Where did all this pi come from?

$\begin{aligned} \sum_{n=1}^\infty \frac{\zeta(3n)-1}{n} & = \sum_{n=1}^\infty \frac{1}{n}\sum_{k=2}^\infty \frac{1}{k^{3n}} \; = \; \sum_{k=2}^\infty \sum_{n=1}^\infty \frac{1}{n\,k^{3n}} \\ & = -\sum_{k=2}^\infty \ln\big(1 - k^{-3}\big) \; = \; -\ln\left(\prod_{k=2}^\infty \big(1 - k^{-3}\big)\right) \\ & = -\ln\left(\frac{\cosh \frac12\pi\sqrt{3}}{3\pi}\right) \; =\; -\ln\left(\frac{e^{\pi\sqrt{3}} + 1}{6\pi e^{\frac12\pi \sqrt{3}}}\right) \\ & = \ln(6\pi) + \tfrac12\pi\sqrt{3} - \ln\big(e^{\pi\sqrt{3}} + 1\big) \; = \; \ln(6\pi) + \pi\cos\tfrac16\pi - \ln\big(e^{\pi\sqrt{3}} + \tan\tfrac14\pi\big) \end{aligned}$ making the answer $6 + 6 + 3 + 4 = \boxed{19}$ .

Let $H(x)$ be the harmonic function, which has the following Taylor series: $\begin{aligned} H(x) & = -\sum_{k=2}^{\infty} \zeta(k)(-x)^{k-1} \\ -x H(-x) & = \sum_{k=2}^{\infty} \zeta(k)x^k \\ \end{aligned}$ Let $w = \dfrac{-1+\sqrt{-3}}{2}$ . Then $1+w+w^2 = 0$ . Therefore, $\begin{aligned} -x H(-x) -wx H(-wx) - w^2x H(-w^2x) & = \sum_{k=2}^{\infty} \left(\zeta(k)x^k+\zeta(k)(wx)^k+\zeta(k)(w^2x)^k\right) \\ -x H(-x) -wx H(-wx) - w^2x H(-w^2x) & = 3\sum_{n=1}^{\infty} \zeta(3n)x^{3n} \\ -x H(-x) -wx H(-wx) - w^2x H(-w^2x) - \dfrac{3x^3}{1-x^3} & = 3\sum_{n=1}^{\infty} \left(\zeta(3n)-1\right)x^{3n} \\ -H(-x) -w H(-wx) - w^2 H(-w^2x) - \dfrac{3x^2}{1-x^3} & = \sum_{n=1}^{\infty} 3\left(\zeta(3n)-1\right)x^{3n-1} \end{aligned}$

Integrating both sides gives us:

$\begin{aligned} \ln (-x)! - \gamma x + \ln(-wx)! - \gamma wx + \ln(-w^2x)! - \gamma w^2x + \ln(1-x^3) & = \sum_{n=1}^{\infty} \dfrac{\left(\zeta(3n)-1\right)x^{3n}}{n} \\ \ln (1-x)! + \ln(1-wx)! + \ln(1-w^2x)! & = \sum_{n=1}^{\infty} \dfrac{\left(\zeta(3n)-1\right)x^{3n}}{n} \end{aligned}$

Now substitute $x = 1$ :

$\begin{aligned} \sum_{n=1}^{\infty} \dfrac{\zeta(3n)-1}{n} & = \ln\left((1-w)! (1-w^2)!\right) \\ & = \ln\left(\left(\dfrac{3+\sqrt{-3}}{2}\right)! \left(\dfrac{3-\sqrt{-3}}{2}\right)! \right) \\ & = \ln\left(\dfrac{3+\sqrt{-3}}{2} * \dfrac{1+\sqrt{-3}}{2} * \dfrac{3-\sqrt{-3}}{2} * \dfrac{1-\sqrt{-3}}{2}\right) + \ln\left(\left(\dfrac{-1+\sqrt{-3}}{2}\right)! \left(\dfrac{-1-\sqrt{-3}}{2}\right)! \right) \end{aligned}$

Now we can use Gauss' theorem which says that $(x-1)!(-x)! = \pi \csc(\pi x)$ :

$\begin{aligned} \sum_{n=1}^{\infty} \dfrac{\zeta(3n)-1}{n} & = \ln(3) +\ln\left(\pi \csc\left(\pi\left(\dfrac{1+\sqrt{-3}}{2}\right)\right) \right) \\ & = \ln(3\pi) - \ln\left(\cos\left(\dfrac{\pi\sqrt{-3}}{2} \right)\right) \\ & = \ln(3\pi) - \ln\left(\cosh\left(\dfrac{\pi\sqrt{3}}{2} \right)\right) \\ & = \ln(3\pi) - \ln\left(\dfrac{e^{\pi\sqrt{3}}+1}{2e^{\sqrt{3}\pi/2}}\right) \\ & = \ln(6\pi) + \dfrac{\sqrt{3}}{2}\pi - \ln(e^{\pi\sqrt{3}}+1) \\ \sum_{n=1}^{\infty} \dfrac{\zeta(3n)-1}{n} & = \ln(6\pi) + \pi \cos\left(\dfrac{\pi}{6}\right) - \ln\left(e^{\pi\sqrt{3}}+\tan\left(\dfrac{\pi}{4}\right)\right) \end{aligned}$

notice that $\sum_{n=1}^\infty \dfrac{\zeta(an)-1}{n}=\sum_{n=1}^\infty \sum_{k=2}^\infty \dfrac{1}{nk^{an}}=-\sum_{k=2}^\infty \ln(1-k^{-a})=\\-\sum_{k=2}^\infty \ln(\prod_{n=1}^a(1-\omega^n k^{-1}))=-\ln\left(\prod_{k=2}^\infty\prod_{n=1}^a(1-\omega^n k^{-1})\right)$ where $\omega$ is an $a^{th}$ primitive root of unity. using wiestrass product $\dfrac{1}{\Gamma(z)}=z(1+z) e^{\gamma z} \prod_{k=2}^\infty e^{-z/k} (1+zk^{-1})\to\dfrac{1}{\Gamma(z+2)}=e^{\gamma z} \prod_{k=2}^\infty e^{-z/k} (1+zk^{-1})$ plugging in $z=-\omega,-\omega^2...-\omega^{a-1},$ into the first one and $z=-1$ in the second one and multiplying $\prod_{n=1}^{a-1} \dfrac{1}{\Gamma(-\omega^n)}=\left(\prod_{n=1}^{a-1} -\omega^n(1-\omega^n) e^{\gamma \omega^n} \left(\prod_{k=2}^\infty e^{-\omega^n/k} (1-\omega^n k^{-1})\right)\right)\left(e^{-\gamma } \prod_{k=2}^\infty e^{1/k} (1-k^{-1})\right)$ by using properties of roots of unity the product ends up being, namely their sums been 0 and product been $(-1)^{a-1}$ , $\prod_{n=1}^{a-1} \dfrac{1}{\Gamma(-\omega^n)}=a\prod_{n=1}^a\prod_{k=2}^\infty (1-\omega^n k^{-1})$ we can manipulate this into $-\ln\left(\prod_{n=1}^a\prod_{k=2}^\infty (1-\omega^n k^{-1})\right)=\ln(a)+\sum_{n=1}^{a-1} \ln(\Gamma(-\omega^n))\to\sum_{n=1}^\infty \dfrac{\zeta(an)-1}{n}=\ln(a)+\sum_{n=1}^{a-1} \ln(\Gamma(-\omega^n))$ the rest is just easy plugging $a=3$