $n(3n + 1)/2$ fall into one of the following four classes modulo $7$ as $n$ takes integer values: $$\begin{aligned}n(3n + 1)/2\,\,&\equiv 0\pmod{7}\text{ if } n &\equiv 0, 2\pmod{7}\\ &\equiv 1\pmod{7}\text{ if } n &\equiv 3, 6\pmod{7}\\ &\equiv 2\pmod{7}\text{ if } n &\equiv 1\pmod{7}\\ &\equiv 5\pmod{7}\text{ if } n &\equiv 4, 5\pmod{7}\end{aligned}$$ and hence, if we put $r = p^{1/7}$, then we see that $$P = f(-r) = A_{0} + A_{1}r + A_{2}r^{2} + A_{3}r^{5}$$ where $A_{i}$ are power series in $p$ and do not involve any fractional powers of $q$. Further, if we replace $r = q^{1/7}$ by $r\zeta^{i}$ for $i = 1, 2, \ldots, 6$, where $\zeta$ is a primitive $7^{\text{th}}$ root of unity, then the product $Q = \prod_{i = 0}^{6}f(-r\zeta^{i})$ can be expressed without any fractional powers of $q$. In fact it is easy to show that $$Q = \prod_{i = 0}^{6}f(-r\zeta^{i}) = \frac{f^{8}(-q)}{f(-q^{7})}$$ The idea is then to find $P/P = \prod_{i = 1}^{6}f(-r\zeta^{i})$, and we try to find the coefficient of $r^{5}$ in $P/P$. Suppose that the coefficient is $R$. Then we have $$\sum_{n = 0}^{\infty}p(n)q^{n/7} = 1/P = (1/Q)(Q/P)$$ and hence $$\sum_{n = 0}^{\infty}p(7n + 5)q^{n} = R/Q$$ Further Update: I recently found a paper by Oddmund Kolberg titled "Some Identities Involving the Partition Function" which provides an elementary proof of Ramanujan's identity for $p(7n + 5)$. La misma prueba se ha presentado en mi blog.
