Given Ramanujan tau function $\tau(n)$, which is the nth Fourier coefficient of the modular discriminant

$\displaystyle \Delta(q)=q\prod_{m=1}^\infty (1-q^m)^{24} = \sum_{n=1}^\infty \tau(n)\,q^n\tag{1a}$

Nothing forbids us from generalizing the generating function to

$\displaystyle \Delta_{k}(q)= \sum_{n=1}^\infty \tau((2k-1)n)\,q^n\tag{1b}$

for integers $k\gt0$

After some experimentation with wolfram mathematica, one discovers the following intriguing patterns




From which we deduce the following conjecture

The nth coefficient $\tau((2k-1)n)$ is odd if and only if the powers of q in the generating $(1b)$ depend on whether $(2k-1)$ is square-free or non square-free as follows

$n\stackrel{\mathrm{def}}{=} \begin{cases}{\color{brown}{(2k-1),9(2k-1),25(2k-1),49(2k-1),...}} & \mbox{ where $2k-1=1,3,5,7,11,13,...$ is any odd square-free integer}\\{\color{blue}{1,9,25,49,81,...}} & \mbox{ where $2k-1=9,25,27,49,81,...$ is any odd non square-free integer} \end{cases}$

Or stated differently

$\displaystyle\sum_{n=1}^\infty \tau((2k-1)(n)) q^{n}\equiv \sum_{m=1}^\infty q^{(2m-1)^2}\bmod 2$

Whence $\tau((2k-1)(n))$ is odd iff the powers of q in $(1b)$ are odd squares $n=(2m-1)^2$ given that $(2k-1)$ is any odd non square-free integer valid for $k\gt0$

And also

$\displaystyle\sum_{n=1}^\infty \tau((2k-1)(n)) q^{n}\equiv \sum_{m=1}^\infty q^{(2k-1)(2m-1)^2}\bmod 2$

That is, $\tau((2k-1)(n))$ is odd iff the powers of q in $(1b)$ are $n=(2k-1)(2m-1)^2$ given that $(2k-1)$ is any odd square-free integer valid for $k\gt0$

Question: How do we prove the conjectured criterion?

Remark: The Möbius function $\mu(n)$ is either $1$ or $-1$ if $n$ is square-free depending on whether it has a number of even prime factors or odd respectively and $\mu(n)=0$ if it is non square-free. It happens to be a very important arithmetic function in number theory. For example, the Dirichlet series that generates the Möbius function is the reciprocal of the Riemann zeta function $\displaystyle \sum_{n=1}^{\infty} \frac{\mu(n)}{n^s}=\frac{1}{\zeta(s)}$. The fact that these conjectured congruences have something to do with the möbius function is exactly what makes them interesting.

  • $\begingroup$ I just came across the related post for the special case $k=1$ $\endgroup$
    – Nicco
    Commented Jun 10, 2020 at 22:16
  • $\begingroup$ $\tau(n)$ is odd if $n$ is an odd square. So $\tau(kn)$ is obviously only odd if $nk$ is an odd square. (Surely you mean to write $\tau(kn)$ rather than $\tau((2k-1)n)$ is otherwise when $k = 2$ the expression is odd when $n = 3$ which is not part of your conjecture). Anyway, $nk$ is an odd square when $k$ is odd and $n$ is $k$ times an odd square. The converse is true when $k$ is squarefree. But if $k$ is not squarefree they are not equivalent. For example, $\tau(9n)$ is odd when $n = 1$. $\endgroup$
    – user760870
    Commented Jun 10, 2020 at 22:25
  • $\begingroup$ @user760870: I'm sure everything is clarified now $\endgroup$
    – Nicco
    Commented Jun 11, 2020 at 1:02
  • $\begingroup$ And there's one more post which mentions that the problem for proving that $\tau(n)$ is odd if and only if $n=(2m+1)^2$ for some $m$ is from Ram Murty's "Problems in the theory of modular forms, exercise 1.5.1" $\endgroup$
    – Nicco
    Commented Jun 11, 2020 at 12:28
  • $\begingroup$ What are you rambling about? The case of $k=1$ is given in the link. The case of $k=5$ is false, since $\tau((2k-1)n) = \tau((2 \times 5 - 1) \times 1) = \tau(9 \times 1) = \tau(9)$ is odd for $n = 1$. The correct statement is also obvious from the case $k=1$; $\tau((2k-1)n)$ is odd iff $(2k-1)n$ is an odd square. $\endgroup$
    – user760870
    Commented Jun 11, 2020 at 17:26

1 Answer 1


What do you get from the linked post (assuming the formula is correct, I didn't find a reference) $$\Delta(q)=q \left(\sum_{k=0}^\infty(-1)^k(2k+1)q^{k(k+1)/2}\right)^8\equiv q \sum_{k=0}^\infty ((-1)^k(2k+1)q^{ k(k+1)/2})^8\bmod 2$$


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