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



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> **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 $n$ .
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The SageMath cell that demonstrates this claim can be found [here](https://sagecell.sagemath.org/?z=eJw9TcsKwjAQvOc_hN0abVPBS9hv0LtUaCWpS23UvC6i32566VyGeTCTBlLNAi1Osxl7BQ7pw9ToTLa_xacvhraFJlLyYdwY75B_KNlCvkxSdZsjlYSJt4tuO0TtTUzeAaP-ikAhzeBKJQ0SYKfwuh5hXbbFy7OLECq1X0VbnbkObx_hUOb-A88xLQ==&lang=gp&interacts=eJyLjgUAARUAuQ==).



> 
> **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 $n$ .
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The SageMath cell that demonstrates this claim can be found [here](https://sagecell.sagemath.org/?z=eJxFjc0KwjAQhO99D2G3RttU6iXsM-hdKrSS1KU2av4uos9uetG5DN_AzMSBZL1IFYdZj30LFunFVKtEpr-Eu8uBMtkmkuKm7RiukD4o2EA6TUJ2qz1RK5h4vXDTISqnQ3QWGNW78OTjDDaX4yAANhLPvyOs8nbxcGwD-FJu__B0AXZYHrlqUH0BCzcxOQ==&lang=gp&interacts=eJyLjgUAARUAuQ==).



Two conjectural infinite series for $\pi$
You recently asked a similar [question](https://math.stackexchange.com/questions/4246647/a-conjectural-infinite-series-for-frac-pi2) for modulus $4$ on math.stackexchange. I just used the exact same technique.




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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](https://math.stackexchange.com/questions/906424/analogue-of-zeta2-frac-pi26-for-dirichlet-l-series-of-mathbbz-3/906443#906443) answer.


