Vonmangoldt sums The dirichlet series for the Vonmangoldt function, $\Lambda(n)$, which is equal to zero when $n$ is not a prime a power, and $\ln(p)$ when it is a prime power say, $n=p^j$, is
$$-\frac{\zeta'(s)}{\zeta(s)}=\sum_{k=1}^{\infty}\frac{\Lambda(k)}{k^s}$$
Where $\zeta(s)$ is the zeta function, and $\zeta'(s)$ is the derivative of the zeta function with respect to $s$,
This can be re-written as $$-\frac{\zeta'(s)}{\zeta(s)}=\sum_{k=1}^{\infty}\frac{\Lambda(k)}{k^s}=\sum_{p}\frac{\ln(p)}{p^s-1},$$ with the last sum ranging over all primes p,
Can someone help me brake this sum, up into prime congruence sums similarly
$$\sum_{k=0}^{\infty}\frac{\Lambda(5k+1)}{(5k+1)^s}$$, ie re-write it as prime sums, where the primes are congruent to some b modulo $5$. I have done it before modulo $4$, and $3$ so I know it can be done, I am just having trouble restricting the powers appropietly to account for cases when $p^a\equiv1$ mod $5$, has no solutions.
 A: See Greg Martin's answer on this thread: Von mangoldt function dirichlet series
In general,
$$\sum_{n=0}^\infty \frac{\Lambda(qn+b)}{(qn+b)^s}=\frac{1}{\phi(q)}\sum_{\chi \pmod{q}}\overline{\chi}(b)\sum_{p}\frac{\chi(p)\log p}{p^{s}-\chi(p)}.$$  We may also write this as $$\frac{-1}{\phi(q)}\sum_{\chi\pmod{q}}\overline{\chi}(b)\frac{L^{'}}{L}(s,\chi).$$ Looking at a specific example, this yields $$\sum_{n\equiv1\ (3)}^{\infty}\frac{\Lambda(n)}{n^{s}}=\sum_{p\equiv1\ (3)}\frac{\log p}{p^{s}-1}+\sum_{p\equiv2\ (3)}\frac{\log p}{p^{2s}-1} .$$    
To see this, notice that $$\sum_{n=0}^\infty \frac{\Lambda(qn+b)}{(qn+b)^s}=\sum_{n\equiv b\pmod{q}} \frac{\Lambda(n)}{n^s}.$$ Using the orthogonality relations of the Dirichlet Characters, that is the fact that  $$\frac{1}{\phi(q)}\sum_{\chi \pmod{q}} \chi(a)=\left\{ \begin{array}{c}
1\ \text{when } a\equiv 1 \pmod{q} \\
0\ \text{otherwise}
\end{array}\right\},$$ it follows that
$$\sum_{n=0}^\infty \frac{\Lambda(qn+b)}{(qn+b)^s}=\frac{1}{\phi(q)}\sum_{\chi\pmod{q}}\overline{\chi}(b)\sum_{n=1}^{\infty}\frac{\Lambda(n)\chi(n)}{n^{s}}$$  Recalling the function $L(s,\chi)$, this last sum then equals $$\frac{-1}{\phi(q)}\sum_{\chi\pmod{q}}\overline{\chi}(b)\frac{L^{'}}{L}(s,\chi).$$ 
From here, we obtain the first equation by expanding we obtain the first equation by using the fact that $$L(s,\chi)=\prod_{p}\left(1-\frac{\chi(p)}{p^{s}}\right)^{-1}.$$
Key Idea: Use Dirichlet characters to isolate an arithmetic progression.  This is how most major results on primes in arithmetic progressions are proven.
