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