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Given an Euler product of the form \begin{align} L(s) = \prod_{p}(1 - a_{p} p^{-s} - b_{p} p^{-2s})^{-1} \end{align} where $a_n$ and $b_n$ are not necessarily a multiplicative arithmetic functions of $n$, is there a prescription for computing the series coefficients, $L(s) = \sum_{n \geq 1} c_n \ n^{-s}$ given $a_p$ and $b_p$?

This example specializes to some familiar examples. For instance, if $b_{p} = 0$ for all primes, then $c_n = \prod_{p \mid n} a_{p}^{\text{ord}_{p}(n)}$. In particular, if $a_p = \chi(p)$, a Dirichlet character, then $L$ is the corresponding Dirichlet $L$-function with $c_n = \chi(n)$.

References are certainly welcome!

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Notice that in your example where $b_p = 0$, the statement "then $c_n = a_n$" actually makes no sense, because all you are being given at first are the numbers $\{a_p,b_p\}$ as $p$ runs over primes. There is no such thing as $a_n$ initially. –  KCd Mar 14 '12 at 15:42
    
Any book that discusses the $L$-series of elliptic curves or modular forms will show you a recursion when $b_p = -p$, and from that you can reverse engineer the work to figure out your answer for any $b_p$. –  KCd Mar 14 '12 at 15:44
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It makes more sense to just let $a_p$ be defined for prime $p$ only; anything more is superfluous, since only values at primes are relevant (and also the remark about complete multiplicativeness is false). –  anon Mar 14 '12 at 17:00
    
Edited. Thanks. –  user02138 Mar 14 '12 at 17:09

1 Answer 1

The recursions should have special properties when you're studying the Euler products of naturally occurring L-functions (or whatever), but in the highest generality on $a_p$ and $b_p$ I suppose what you can do is factor the quadratic Euler factors (via quadratic formula), obtaining two local products

$$L(s)=\prod_{p} \left(1-\frac{a_p+\sqrt{a_p^2+4b_p}}{2}p^{-s}\right)^{-1} \prod_{p} \left(1-\frac{a_p-\sqrt{a_p^2+4b_p}}{2}p^{-s}\right)^{-1}$$

Expanding out and then Dirichlet-convoluting the two local functions gives the general formula

$$c_n=\sum_{d|n} \prod_{p|n} \left(\frac{a_p+\sqrt{a_p^2+4b_p}}{2}\right)^{\operatorname{ord}_p(d)} \left(\frac{a_p-\sqrt{a_p^2+4b_p}}{2}\right)^{\operatorname{ord}_p(n/d)}.$$

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Should the product be taken over primes dividing $d$ rather than $n$ in your last equation? –  user02138 Mar 14 '12 at 17:54
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@user02138: Technically we would first get the summand $$\prod_{p|n}\big(\cdot\big)^{\operatorname{ord}_p(d)} \times \prod_{p|n/d}\big(\cdot\big)^{\operatorname{ord}_p(n/d)}.$$ These can be subsumed into a single product as I have written it; do you see why? However, simply writing $p|d$ under a single $\prod$ symbol would not necessarily account for all of the factors needed (take e.g. $d=1$ and get an empty product). You could even do the product over all primes, because $\text{ord}_p(a)=0$ when $(a,p)=1$ after all. –  anon Mar 14 '12 at 18:09

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