Let $A$ be the matrix defined as:
$$A(n,k)= \text{If } k \text{ divides } n \text{ then } k \text{ else } 0$$
and let $B$ be the matrix defined as:
$$B(n,k)= \text{If } n \text{ divides } k \text{ then } \mu(n) \text{ else } 0$$
where $\mu(n)$ is the Möbius function.
Then repeatedly take the matrix exponential of A and after that multiply it with matrix $B$ to get matrix $M$:
$$\Large M=e^{e^{e^{.^{.^{.^{e^{A}}}}}}}.B$$
Then define the Greatest Common Divisor related polynomial $T$ as:
$$\displaystyle T(n,1)=0, T(1,k)=0, n \geq k: x -\sum\limits_{i=1}^{k-1} T(n-i,k), n<k: x -\sum\limits_{i=1}^{n-1} T(k-i,n)$$
Then add the polynomial $T$ to the matrix $M$ and take the determinant equal to zero and solve for x, like in an eigenvalue characteristic polynomial matrix:
$$|M+T|=0$$
The first few zeros are:
$$\left\{0,0,0,0,e^{e^{e^{e^2}}},e^{e^{e^{e^3}}},e^{e^{e^{e^5}}},\frac{e^{e^{e^{e^6}}}}{4},e^{e^{e^{e^7}}},\frac{e^{e^{e^{e^{10}}}}}{6},e^{e^{e^{e^{11}}}}\right\}$$
This is the same sequence as in Eric Naslund's answer with a small modification:
$$\frac{e^{e^{e^{e^n}}} \mu (n)^2}{n-\phi (n)}$$
which for $n>1$ gives the sequence:
$$\left\{e^{e^{e^{e^2}}},e^{e^{e^{e^3}}},0,e^{e^{e^{e^5}}},\frac{e^{e^{e^{e^6}}}}{4},e^{e^{e^{e^7}}},0,0,\frac{e^{e^{e^{e^{10}}}}}{6},e^{e^{e^{e^{11}}}},0\right\}$$
If instead we set $$\Large M=A^s.B$$ where $A^s$ stands for matrix $A$ raised to the matrix power $s$, and let the polynomial $T$ be defined as before and again solve the determinant equation:
$$|M+T|=0$$
Then we apparently get the sequence:
$$\frac{\mu (n)^2 n^s}{n-\phi (n)}$$
But what is the Dirichlet generating function for:
$$\sum\limits_{n \geq 1}\frac{\mu (n)^2 n^s}{n-\phi (n)}=?$$
where $\phi (n)$ is the Euler totient function.
Associated Mathematica code: http://pastebin.com/p1EnCmEZ
