TITLE: Are L-functions uniquely determined by their values at negative integers?
QUESTION [12 upvotes]: Are L-functions uniquely determined by their values at negative integers? In another words, is there a sequence of integers $a_1, a_2, a_3, \cdots$ such that 

the corresponding L-function
$$L_{\{a_n\}}(s):=\sum_{n=1}^{\infty}\frac{a_n}{n^s}$$
converges well for $\text{Re}(s) > M$ for some $M\in \mathbb{R}$
$L_{\{a_n\}}(s)$ has analytic continuation to a meromorphic function on the whole complex plane
$L_{\{a_n\}}(n)=0$ for all negative integers $n$
not all of $a_n$ are zero? 

Added : It was suggested in the answers that I should have used the term "Dirichlet series of integer sequences" instead of "L-function" as it lacks Euler product. I apologize for the confusion :)

REPLY [4 votes]: Let $F(s) =\sum_{n=1}^\infty a_n n^{-s}= \sum_{j=1}^J c_j F_j(s)$ be a linear combination of L-functions $F_j$ of degree $\le d$, that is $(s-1)^r F_j(s)$ entire and its functional equation contains at most $d$ gamma factors $\Gamma(s/2+a)$.
Then for $m > d$ or for $m=d,|x| < C$, $\Gamma(ms)F(s)x^{-s}$ decays fast enough to apply the residue theorem $$\begin{eqnarray}f_m(x)&=&\frac{1}{2i\pi} \int_{2-i\infty}^{2+i\infty} F(s) \Gamma(ms) x^{-s}ds \\ &=&  \sum Res(F(s) \Gamma(ms) x^{-s}) \\ &=& \sum Res_{s=1}(F(s) \Gamma(ms)x^{-s}) + \sum_{k=0}^\infty \frac{(-1)^k}{k!m} F(-k/m) x^{k/m}\end{eqnarray}$$
(for $m = d$ it is valid only for $|x| < C$ but the exponential decay of $\Gamma(ms)F(s)$ on $\Im(s)=2$ implies $f_m$ is analytic so it is determined by $x \in (0,C)$)
And hence $$F(s) = \frac{1}{\Gamma(ms)} \int_0^\infty f_m(x) x^{s-1}dx$$ 
is fully determined by its principal part at $1$ and its values at $s=-k/m$.
In other words your claim holds only for L-functions of degree $d=1$.