Mathematica knows that: $$\lim_{s\to 1} \, \zeta (s)\left(-2^{1-s}-3^{1-s}+6^{1-s}+1\right)=\sum _{n=0}^{\infty } \left(\frac{1}{6 n+1}+\frac{-1}{6 n+2}+\frac{-2}{6 n+3}+\frac{-1}{6 n+4}+\frac{1}{6 n+5}+\frac{2}{6 n+6}\right)=0$$
And the Riemann zeta function:
$$\zeta(\rho_1)= \sum\limits_{n=1}^{\infty} \frac{1}{n^{\rho_1}}=0$$
is zero for $\rho_1$=Riemann zeta zero, for example $\rho_1=0.5 + i14.1347...$
Therefore I thought that solving the following equations: $$\frac{1}{6 n+1}=\frac{1}{(6 n+1)^{x_1}}$$ $$-\frac{1}{6 n+2}=\frac{1}{(6 n+2)^{x_2}}$$ $$-\frac{2}{6 n+3}=\frac{1}{(6 n+3)^{x_3}}$$ $$-\frac{1}{6 n+4}=\frac{1}{(6 n+4)^{x_4}}$$ $$\frac{1}{6 n+5}=\frac{1}{(6 n+5)^{x_5}}$$ $$\frac{2}{6 n+6}=\frac{1}{(6 n+6)^{x_6}}$$ would be a good idea, because then I would have the all ones sequence $1,1,1,1,1,1,...$ in the numerator like in the Riemann zeta function.
Solving out the x-es Mathematica gives me in simplified form:
$$x_1=-\frac{\log \left(\frac{1}{6 n+1}\right)}{\log (6 n+1)}$$ $$x_2=-\frac{\log \left(-\frac{1}{6 n+2}\right)}{\log (6 n+2)}$$ $$x_3=-\frac{\log \left(-\frac{2}{6 n+3}\right)}{\log (6 n+3)}$$ $$x_4=-\frac{\log \left(-\frac{1}{6 n+4}\right)}{\log (6 n+4)}$$ $$x_5=-\frac{\log \left(\frac{1}{6 n+5}\right)}{\log (6 n+5)}$$ $$x_6=-\frac{\log \left(\frac{2}{6 n+6}\right)}{\log (6 n+6)}$$
this when discarding adding integer multiples of $2\pi\,i$ which you get from the Reduce[] command.
For a zeta zero I then would like to set $x_1=x_2=x_3=x_4=x_5=x_6$ like this:
$$-\frac{\log \left(\frac{1}{6 n+1}\right)}{\log (6 n+1)}=-\frac{\log \left(-\frac{1}{6 n+2}\right)}{\log (6 n+2)}=-\frac{\log \left(-\frac{2}{6 n+3}\right)}{\log (6 n+3)}=-\frac{\log \left(-\frac{1}{6 n+4}\right)}{\log (6 n+4)}=-\frac{\log \left(\frac{1}{6 n+5}\right)}{\log (6 n+5)}=-\frac{\log \left(\frac{2}{6 n+6}\right)}{\log (6 n+6)}$$
But I can not think of any operation on the equations above that would make them equal, except multiplying all sides:
$$\frac{\text{Log}\left[\frac{1}{1+6n}\right]\text{Log}\left[-\frac{1}{2+6n}\right]\text{Log}\left[-\frac{2}{3+6n}\right]\text{Log}\left[-\frac{1}{4+6n}\right]\text{Log}\left[\frac{1}{5+6n}\right]\text{Log}\left[\frac{2}{6+6n}\right]}{\text{Log}[1+6n]\text{Log}[2+6n]\text{Log}[3+6n]\text{Log}[4+6n]\text{Log}[5+6n]\text{Log}[6+6n]}$$
But it is not a constant function unsurpisingly.
And no combination of $x_1=x_2=x_3=x_4=x_5=x_6$ is equal when trying to solve it in Mathematica.
Where is the fault in all this? At the beginning?
Edit:
Evaluating:
$$f_1(n)=\frac{\text{Log}\left[\frac{1}{1+6n}\right]\text{Log}\left[-\frac{1}{2+6n}\right]\text{Log}\left[-\frac{2}{3+6n}\right]\text{Log}\left[-\frac{1}{4+6n}\right]\text{Log}\left[\frac{1}{5+6n}\right]\text{Log}\left[\frac{2}{6+6n}\right]}{\text{Log}[1+6n]\text{Log}[2+6n]\text{Log}[3+6n]\text{Log}[4+6n]\text{Log}[5+6n]\text{Log}[6+6n]}$$
for $n=1,2,3,4,5,...\infty$ and knowing that $\lim_{s\to 1} \, f(n) = 1 $ one can do: $$f_1(1)=f_1(2)=f_1(3)=f_1(4)=f_1(5)=...=f_1(\infty)$$ Then multiplying all sides sides: $$f_2=f_1(1)*f_1(2)*f_1(3)*f_1(4)*f_1(5)*...*f_1(\infty)$$ one should get a new function. It would be interesting to know if f_2 ever converges to a constant or if it is a new not constant function.
Some Mathematica code:
Clear[x1, x2, x3, x4, x5, x6, n, k]
Sum[1/(6*n + 1) - 1/(6*n + 2) - 2/(6*n + 3) - 1/(6*n + 4) +
1/(6*n + 5) + 2/(6*n + 6), {n, 0, Infinity}]
Print["x1"]
x1 = x1 /. Solve[1/(6*n + 1) == 1/(6*n + 1)^x1, x1]
Print["x2"]
x2 = x2 /. Solve[-1/(6*n + 2) == 1/(6*n + 2)^x2, x2]
Print["x3"]
x3 = x3 /. Solve[-2/(6*n + 3) == 1/(6*n + 3)^x3, x3]
Print["x4"]
x4 = x4 /. Solve[-1/(6*n + 4) == 1/(6*n + 4)^x4, x4]
Print["x5"]
x5 = x5 /. Solve[1/(6*n + 5) == 1/(6*n + 5)^x5, x5]
Print["x6"]
x6 = x6 /. Solve[2/(6*n + 6) == 1/(6*n + 6)^x6, x6]
Print["setting x1 equal to one which it essentially is"]
x1 = 1;
f[n_] = x1*x2*x3*x4*x5*x6
Product[f[n], {n, 0, Infinity}]
N[%]
Limit[f[n], n -> Infinity]