11
$\begingroup$

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.

$\endgroup$
23
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ Is there a conceptual explanation of why this is true? $\endgroup$ – Kevin Casto Sep 25 '16 at 16:46
  • 1
    $\begingroup$ 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.) $\endgroup$ – Noam D. Elkies Sep 25 '16 at 16:54
  • $\begingroup$ Perhaps one should notice that the Bernoulli number Noam refers to is $B_{12}$. $\endgroup$ – Sylvain JULIEN Sep 25 '16 at 19:37
  • 2
    $\begingroup$ $E_6(z + 1) - E_6(z)$ is identically zero. Do you maybe mean $E_6(z + \tfrac{1}{2})$? $\endgroup$ – David Loeffler Sep 26 '16 at 8:36
  • $\begingroup$ 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. $\endgroup$ – Noam D. Elkies Sep 26 '16 at 19:24
4
$\begingroup$

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

$\endgroup$
  • 2
    $\begingroup$ 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. $\endgroup$ – Noam D. Elkies Sep 25 '16 at 22:05
  • $\begingroup$ @NoamD.Elkies: I edited the phrasing on "tradition". Also, the two claims now have serial number "III" and "IV". $\endgroup$ – T. Amdeberhan Sep 25 '16 at 22:09
  • $\begingroup$ 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. $\endgroup$ – Noam D. Elkies Sep 26 '16 at 2:45
  • $\begingroup$ @T.Amdeberhan: I've made the argument of the Dedekind eta function for I, II, III, IV consistent with q. $\endgroup$ – Tito Piezas III Sep 26 '16 at 14:04
  • $\begingroup$ Any clue about the plausible link between the numerator of $\zeta(12)/\pi^{12}$ and the Ramanujan congruence? $\endgroup$ – Sylvain JULIEN Sep 26 '16 at 19:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.