Let $s=α+iβ$ be a complex number. Consider the Dirichlet series of the form $$f(s)=∑\_{n=1}^{∞}(a\_{n})/n^{s}$$
where $(a\_{n})\_{n≥1}$ is a real sequence. 


We consider the class of Dirichlet series satisfying the following conditions:


(1) The real sequence $(a\_{n})\_{n≥1}$ is not identically zero.


(2) The function $f(s)$ is analytic for all $s=α+iβ∈ℂ$ such that $α>0$.


(3) The first term of $f(s)$ is not zero, i.e., $a₁≠0$.


(4) There exist a complex zero $a=1/2+iβ₁$ of $f $ in the critical strip: $s=α+iβ∈ℂ$ with $0<α<1$.


(5) $f $ has infinitely many non trivial zeros.


(6) For all $α∈(0,1)$ with $α≠1/2$ we have $f(α+iβ₁)≠0$.


**My question is**: Does there exist a known Dirichlet series verifying all these conditions and have non trivial zeros off the critical line.



Does there exist a known Dirichlet series verifying all these conditions and have non trivial zeros off the critical line
What's wrong with the most obvious choice for a Dirichlet series that converges for $\text{Re}(s)>0$?  

$$
f(s)=(1-2^{1-s})\zeta(s)=\sum\_{n=1}^\infty \frac{(-1)^{n-1}}{n^s}
$$
converges for real $s>0$ by [Dirichlet's test](https://en.wikipedia.org/wiki/Dirichlet's_test)
and in the half plane by standard facts about the convergence of Dirichlet series. Any nontrivial zero of $\zeta(s)$ is a zero of $f(s)$. The only non obvious fact is your 6) above, but this could be verified for the zero at $1/2+i\cdot 14.1347\ldots$ via the [Argument Principle](https://en.wikipedia.org/wiki/Argument_principle)
and a numerical estimate of the integral. Since it's an integer, careful bounds on the error will give you a rigorous answer.


The term $(1-2^{1-s})$ is zero for $(1-s)\log(2)=2\pi i\cdot k$.


