Let $q = e^{2\pi i\,z}$.



> 
> **I. 24th power**
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> 


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](https://oeis.org/A000735).



Ramanujan's tau function, $691$ congruence, and $\eta(z)^{12}$
Yes, this is true, as a consequence of an identity in a space of
modular forms of weight $6$.


The form $\eta(2z)^{12}$ is in this space; and $\sigma\_5(n)$ for $n$ odd
are the coefficients of the weight-$6$ form
$$
\frac1{1008} \Bigl(E\_6(z+\frac12) - E\_6(z)\Bigr) =
q + 244 q^3 + 3126 q^5 + 16808 q^7 + 59293 q^9 + \cdots.
$$
The difference is
$$
256(q^3 + 12 q^5 + 66 q^7 + 232 q^9 + 627 q^{11} + 1452 q^{13} + \cdots);
$$
after a bit of experimentation we recognize this as $256q^3$ times the
$12$-th power of $1+q^2+q^6+q^{12}+q^{20}+\cdots$, which is to say
$256$ times the $12$-th power of the weight-$1/2$ modular form
$\sum\_{k=0}^\infty q^{(2k+1)^2/4}$. Such a formula, once surmised,
can be proved by comparing initial segments of the $q$-expansions;
I did this to $O(q^{61})$, which is way more than enough.
The congruence mod $256$ follows because the $12$-th power of
$1+q^2+q^6+q^{12}+q^{20}+\cdots$ clearly has integral coefficients.


*Added later*:
This identity (and thus the congruence that **Tito Piezas III** asked for)
gives a formula $(\sigma\_5(n) - \rho(n)) / 256$ for the number of
representations of $4n$ as the sum of $12$ odd squares, or equivalently of
$(n-3)/2$ as the sum of $12$ triangular numbers. Following this lead,
I soon found both the formula and the congruence in the paper



> 
> Ken Ono, Sinai Robins, and Patrick T. Wahl: On the representation of integers as sums of triangular numbers, *Aequationes Math.* **50** (1995) #1, 73-94
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available
[on
 Ken Ono's page](http://www.mathcs.emory.edu/~ono/publications-cv/pdfs/006.pdf),
where the enumeration is given (in the triangular-number form) as Theorem 7,
and the congruence as a "simple consequence" of that theorem.


