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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 $c$ is there always a sequence $\{\textbf{a}\}$ with this property?

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  • $\begingroup$ you can always take the taylor series of a possibly rotated sine function and have all the zeros be on some Re = c. $\endgroup$ – sku Oct 26 '17 at 2:17
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Yes, for example the sequence $a_n = 2^n$ gives the complex function $F(z) = \sum_{n \ge 0} 2^{-nz}$, which converges for $\Re(z) > 0$ to $F(z) = (1 - 2^{-z})^{-1}$, whose continuation on $\Bbb C \setminus \{2ik\pi/\log2 \mid k\in \Bbb Z\}$ doesn't have any zero at all, so those zeroes all are on all the lines $\Re(z)=c$, for every $c$ at once.

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  • $\begingroup$ I have thought this example shows that the nonexisting nonzeros all fall on a cardioid and not a line but seriously, can you also give an example of an infinite sequence that actually has more than one zeros (for they must come in conjugate pairs) of the right half plane? $\endgroup$ – hyportnex Oct 20 '17 at 12:34

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