The well known Ramanujan tau function $\tau(n)$ is defined as the nth Fourier coefficient of the modular discriminant
$\displaystyle \Delta(q)=q\prod_{m=1}^\infty (1-q^m)^{24} = \sum_{n=1}^\infty \tau(n)\,q^n\tag{1a}$
In the spirit of my previous posts( here (1) and here (2) for example) concerning the parity of certain coefficients of modular forms, I was naturally led to investigate this subsequent conjecture
Let us introduce the following convolution identity, see here (3)
$\displaystyle\Big(\sum_{n=1}^{\infty} \tau(n) q^n\Big)^2+\sum_{n=1}^{\infty} \tau^2(n) q^{2n}=2\sum_{n=0}^{\infty} \sum_{k=0}^{n} \tau(k+1) \Big(\tau(2n-k+1) q^{2n+2}+\tau(2n-k+2) q^{2n+3}\Big) \tag{1b}$
Think of it as a generalisation of the cauchy product and power sum symmetric polynomials. In generalc consider a sequence of complex terms $a_0, a_1, a_2,...$, then the ff convolution identity holds $\displaystyle\Big(\sum_{n=1}^{\infty} a_{n}\Big)^2+\sum_{n=1}^{\infty} a^2_{n}=2\sum_{n=0}^{\infty} c_{n}$
where $\displaystyle c_{n}=\sum_{k=0}^{n} a_{k+1} \Big(a_{2n-k+1}+a_{2n-k+2}\Big)$
given of course that the sums $\displaystyle\Big(\sum_{n=1}^{\infty} a_{n}\Big)^2$ and $\displaystyle\sum_{n=1}^{\infty} a^2_{n}$ converge.
A little bit of experimentation leads one to the ff conjecture
The nth coefficient in the partial sum of the series $\displaystyle \sum_{n=0}^{\infty} \sum_{k=0}^{n} \tau(k+1)\Big(\tau(2n-k+1) q^{2n+2}+\tau(2n-k+2) q^{2n+3}\Big)\tag{1c}$ is odd if and only if the power of its last term is $\stackrel{\mathrm{def}}{=}2z$ where $z\equiv 1\pmod{4}$ with a unique partition as a sum of two squares $x^2+y^2$, a sequence of integers $\color{orange}{1}, \color{orange}{5}, \color{orange}{9}, \color{orange}{13}, \color{orange}{17}, \color{orange}{29}, \color{orange}{37},....$ which can readily be found in this oeis entry
$\displaystyle \sum_{n=0}^{\infty} \sum_{k=0}^{n} \tau(k+1)\Big(\tau(2n-k+1) q^{2n+2}+\tau(2n-k+2) q^{2n+3}\Big)=q^{2×\color{orange}{1}}-24q^3+828q^4-7520q^5+103662q^6-492912q^7+3512352q^8-8147520q^9+25870905q^{2×\color{orange}{5}}+19335800q^{11}-193971120q^{12}+6341175146q^{13}-465787812q^{14}-2153273040q^{15}+12714981120q^{16}-11761115712q^{17}-6692143617q^{2×\color{orange}{9}}+68609297160q^{19}-....$
Question: How do we prove it?
Modified conjecture
Following Somos' proof, the original conjecture may be modified as follows
The nth coefficient of the partial series of $(1c)$ is odd if and only if $r(n)$ the number of integer solutions $n=x^2+y^2$ where $0< x \le y$ is odd since
$\displaystyle \sum_{n=0}^{\infty} \sum_{k=0}^{n} \tau(k+1)\Big(\tau(2n-k+1) q^{2n+2}+\tau(2n-k+2) q^{2n+3}\Big) \equiv \sum_{n=1}^\infty r(n)q^n \pmod 2$
By using the RHS of our convolution identity (1b) we can determine every nth coefficient of the LHS with ease. For example let us determine the counterexample( spotted by Somos) to the unmodified stated conjecture
$\displaystyle \sum_{k=0}^{424} \tau(k+1)\tau(849-k)q^{850}=-90512663966015166416921687775q^{850}$
The question asks about a conjecture that is almost correct.
A key idea is the relation between two $q$-series expansions as follows.
Define the Ramanujan $\,\psi\,$ function (generating function of OEIS sequence A010054)
$$ \psi(q) := \sum_{k=0}^\infty q^{k(k+1)/2} = \prod_{k=1}^\infty \frac{(1 - q^{2k})^2}{1-q^k} = 1 + q + q^3 + q^6 + \cdots. \tag1 $$
Compare this to $\,f(-q)^3\,$ (generating function of OEIS sequence A010816) defined by
$$ f(-q)^3 := \prod_{k=1}^\infty (1-q^k)^3 = \sum_{k=1}^\infty (-1)^k (2k\!+\!1)\,q^{k(k+1)/2} = 1 - 3q + 5q^3 - 7q^6 + \cdots. \tag2 $$
Verify that the coefficients of these two $q$-series have the same parity.
Now inflate both series using $\,q\to q^8\,$ as follows
$$ T(q) := q\,\psi(q^8) = \sum_{k=0}^\infty q^{(2k+1)^2} = \sum_{n=1}^\infty t(n)\,q^n = q + q^9 + q^{25} + q^{49} + \cdots. \tag3 $$
and similarly
$$ q\,f(-q^8)^3 = q\prod_{k=1}^\infty (1-q^{8k})^3 = q - 3q^9 + 5q^{25} - 7q^{49} + \cdots. \tag4 $$
As before, the coefficients of these two $q$-series have the same parity.
Now define $\,\Delta(q)\,$ (generating function of OEIS sequence A000594) with
$$ \Delta(q) := q\,f(-q)^{24} = q\,(f(-q)^8)^3 = \sum_{n=1}^\infty \tau(n)\,q^n = q - 24q^2 + 252q^3 - \cdots. \tag5 $$
Verify that the coefficients of the polynomials $\,(1-q)^8\,$ and $\,(1-q^8)\,$ have the same parity.
Combine all previous parity results.
They imply that $\,T(q)\,$ (from equation $(3)$) and $\,\Delta(q)\,$ (from equation $(5)$) have the same parity.
The question links to MSE question 2408529 and my answer to that question is relevant here.
Lemma: Given a sequence $\,a_1,a_2,a_3,\dots,\,$ and that the power sums
$$ S_1:=\sum_{i}a_i,\;\; S_2:=\sum_ia_i^2\;\; \text{ converge, then }\;\; S_1^2+S_2=2\sum_{i\le j}a_ia_j. \tag6 $$
Let $\,r(n)\,$ be the number of integer solutions $\,n = x^2 + y^2\,$ where $\,0 < x \le y.\,$ Its g.f. is
$$ R(q) := \Big(T(q)^2 + \sum_{n=1}^\infty t(n)^2q^{2n}\Big)/2 = \sum_{n=1}^\infty r(n)\,q^n. \tag7 $$
Apply the equation $(6)$ lemma to the series in equations $(3)$ and $(5)$ to get the result that
$$ \Big(\Delta(q)^2 + \sum_{n=1}^\infty \tau(n)^2q^{2n}\Big)/2 \equiv R(q) \pmod 2 . \tag8 $$
Notice that the coefficient of $q^{2\cdot 425}$ on the left side of equation $(8)$ is $-90512663966015166416921687775$ which is odd while $\,r(2\cdot 425) = 3 > 1\,$ because $425 = 5^2 + 20^2 = 8^2 + 19^2 = 13^2 + 16^2$. Thus $\,425\,$ does not have a unique partition into sum of two squares. This is a counterexample to the unmodified stated conjecture.