*(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][1] $j_{2A}(\tau)$ 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][2] $\lambda(\tau)=\lambda$ 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. [1]: https://mathoverflow.net/questions/186336/ [2]: https://en.wikipedia.org/wiki/Modular_lambda_function