# Ramanujan's tau function, $691$ congruence, and $\eta(z)^{12}$

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.

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

available on Ken Ono's page, where the enumeration is given (in the triangular-number form) as Theorem 7, and the congruence as a "simple consequence" of that theorem.

• Is there a conceptual explanation of why this is true? – Kevin Casto Sep 25 '16 at 16:46
• As with the Ramanujan identity in weight $12$, the space is of low enough dimension that if you've constructed enough forms there must be a linear relation. Why it's this particular relation I don't know. (The $691$ famously arises as the numerator of a Bernoulli number.) – Noam D. Elkies Sep 25 '16 at 16:54
• Perhaps one should notice that the Bernoulli number Noam refers to is $B_{12}$. – Sylvain JULIEN Sep 25 '16 at 19:37
• $E_6(z + 1) - E_6(z)$ is identically zero. Do you maybe mean $E_6(z + \tfrac{1}{2})$? – David Loeffler Sep 26 '16 at 8:36
• If anything it would have been $E_6(\frac{z+1}{2}) - E_6(\frac{z}{2})$ as the question was originally posed (in terms of $\eta(z)^{12}$). Now that it's $\eta(2z)^{12}$, yes, I'll have to change this too. – Noam D. Elkies Sep 26 '16 at 19:24

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 $$b(n)\equiv \sigma_3(n) \mod 4.$$

• I think the "tradition" here is to ask such variations as a new question. At any rate, the first one of these ("II", for $\eta^8$) has an answer similar to what I gave for Tito Piezas III's question: the difference is $81$ times the fourth power of the theta function of $r+3A_2$ where $r$ is any of the minimal vectors of the $A_2$ lattice. – Noam D. Elkies Sep 25 '16 at 22:05
• @NoamD.Elkies: I edited the phrasing on "tradition". Also, the two claims now have serial number "III" and "IV". – T. Amdeberhan Sep 25 '16 at 22:09
• OK, so I, II, III, IV correspond to $\eta^{24/m}$ with $m=1,2,3,4$. But the IV you suggest is not of the same kind, because you're comparing modular forms of different weights (and the common factor is much smaller). Also the $q^n$ coefficient of $\eta^6$ can be computed from the representation(s) of $n$ as a sum of two squares (as you can surmise from the vanishing of the $q^{21}$ and $q^{33}$ coefficients), which should let you prove the congruence mod $4$ directly. – Noam D. Elkies Sep 26 '16 at 2:45
• @T.Amdeberhan: I've made the argument of the Dedekind eta function for I, II, III, IV consistent with q. – Tito Piezas III Sep 26 '16 at 14:04
• Any clue about the plausible link between the numerator of $\zeta(12)/\pi^{12}$ and the Ramanujan congruence? – Sylvain JULIEN Sep 26 '16 at 19:33