TITLE: Ramanujan's tau function, $691$ congruence, and $\eta(z)^{12}$
QUESTION [11 upvotes]: Let $q = e^{2\pi i\,z}$.

I. 24th power

The Ramanujan tau function $\tau(n)$ is given by the expansion of the Dedekind eta function $\eta(z)$'s $\text{24th}$ power. Then
$$\begin{aligned}\eta(z)^{24} &= \sum_{n=1}^\infty\tau(n)q^n\\&=q - 24q^2 + 252q^3 - 1472q^4 + 4830q^5 - 6048q^6 - 16744q^7 + \dots\end{aligned}$$
Ramanujan observed that 
$$\tau(n)\equiv\sigma_{11}(n)\ \bmod\ 691\tag1$$

II. 12th power

Assume the rho function $\rho(n)$ as,
$$\begin{aligned}\eta(2z)^{12} &= \sum_{n=1}^\infty\rho(n)q^n\\&=q - 12q^3 + 54q^5 - 88q^7 -99q^9 +540q^{11} - 418q^{13} -648q^{15} + \dots\end{aligned}$$
Note the odd powers. Is it true that
$$\rho(n)\equiv\sigma_{5}(n)\ \bmod\ 2^8\tag2$$
analogous to $(1)$? 
P.S. It's true for the first 10000 coefficients in OEIS A000735.

REPLY [4 votes]: To keep the discussion alive and local, I add here further manifestations of the above behavior. Let $q = e^{2\pi i z}$,

III. 8th power

Define the numbers $a(n)$ according to
$$\begin{aligned}\eta(3z)^8 &= \sum_{n=1}^\infty a(n)q^n\\
&=q - 8q^4 + 20q^7 - 70q^{13}+64q^{16} +56q^{19} - 125q^{25} -160q^{28} + \dots\end{aligned}$$
Then I claim that
$$a(n)\equiv \sigma_3(n) \mod 81.$$

IV. 6th power

Define the numbers $b(n)$ according to
$$\begin{aligned}\eta(4z)^6 &= \sum_{n=1}^\infty b(n)q^n\\
&=q - 6q^5 + 9q^9+10q^{13}-30q^{17} +11q^{25}+42q^{29}-70q^{37} + \dots\end{aligned}$$
Then I claim that
$$b(n)\equiv \sigma_3(n) \mod 4.$$