# A curious property of Ramanujan's function $\tau(n)$

As it is well known, Ramanujan's $\tau(n)$ function can be defined through the power series expansion of the modular discriminant: $$\Delta(q)=q\prod\limits_{n=1}^\infty (1-q^n)^{24}=\sum \limits_{n=1}^\infty \tau(n)q^n=q-24q^2+252q^3-1472q^4+4830q^5+\ldots.$$ In the short paper http://arxiv.org/abs/1408.2083 (Moonshine and the Meaning of Life, by Yang-Hui He and John McKay) a curious observation was made that $$\sum \limits_{n=1}^{24}\tau(n)^2\equiv 42 \;\; (\mathrm{mod} \;70).$$ Another observation of the same paper is that $$\sum \limits_{n=1}^{24}c(n)^2\equiv 42 \;\; (\mathrm{mod} \;70),$$ where $c(n)$ are defined through the power series expansion of the $SL_2(\mathbb{Z})$ elliptic modular function $j(q)$: $$j(q)=\sum \limits_{n=-1}^\infty c(n)q^n=q^{-1}+744 + 196884q + 21493760q^2+\ldots.$$

In the abstract, the authors write that "The observation is purely for the sake of entertainment and could be of some diversion to a mathematical audience". Nevertheless, is there any deep mathematics behind these curious observations?

• In the first statement, this is just a fixed sum, right? So, it must be some number, which happens to be 42... – Per Alexandersson Oct 31 '14 at 11:32
• Right. But the same number appears in the second statement indicating, perhaps, that 42 should have some meaning and is not merely an accident. – Zurab Silagadze Oct 31 '14 at 12:58
• "42 is the Answer to the Ultimate Question of Life, the Universe and Everything. This Answer was first calculated by the supercomputer Deep Thought after seven and a half million years of thought. This shocking answer resulted in the construction of an even larger supercomputer, named Earth, which was tasked with determining what the question was in the first place." – Joël Oct 31 '14 at 14:11
• "the number 70, which is chosen a bit ad hoc" -- but how about $1^2+2^2+\cdots+24^2=70^2$? – Zurab Silagadze Oct 31 '14 at 14:41
• The combination (24,70) is somewhat unique. See mathworld.wolfram.com/CannonballProblem.html – Zurab Silagadze Oct 31 '14 at 14:50

## 1 Answer

(Too long for a comment.)

Since the authors point out their two observations in a "jocund" air, we can match their observations with another pair.

I. The Baby monster

Just like the j-function $j(\tau)$ above and the Monster, the modular function $j_{2A}(\tau)$ that can be related to the Baby monster,

\begin{aligned}j_{2A}(\tau) &=\Big(\big(\tfrac{\eta(\tau)}{\eta(2\tau)}\big)^{12}+2^6 \big(\tfrac{\eta(2\tau)}{\eta(\tau)}\big)^{12}\Big)^2 \\ &=\sum\limits_{n=-1}^{\infty}a(n)q^n\\&= q^{-1} + \color{blue}{104} + 4372q + 96256q^2 + 1240002q^3+\cdots \end{aligned}

also has,

$$\sum \limits_{n=1}^{24}a(n)^2\equiv 42 \;\; (\mathrm{mod} \;70),$$

Recall that $e^{\pi\sqrt{58}} =396^4-\color{blue}{104}.00000017\dots$.

II. Modular lambda function

Given the modular lambda function $\lambda(\tau)=\lambda$ such that,

$$j(\tau) = \frac{256(1-\lambda+\lambda^2)^3}{\lambda^2 (1-\lambda)^2}$$

and,

\begin{aligned}\lambda(\tau) &= \Big(\tfrac{\sqrt{2}\,\eta\big(\tfrac{\tau}{2}\big)\eta^2(2\tau)}{\eta^3(\tau)}\Big)^8\\ &=\sum\limits_{n=1}^{\infty}b(n)q^n\\ &=16q - 128q^2 + 704 q^3 - 3072q^4 + 11488q^5 - 38400q^6 + \dots \end{aligned}

then,

$$\sum \limits_{n=1}^{24}b(n)^2\equiv 42 \;\; (\mathrm{mod} \;70),$$

Of course, it is well-known that,

$$1^2+2^2+3^2+\dots+24^2 = 70^2$$

It should be interesting if, for these four related functions, there is a reason for the congruences other than whimsy.

$\color{red}{Update}$

P.S. By a really serendipitous error almost straight from the pages of the Hitchhiker's Guide, it turns out the same $24$ coefficients $s(n)$ of all four sequences also obey,

$$\sum_{n=1}^{24} \frac{s(n)^2+42}{70} = \,\text{Integer}$$

The discussion below should illustrate how the error was found.

• Interesting! However, I have checked that in all four cases the correct number is 42, not -42. Note the definition of $a\equiv b (\mathrm{mod}\;\;n)$: $a-b$ is divisible by $n$. Please check and correct. Also I noticed that all four sums are congruent to zero (mod 14)! I suggest you to publish your findings on arxiv. – Zurab Silagadze Apr 30 '15 at 12:18
• @ZurabSilagadze: Regarding the mod, for the j-function, I have $$\frac{\sum_{n=1}^{24}c(n)^2-(\color{red}{-}42)}{70} =19344611534578279675385464545221922990834243736585$$ whereas, $$\frac{\sum_{n=1}^{24}c(n)^2-(42)}{70} =\frac{96723057672891398376927322726109614954171218682781}{5}$$hence $$\sum_{n=1}^{24}c(n)^2 \equiv -42 \;\; (\mathrm{mod} \;70)$$ One must add (not subtract) $42$ to the sums, correct? (I've double-checked my results with Mathematica.) – Tito Piezas III Apr 30 '15 at 12:56
• No. I have subtracted 42. If I add, the result is not a multiple of 70. I used maxima. – Zurab Silagadze Apr 30 '15 at 13:02
• If I subtract 42, I get a rational number. We cannot both be correct. You started with $c(1) = 196884$? – Tito Piezas III Apr 30 '15 at 13:04
• I have checked again. If I subtract 42, maxima gives 19344611534578279675385464545221922990834243736570 and if I add 96723057672891398376927322726109614954171218682856/5. – Zurab Silagadze Apr 30 '15 at 13:08