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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.

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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.

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    @Ethan: It just means over those $a$ which are relatively prime to $b$, but you only hit each $a$ once. For example, if $b$ were $6$, we would be summing over $a=1$ and $a=5$. If $b$ were$a$prime $p$, we sum over $a=1,2,3,\dots,p-1$. In other words, sum over 1\leq a such that $\gcd(a,b)=1$.2012-12-23