# Two conjectural infinite series for $\pi$

I am looking for a proofs of the following two claims:

Claim 1. $$\frac{2\pi}{\sqrt{3}}=\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{\Omega_1(n)}}{n}$$ where $$\Omega_1(n)$$ is the number of prime factors of the form $$p \equiv 1 \pmod{6}$$ of $$n$$ .

The SageMath cell that demonstrates this claim can be found here.

Claim 2. $$\frac{\sqrt{3}\pi}{2}=\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{\Omega_5(n)}}{n}$$ where $$\Omega_5(n)$$ is the number of prime factors of the form $$p \equiv 5 \pmod{6}$$ of $$n$$ .

The SageMath cell that demonstrates this claim can be found here.

• you count prime divisors with multiplicity, right? Commented Sep 16, 2021 at 8:40
• The first one doesn't seem to be particularly close from the SageMath cell linked.
– gmvh
Commented Sep 16, 2021 at 8:45
• @FedorPetrov Right. Commented Sep 16, 2021 at 8:48
• @gmvh Perhaps it converges very slowly. Commented Sep 16, 2021 at 8:49
• Isn't the second series just $3\cdot L(1,\chi)$, where $\chi$ is the non-trivial Dirichlet character modulo $6$? And the product of the two series is $9\cdot \prod_{p\ge 5}(1-p^{-2})^{-1}=6\zeta(2)=\pi^2$? Commented Sep 16, 2021 at 9:18

You recently asked a similar question for modulus $$4$$ on math.stackexchange. I just used the exact same technique.
For $$i\in\{1,5\}$$, define $$f_i:\mathbb{Z}_{\ge 1}\to \mathbb{C}^\times$$ by $$f_i(n)=(-1)^{\Omega_i(n)}$$, then $$f_i$$ is completely multiplicative and $$L(s,f_i)=\left(1-2^{-s}\right)^{-1}\left(1-3^{-s}\right)^{-1}\prod_{p\equiv i\pmod 6}\left(1+p^{-s}\right)^{-1}\prod_{p\equiv -i\pmod 6}\left(1-p^{-s}\right)^{-1}.$$ Hence, $$L(s,f_1)L(s,f_5)=\left(1-2^{-s}\right)^{-2}\left(1-3^{-s}\right)^{-2}\prod_{p\ge 5}\left(1-p^{-2s}\right)^{-1}=\frac{2^{s}+1}{2^s-1}\cdot\frac{3^s+1}{3^s-1}\cdot \zeta(2s).$$ Therefore, if both $$L(1,f_1)$$ and $$L(1,f_5)$$ converge, their product is $$6\zeta(2)=\pi^2$$. Now, let $$\chi$$ be the non-trivial Dirichlet character modulo $$6$$, then $$L(1,\chi)=\prod_{ p\equiv 1\pmod 5}(1-p^{-1})^{-1}\prod_{p\equiv 5\pmod{6}}\left(1+p^{-1}\right)^{-1}=\frac13L(1,f_5).$$ Therefore, it suffices to show that $$L(1,\chi)=\frac{\pi}{2\sqrt{3}}.$$
Let $$\chi_2$$ be the non-trivial Dirichlet character modulo $$3$$, then $$\chi(p)=\chi_2(p)$$ for all primes $$p\neq 2$$ and $$L(1,\chi)=\frac{L(1,\chi_2)}{(1+2^{-1})^{-1}}=\frac32 \cdot \frac{\pi}{3\sqrt 3}=\frac{\pi}{2\sqrt{3}}$$ Where we use this answer.