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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

$\Delta_{1}(q)=q^{\color{blue}{1}}-24q^2+252q^3-1472q^4+4830q^5-6048q^6-16744q^7+84480q^8-113643q^{\color{blue}{9}}-115920q^{10}+543612q^{11}-\cdots$

$\Delta_2(q)=252q-6048q^2-113643q^{\color{brown}{3}}-370944q^4+1217160q^5+2727432q^6-4219488q^7+21288960q^8-73279080q^9-29211840q^{10}+134722224q^{11}+167282496q^{12}-145589976q^{13}+101267712q^{14}-548895690q^{15}+248758272q^{16}-1740295368q^{17}+1758697920q^{18}+2686677840q^{19}-1791659520q^{20}+1902838392q^{21}-3233333376q^{22}+4698104544q^{23}-9600560640q^{24}-6425804700q^{25}+3494159424q^26+1665188361q^{\color{brown}{27}}+\cdots$

$\vdots$

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 partial sum of $(1b)$ 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}$

Question: How do we prove the conjectured criterion?

Motivation: 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 $\displaystyle \sum_{n=1}^{\infty} \frac{\mu(n)}{n^s}=\frac{1}{\zeta(s)}$

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  • $\begingroup$ I just came across the related post for the special case $k=1$ $\endgroup$
    – Nicco
    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
    Jun 10, 2020 at 22:25
  • $\begingroup$ @user760870: I'm sure everything is clarified now $\endgroup$
    – Nicco
    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
    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
    Jun 11, 2020 at 17:26

1 Answer 1

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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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