The below is part of a formula I devised to calculate the sum of the squared reciprocals of all primes and pseudo-primes (pseudo according to Fermat's Little theorem.)
However, the 4 sums are very convoluted, if I could eliminate the sum over at least one of the indexes I would be happy about the final result, plus it would make computation much easier. That is, I want to collapse at least one of the sums over $i,j,k$ or $p$. But it's so difficult that I threw the towel. Perhaps one of you can. There's another part but it's much simpler.
Btw, I also need help calculating these sums to ensure I didn't make any mistake. Here $m=\log{2}$. On my computer it's taking too long.
For context, these 4 formulae correspond to the last sub-sum ($\sum_{}2^{p n}/n^{2j+2}$) in the equation below (where $n$ prime also means pseudo-prime, excluding 1):
\begin{multline} \sum _{n\text{ prime}}^{}\frac{1}{n^2}=\frac{1}{4}+\frac{1}{2}\sum _{k=1}^{\infty } (-1)^k(2\pi )^{2k}\sum _{p=1}^{2k} \frac{(-2)^{2k-p}}{p!(2k-p)!}\sum_{n=1}^{\infty } \frac{2^{p n}}{n^3}+\\ \sum_{k=1}^{\infty } (-1)^k(2\pi )^{2k}(2k)!\sum _{j=0}^k \frac{\text{BernoulliB}[2j]}{(2k+1-2j)!(2j)!}\sum _{p=1}^{2k} \frac{(-2)^{2k-p}}{p!(2k-p)!}\sum_{n=1}^{\infty } \frac{2^{p n}}{n^{2j+2}} \end{multline}
The formulae are based on an analytic continuation of $\sum_{n=1}^{\infty}e^{mp n}/n^{2j+2}$ for $m>0$ -- the sum is obviously finite if $m<0$:
And these are the sums I need simplified (I'm not sure if the below really converges due to being an analytical continuation for positive $m=\log{2}$, but I have an alternative to that anyway, if it doesn't):
1:
\begin{multline} -\sum _{k=1}^{\infty } (-1)^k(2\pi )^{2k}(2k)!\sum _{j=0}^k \frac{\text{BernoulliB}[2j]}{(2k+1-2j)!(2j)!}\sum _{p=1}^{2k} \frac{(-2)^{2k-p}}{p!(2k-p)!}\frac{(m p)^{2 j+1}\text{Log}\left[\frac{m p}{2\pi }\right]}{(2 j+1)!} \end{multline}
2:
\begin{multline}\sum _{k=1}^{\infty } (-1)^k(2\pi )^{2k}(2k)!\sum _{j=0}^k \frac{\text{BernoulliB}[2j]}{(2k+1-2j)!(2j)!}\sum _{p=1}^{2k} \frac{(-2)^{2k-p}}{p!(2k-p)!}\sum _{i=0}^{j+1} \frac{(m p)^{2j+2-2i}\text{Zeta}[2i]}{(2j+2-2i)!}+\sum _{i=1}^j \frac{(m p)^{2j+1-2i}\text{Zeta}[2i+1]}{(2j+1-2i)!} \end{multline}
3:
\begin{multline}\sum _{k=1}^{\infty } (-1)^k(2\pi )^{2k}(2k)!\sum _{j=0}^k \frac{\text{BernoulliB}[2j]}{(2k+1-2j)!(2j)!}\sum _{p=1}^{2k} \frac{(-2)^{2k-p}}{p!(2k-p)!}\sum _{i=1}^j \frac{(m p)^{2j+1-2i}\text{Zeta}[2i+1]}{(2j+1-2i)!} \end{multline}
4:
\begin{multline} -\frac{1}{2}\sum _{k=1}^{\infty } (-1)^k(2\pi )^{2k}(2k)!\sum _{j=0}^k \frac{\text{BernoulliB}[2j]}{(2k+1-2j)!(2j)!}\sum _{p=1}^{2k} \frac{(-2)^{2k-p}}{p!(2k-p)!}\\ \frac{(m p)^{2 j+2}}{(2 j+1)!}\int _0^1 \left((1-u)^{2j+1}\text{Coth}\left[\frac{m p u}{2}\right]-\frac{2\pi }{m p} (1-u) \text{Cot}[\pi u]\right)du \end{multline}
And to make things more interesting, if $x\geq 2$ then $\pi^*(x)$ (i.e., including pseudo-primes) is given by:
\begin{multline} \pi^*(x)=2+\sum _{k=1}^{\infty } (-1)^k(2\pi )^{2k}(2k)!\sum _{j=1}^k \frac{\text{BernoulliB}[2j]}{(2k+1-2j)!(2j)!}\sum _{p=1}^{2k} \frac{(-2)^{2k-p}}{p!(2k-p)!}\sum _{n=1}^x \frac{2^{p n}}{n^{2j}} \end{multline}
Again, I'd be able to replace the sum over $n$ with formulae, but it'd not be simple and elegant, too many indexes to sum over.
Now, back to the original question, this will only be solved if there's some genius reading. Mathematica's input formulae if you wanna try:
1:
x1 = -Sum[(-1)^k*(2*Pi)^(2*k)*(2*k)!*Sum[(BernoulliB[2*j]/((2*k + 1 - 2*j)!*(2*j)!*(2*j + 1)!))*Sum[(((-2)^(2*k - p)*(m*p)^(2*j + 1))/(p!*(2*k - p)!))*Log[(m*p)/(2*Pi)], {p, 1, 2*k}], {j, 0, k}], {k, 1, M}]
2:
x2 = Sum[(-1)^k*(2*Pi)^(2*k)*(2*k)!*Sum[(BernoulliB[2*j]/((2*k + 1 - 2*j)!*(2*j)!))*Sum[((-2)^(2*k - p)/(p!*(2*k - p)!))*Sum[((m*p)^(2*j + 2 - 2*i)*Zeta[2*i])/(2*j + 2 - 2*i)!, {i, 0,j + 1}],{p, 1, 2*k}], {j, 0, k}], {k, 1, M}]
3:
x3 = Sum[(-1)^k*(2*Pi)^(2*k)*(2*k)!*Sum[(BernoulliB[2*j]/((2*k + 1 - 2*j)!*(2*j)!))*Sum[((-2)^(2*k - p)/(p!*(2*k - p)!))*Sum[((m*p)^(2*j + 1 - 2*i)*Zeta[2*i + 1])/(2*j + 1 - 2*i)!, {i, 1, j}],{p, 1, 2*k}], {j, 0, k}], {k, 1, M}]
4:
x4 = (-(1/2))*Sum[(-1)^k*(2*Pi)^(2*k)*(2*k)!*Sum[(BernoulliB[2*j]/((2*k + 1 - 2*j)!*(2*j)!*(2*j + 1)!))*Sum[(((-2)^(2*k - p)*(m*p)^(2*j + 2))/(p!*(2*k - p)!))*Integrate[(1 - u)^(2*j + 1)*Coth[(m*p*u)/2] -((2*Pi)/(m*p))*(1 - u)*Cot[Pi*u], {u, 0,1}], {p, 1, 2*k}], {j, 0, k}], {k, 1, M}]