Apply the <Serre derivative> in weight twelve to the <modular discriminant>. The preceding result makes $D_{12}\Delta=D\Delta-E_2\Delta$ a weight-fourteen <cusp form>. The <vanishing of weight-fourteen level-one cusp forms> follows directly from the <valence formula for the modular group>: a nonzero such form has order at infinity at least one, and $f(i)=i^{14}f(i)=-f(i)$ forces order at $i$ at least one. Thus its weighted number of zeros would be at least $1+1/2>14/12$, a contradiction. Therefore $D\Delta=E_2\Delta$.

Compare the <Fourier coefficients> at positive indices using $\Delta=\sum_{n\geq1}\tau(n)q^n$ and $E_2=1-24\sum_{r\geq1}\sigma_1(r)q^r$. This gives $n\tau(n)=\tau(n)-24\sum_{r=1}^{n-1}\sigma_1(r)\tau(n-r)$. For the <Ramanujan tau function>, the concise recurrence is
$$
\boxed{(1-n)\tau(n)=24\sum_{r=1}^{n-1}\sigma_1(r)\tau(n-r)\qquad(n\geq1).}
$$
At $n=1$ the sum is empty and both sides are zero; normalization supplies $\tau(1)=1$. For instance the next coefficient is $\tau(2)=-24$.

Solution

Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 137 / 3 / c / Solution

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{title2=$\tau(n)$}
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The Ramanujan tau function gives the <Fourier coefficients> at positive indices of the <modular discriminant>: $\Delta(z)=\sum_{n\geq1}\tau(n)q^n$, with $\tau(1)=1$. The <Serre derivative> and <vanishing of weight-fourteen level-one cusp forms> give $q\,d\Delta/dq=E_2\Delta$ and hence the convolution recurrence
$$
(1-n)\tau(n)=24\sum_{r=1}^{n-1}\sigma_1(r)\tau(n-r).
$$

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