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While studying Ramanujan's tau function, I observed that the function satisfies a beautiful identity that I had not seen previously in the literature.

Let $\tau(n)$ be Ramanujan's tau function, such that $$q\prod_{n=1}^\infty (1-q^n)^{24} = \sum_{n=1}^\infty \tau(n)\,q^n,$$ then $$\tau(2)=8\dfrac{\tau(8)-2\tau(6)}{\tau(4)-15 \cdot 2^{11}}$$

I believe it is well known due to the popularity of the function. Can anyone prove the identity? The identity implies the congruence property $$\tau(2)\equiv0\pmod{8}$$ Thus proving the identity is akin to proving the congruence property.

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  • $\begingroup$ let me know if I edited wrong! :) $\endgroup$ – Ant Jul 16 '15 at 20:08
  • $\begingroup$ You edited nicely. $\endgroup$ – Nicco Jul 16 '15 at 20:13
  • $\begingroup$ if you are referring to en.wikipedia.org/wiki/Ramanujan_tau_function#Values the values at integers are integers, and tabulated for small $n.$ $\endgroup$ – Will Jagy Jul 16 '15 at 20:22
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This answer is meant to demonstrate some ways of calculating $\tau(n)$, ranging from elegant via straight to weird.

Elegant: Ramanujan's $\tau(n)$ is a multiplicative arithmetic function; so $\tau(mn) = \tau(m)\,\tau(n)$ whenever $\gcd(m,n)=1$. However, its multiplicativity properties are also known for non-coprime $m,n$: $$\begin{align} \tau(m)\,\tau(n) &= \sum_{d\mid\gcd(m,n)} d^{11}\tau\!\left(\frac{mn}{d^2}\right) \\ \tau(mn) &= \sum_{d\mid\gcd(m,n)} \mu(d)\,d^{11}\tau\!\left(\frac{m}{d}\right)\tau\!\left(\frac{n}{d}\right) \end{align}$$ where $\mu$ is the Möbius function. Since $\tau$ is not the zero function, its multiplicativity implies $\tau(1)=1$, which I will implicitly use from now on. From the special multiplicativity properties above we get $$\begin{align} \tau(4) &= \tau(2\cdot 2) = \tau(2)^2 - 2^{11} \\ \tau(6) &= \tau(2)\,\tau(3) \\ \tau(8) &= \tau(2\cdot 4) = \tau(2)\,\tau(4) - 2^{11}\tau(2) = \tau(2)^3 - 2^{12}\tau(2) \end{align}$$ Plugging in and assuming $\tau(2)\neq0$, which will be justified later, we find that your claim is equivalent to $$7\,\tau(2)^2 = 16\,\tau(3) \tag{*}$$

Straight: So let us work out $\tau(2),\tau(3)$ now. Denoting $$\begin{align} \operatorname{D}(q) &= q\prod_{n=1}^\infty(1-q^n)^{24} = \sum_{n=1}^\infty\tau(n)\,q^n && \text{(modular discriminant)} \\ \operatorname{P}(q) &= 1 - 24\sum_{n=1}^\infty \sigma(n)\,q^n && \text{(quasi-modular Eisensein series)} \\ \text{where}\quad \sigma(n) &= \sum_{d\mid n}d && \text{(divisor sum)} \end{align}$$ we have $$q\frac{\mathrm{d}\operatorname{D}(q)}{\mathrm{d}q} = \operatorname{P}(q) \operatorname{D}(q)$$ which after expansion into $q$-series and comparing coefficients yields the recurrence relation $$(n-1)\,\tau(n) = -24\sum_{k=1}^{n-1}\sigma(k)\,\tau(n-k)$$ In particular, $$\begin{align} \tau(2) &= -24 \\ \tau(3) &= -12\left(3 + \tau(2)\right) = 2^2 3^2 7 = 252 \end{align}$$ and that fulfills $(*)$.

Weird: In fact, $(*)$ is a direct consequence of another general recurrence which I had worked out years ago just for fun, without finding any use for it.

Namely, $\operatorname{P}(q)$ fulfills a differential equation of order 3 and Chazy type III: $$2\operatorname{P}'''\!{} - 2\operatorname{P}\operatorname{P}''\!{} + 3\operatorname{P}'^2 = 0 \quad\text{where}\quad (\,)' = q\frac{\mathrm{d}(\,)}{\mathrm{d}q}$$ Plugging in $\operatorname{P} = \frac{\operatorname{D}'}{\operatorname{D}}$ gives a fourth-order differential equation for $\operatorname{D}$: $$2\operatorname{D}^3\operatorname{D}''''\!{} - 10\operatorname{D}^2\operatorname{D}'\operatorname{D}'''\!{} - 3\operatorname{D}^2\operatorname{D}''^2\!{} + 24\operatorname{D}\operatorname{D}'^2\operatorname{D}''\!{} - 13\operatorname{D}'^4 = 0$$ Plugging in the $q$-series for $\operatorname{D}$ then results in a nonlinear recurrence relation for $\tau(n)$ that does not use other arithmetic functions: $$\begin{align} 2(n-2)(n-1)^3\tau(n) &= \sum_{\substack{a,b,c,d=1,\ldots,n-1\\a+b+c+d=n+3}} f(a,b,c,d)\,\tau(a)\,\tau(b)\,\tau(c)\,\tau(d) \\ \text{with}\quad f(a,b,c,d) &= -2 a^4 + 10 a^3 b + 3 a^2 b^2 - 24 a^2 b c + 13 a b c d \end{align}$$ Surely you want to quit, but we are now ready to apply this for $n=3$. Then $(a,b,c,d)$ runs through the six distinct permutations of $(1,1,2,2)$, and we get $$16\,\tau(3) = (15+34+16-14-80+36)\,\tau(2)^2 = 7\,\tau(2)^2$$ which confirms $(*)$ without even needing an explicit value for $\tau(2)$.

As mentioned before, I do not believe such relations to be of great interest; but if you are in a playful mood, working out such stuff might be fun.

By the way, one can work out an order-$3$ differential equation for $\operatorname{D}$ (which unfortunately is even more complex) and try to work out a recurrence relation for $\tau(n)$ from that. Interested?

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The identity follows immediately from the fact that 1728 can be expressed in terms of ramanujan tau function in two different ways,namely

$$2\tau(2){(\tau(3)-2^{11})}-\tau(8)=12^{3}$$

And

$$-\dfrac{\tau(2){(\tau(4)+2^{11})}}{8}=12^{3}$$

After equating the two identities and using the multiplicative property of the function, $\tau(mn)=\tau(m)\tau(n)$ for $\gcd(m,n)=1$, Then we formally obtain the identity after some simple algebraic manipulation.

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  • $\begingroup$ Quite witty and observant! $\endgroup$ – T.A.Tarbox Mar 17 '17 at 3:19
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Looking at the link,

$\tau(2)=8\dfrac{\tau(8)-2\tau(6)}{\tau(4)-15 \cdot 2^{11}} $

is $-24 =8\frac{84480-2\ (-6048)}{-1472-15\cdot 2048} =-24 $

so it checks.

How did you find it?

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  • $\begingroup$ I found it by simple observation $\endgroup$ – Nicco Jul 16 '15 at 20:41

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