zeros of a complex function defined by integers 
This is a crosspost from Math.SE. 
Is there a known increasing sequence of positive integers $\{\textbf{a}\} = a_0<a_1<a_2<.....$ such that all the zeros $z_k$ on $\Re[z]>0$ of the complex function $F(z;\{\textbf{a}\})= \frac{1}{a_0^z}+\frac{1}{a_1^z}+\frac{1}{a_2^z}+.......$ with $F(z_k;\{\textbf{a}\})=0$ are such that $\Re[z_0]=\Re[z_1]=\Re[z_2]=... =c$ for some real number $c$? Also, if we are given an arbitrary real $c>0$ is there always a sequence $\{\textbf{a}\}$ with this property? Thank you.
 A: An example which is almost trivial: take the sequence $\{1,4,8,16,32\ldots\}$, i.e. powers of $2$ omitting two itself.  For real part of $s>0$, the Dirichlet series is a convergent geometric series, summing as
$$
\frac{1-2^s+4^s}{1-2^s}
$$
which has a meromorphic continuation.  The zeros are all purely imaginary, at
$$
\frac{\pi i(6n\pm1)}{3\log(2)},\qquad \text{integer } n
$$
if I've done the calculation correctly.
Building on this, one can sum
$$
1+2^{-s}+\sum_{k=3}^\infty 2^{-ks}=\frac{-2^{-2 s} \left(1-2^s+2^{3 s}\right)}{1-2^s}.
$$
The numerator has a term which is a cubic in 2^s.  A Mathematica computation then gives three infinite sets of zeros, one along the line with real part equal
$$
\frac{\log \left(\sqrt[3]{\frac{2}{3 \left(9-\sqrt{69}\right)}}+\frac{\sqrt[3]{\frac{1}{2} \left(9-\sqrt{69}\right)}}{3^{2/3}}\right)}{\log (2)}\approx 0.405685
$$
A: Can we not proceed in a manner similar to how Riemann did with the Mellin transform? 
For example, let me define $\theta(u) = \sum_{n=0}^{\infty}e^{-\pi a_n^2u}$
$$\int_0^{\infty} \theta(y) y^{s/2} dy/y = \int_0^{\infty}\sum_{n=0}^{\infty}e^{-\pi a_n^2y} = \sum_{n=0}^{\infty} \int_0^{\infty}e^{-\pi a_n^2y} y^{s/2} dy/y $$
$$= \pi^{-s/2} \sum_{n \ge 0} \frac{1}{a_n^s} \int_0^{\infty}e^{-u} u^{s/2} du/u$$ (after substituting $\pi a_n^2 y = u$)
$$= \pi^{-s/2} \Gamma(s/2) F(s)$$
I don't know if Jacobi's identity $\theta(u) = u^{-1/2}\theta(1/u)$ holds or not. It is probable that the identity holds, and then the functional equation will look similar to RFE and then one can probably go on to guess all zeros have real part $1/2$. 
A: Just a remark, this is true for any two element sequence (no, it's not infinite). 
For example
$$1 + \frac{1}{2^s}.$$ All zeros have real part zero.
A less obvious experimental fact is that the same (the zeros lie on a vertical line, not zero) is true for
$$1 + \frac{1}{2^s} + \frac{1}{3^s}.$$
Here is the picture from mathematica (contour lines of $|f| = 0.2,$ in case you are wondering).

The next picture is what happens when you have the $\sum_{k=1}^{10} k^{-s}.$

