[Extracted from the manuscripts of the author by G. H. Hardy]¹

1. Let

Then it is well known that Further, let where $\sigma_k (n)$ is the sum of the $k$th powers of the divisors of $n$; so that and in particular

Then [it may be deduced from the theory of the elliptic modular functions, and has been shewn by the author in a direct and elementary manner², that, when $r+s$ is odd, and $r \lt s, \Phi_{r,s}(x)$ is expressible as a polynomial in $P, Q$, and $R$, in the form $$ \Phi_{r,s}(x) = \sum k_{l, m, n} P^l Q^m R^n, $$ where $$ l - 1 \leq \: {\rm Min}\: (r,s), \: 2l + 4m + 6n = r+s + 1.$$ In particular³]

Modulus 5

2. We denote generally by $J$ an integral power-series in $x$ whose coefficients are integers. It is obvious from (1.12) that $$ Q = 1 + 5J.$$ Also $n^5-n\equiv 0\pmod{5}$, and so, from (1.11) and (1.13), $$ R = P + 5J.$$ Hence $$Q^3 - R^2 = Q(1+5J)^2 - (P+5J)^2 = Q - P^2 + 5J.$$ Using (1.4), (1.71), and (1.51), we obtain

Also $$(1-x)^{25} = 1-x^{25} + 5J, $$ and so $$ (f(x))^{25} = f(x^{25}) + 5J,$$ But $$ \frac{1}{f(x)} = 1 + p(1) x+ p(2) x^2 + \cdots, $$ and therefore, by (2.1) and (2.2),

Multiplying by 2, rejecting multiples of 5, and replacing $f(x^{25})$ by its expansion given by (1.3), we obtain \begin{eqnarray*} (x - x^{26} - x^{51} + x^{126} + \cdots ) ( 1+ p(1) x + p(2) x^2 + \cdots) = \sum^\infty_{n=1} n \sigma_1 (n) x^n + 5J. \end{eqnarray*}

Hence

the numbers 1, 26, 51,{\ldots} being the numbers of the forms $$ \frac{25}{2} n (3n-1) + 1, \: \frac{25}{2} n (3n+1) + 1, $$ or, what is the same thing, of the forms $$\frac{1}{2} (5n-1) (15n - 2), \: \frac{1}{2} (5n+1)(15n+2).$$ In particular it follows from (2.3) that

Modulus 7

It is obvious from (1.13) that $$ R = 1 + 7J.$$ Also $n^7 - n \equiv 0 \pmod{7}$, and so, from (1.11) and (1.61), $$ Q^2 = P + 7J.$$ Hence \begin{align*} (Q^3 - R^2)^2 = (PQ - 1 + 7J)^2 &= P^2Q^2 - 2PQ + 1 + 7J \\ &= P^2 - 2PQ + R + 7J. \end{align*} But, from (1.72) and (1.81), \begin{align*} P^3 - 2PQ + R & = 144 \sum^\infty_{n=1} (5n \sigma_3 (n) - 12n^2 \sigma_1 (n) ) x^n\\ & = \sum^\infty_{n=1} (n^2 \sigma_1 (n) - n \sigma_3 (n)) x^n + 7J. \end{align*} And therefore

Again (by the same argument which led to (2.2)) we have Combining (3.1) and (3.2), we obtain From (3.3) it follows (just as (2.4) and (2.5) followed from (2.3)) that the numbers, 2, 51, 100, {\ldots} being those of the forms $$ \frac{1}{2} (7n - 1) (21n - 4), \:\:\frac{1}{2} (7n+1) (21n + 4); $$ and that

Modulus 11

It is obvious from (1.62) that

Also $n^{11} - n \equiv 0\pmod{11}$, and so, from (1.11) and (1.63), It is easily deduced that [For \begin{align*} &\quad (Q^3 - 3R^2)^5 - Q(Q^3-3R^2)^3 - R(Q^3-3R^2)^2 + 6QR \\ &= (Q^3 -3R^2)^5 - Q^3R^2(Q^3-3R^2)^3 - Q^3R^4(Q^3-3R^2)^2 + 6Q^6R^6 + 11J \\ &= Q^{15} - 16Q^{12}R^2 + 98 Q^9 R^4 - 285 Q^6 R^6 + 423 Q^3 R^8 - 243 R^{10} + 11J\\ &= (Q^3 - R^2)^5 + 11J \end{align*} by (4.1), and (4.3) then follows from (4.2).]

Again, [if we multiply (1.74), (1.83), (1.92), and (1.93) by $-1$, 3, $-4$, and $-1$, and add, we obtain, on rejecting multiples of 11,] $$ P^5 - 3P^3Q - 4P^2R + 6QR = - 5\Phi_{1,8} + 3 \Phi_{2,7} + 3\Phi_{3,6} - \Phi_{4,5} + 11J;$$ and from this and (4.3) follows

But (by the same argument which led to (2.2) and (3.2)) we have From (4.4) and (4.5) \begin{align*} x^5 \frac{f(x^{121})}{f(x)} &= x^5 (f(x))^{120} + 11J = 1728^5 x^5 (f(x))^{120} + 11J \\ &= (Q^3 - R^2)^5 + 11J \\ &= - \sum^\infty_{n=1} (5n \sigma_7 (n) - 3n^2 \sigma_5 (n) - 3n^3 \sigma_3 (n) + n^4 \sigma_1 (n)) x^n + 11J. \end{align*} It now follows as before that 5, 126, 247, \ldots being the numbers of the forms $$ \frac{1}{2} (11n - 2) (33n - 5), \: \frac{1}{2}(11n + 2) (33n + 5); $$ and in particular that

5. If we are only concerned to prove (4.7), it is not necessary to assume quite so much.

Let us write $\vartheta$ for the operation $x \frac{d}{dx}$. Then⁴ we have

From these equations we deduce [by straightforward calculation \begin{align*} 864 \vartheta^4 P &= P^5-10 P^3Q - 15PQ^2 + 20P^2R + 4QR, \\ 72\vartheta^3 Q &= 5P^3Q + 15PQ^2 - 15P^2R - 5QR, \\ 24\vartheta^2 R &= -14PQ^2 + 7P^2 R + 7QR. \end{align*} The left-hand side of each of these equations is of the form $$x \frac{dJ}{dx}.$$ Multiplying by 1, 8, and 2, adding and rejecting multiples of 11, we find We have also, by (5.11), $$ 6P^2 R - 6QR = 72x R \frac{dP}{dx}. $$ But, differentiating (4.2), and using (4.1), we obtain \begin{align*} 72 x R \frac{dP}{dx} &= 36x R \left(-3Q^2 \frac{dQ}{dx} + 6R \frac{dR}{dx} \right) + 11J \\ & = - 108x Q \frac{dQ}{dx} + 216x R^2 \frac{dR}{dx} + 11J\\ &= x \frac{dJ}{dx} + 11J. \end{align*} Hence From (5.2) and (5.3) we deduce $$ P^5 - 3P^3Q - 4P^2R + 6QR = x \frac{dJ}{dx} + 11J,$$ and from (4.3)] Finally, from (4.5) and (5.4), \begin{align*} x^5 \frac{f(x^{121})}{f(x)} &= x^5 (f(x))^{120} + 11J = (Q^3 - R^2)^5 + 11J\\ &= x \frac{dJ}{dx} + 11J. \end{align*} As the coefficient of $x^{11m}$ on the right-hand side is a multiple of 11, (4.7) follows immediately.

²Ramanujan, p. 165 [pp. 181 – 183].

³ Ramanujan, pp. 163 – 165 [pp. 180 – 181] (Tables I to III). Ramanujan carried the calculation of formulæ of this kind to considerable lengths, the formula of Table I being \begin{eqnarray*} 7 709 321 041 217 + 32 640 \Phi_{0,31} (x) = 764 412 173 217 Q^8 \\ + 5 323 905 468 000 Q^5 R^2 + 1 621 003 400 000 Q^2 R^4. \end{eqnarray*} It is worth while to quote one such formula; for it is impossible to understand Ramanujan without realising his love of numbers for their own sake.

⁴Ramanujan, p.165 [pp. 181].