Let $\mu(n)$ be the Möbius function and let $a(n)$ be the Dirichlet inverse of the Euler totient function:
$$a(n) = \sum\limits_{d|n} d \cdot \mu(d)$$
Let the matrix $T$ be defined as:
$$T(n,k)=a(GCD(n,k))$$
where GCD = the Greatest Common Divisor.
This matrix $T$ starts:
$$T = \left( \begin{array}{ccccccc} +1&+1&+1&+1&+1&+1&+1&\cdots \\ +1&-1&+1&-1&+1&-1&+1 \\ +1&+1&-2&+1&+1&-2&+1 \\ +1&-1&+1&-1&+1&-1&+1 \\ +1&+1&+1&+1&-4&+1&+1 \\ +1&-1&-2&-1&+1&+2&+1 \\ +1&+1&+1&+1&+1&+1&-6 \\ \vdots&&&&&&&\ddots \end{array} \right)$$
There is then the symmetry that the von Mangoldt function can be computed as: $$\Lambda(n) = \sum\limits_{k=1}^{k=\infty}\frac{T(n,k)}{k}$$
$$\Lambda(k) = \sum\limits_{n=1}^{n=\infty}\frac{T(n,k)}{n}$$
In the language of Dirichlet series this is equivalent to: $$\Lambda(n)=\lim\limits_{s \rightarrow 1} \zeta(s)\sum\limits_{d|n} \frac{\mu(d)}{d^{(s-1)}}$$
Can we then say that the Dirichlet series generating function for the whole matrix $T$ is:
$$\sum\limits_{k=1}^{\infty}\sum\limits_{n=1}^{\infty} \frac{T(n,k)}{n^c \cdot k^z} = \sum\limits_{n=1}^{\infty} \frac{\lim\limits_{s \rightarrow z} \zeta(s)\sum\limits_{d|n} \frac{\mu(d)}{d^{(s-1)}}}{n^c} = \frac{\zeta(z) \cdot \zeta(c)}{\zeta(c + z - 1)}$$?
For $c$ a value close to one such as $c=1+\frac{1}{30}$ we have this plot for $\frac{\zeta (c) \zeta (\frac{1}{2}+i t)}{\zeta (c+\frac{1}{2}+i t-1)}$ on the critical line in the complex plane:
where the downward spikes are at Riemann zeta zeros.
(*Mathematica 8 start*)
Clear[s, z, c]
z = 9
c = 7
N[Zeta[z]*Zeta[c]/Zeta[c + z - 1], 20]
N[Sum[Limit[
Zeta[s] Total[1/Divisors[n]^(s - 1)*MoebiusMu[Divisors[n]]],
s -> z]/n^c, {n, 1, 64}], 20]
Clear[c, ss, t]
ss = 30
c = 1 + 1/ss;
sigma = 1/2;
Plot[Re[N[Zeta[sigma + I*t]*Zeta[c]/Zeta[c + sigma + I*t - 1],
20]], {t, 0, 60}, PlotRange -> {0, ss + 2}]
(*Mathematica 8 end*)