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


Where $\zeta(s)$ is the zeta function, and $\zeta'(s)$ is the derivative of the zeta function with respect to s,

But I want to know if there is a closed form expression for the sum $$\sum_{k=1}^{\infty}\frac{\Lambda(3k-2)}{(3k-2)^s}$$

Perhaps in term of polylogarithms or hurrwiz zeta funtions, or some other special funtion, moreover are there even closed form expressions for general sums of the form? $$\sum_{k=1}^{\infty}\frac{\Lambda(ak+b)}{(ak+b)^s}$$

ie can they be expressed in terms of known special functions like the hurrwitz zeta function or polylogarithms and or their derivatives.

It seems like it would be hard to do, because you would essentially then have expressions for functions that sum over primes broken up into congrunce classes, which on its own seems very 'unnatral'.


If $\chi$ is a character modulo $q$, then the logarithmic derivative of the associated Dirichlet $L$-function satisfies $$ - \frac{L'(s,\chi)}{L(s,\chi)} = \sum_{n=1}^\infty \frac{\Lambda(n)\chi(n)}{n^s}. $$ Then the same linear combination used to express $L(s,\chi)$ in terms of Hurwitz zeta functions can be used here to isolate the desired reside class (when $(q,b)=1$ at least - otherwise it can be simplified by hand): $$ \sum_{k=0}^\infty \frac{\Lambda(qk+b)}{(qk+b)^s} = \sum_{n\equiv b\pmod q} \frac{\Lambda(n)}{n^s} = -\frac1{\phi(q)} \sum_{\chi\pmod q} \bar\chi(b) \frac{L'(s,\chi)}{L(s,\chi)}. $$

  • $\begingroup$ So it can be given in terms of the hurwitz zeta function? also im not sure what dirichlet characters or L functions are $\endgroup$ – Ethan Dec 13 '12 at 1:25
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    $\begingroup$ This is a great opportunity to look them up then! I don't think the answer can be expressed in terms of Hurwitz zeta functions, because those functions don't have the Euler products necessary for the logarithmic derivative to have a nice form. $\endgroup$ – Greg Martin Dec 13 '12 at 6:58
  • $\begingroup$ It's okay if their derivatives of the zeta function $\endgroup$ – Ethan Dec 13 '12 at 16:11

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