# Is this sum equal to this convolution?

Consider the matrix $T$ defined by:

$$T=a(GCD(n,k))$$

where $GCD(n,k)$ is the Greatest Common Divisor of row index $n$ and column index $k$, and $$a(n) = \lim\limits_{s \rightarrow 1} \zeta(s)\sum\limits_{d|n} \mu(d)(e^{d})^{(s-1)}$$

is the Dirichlet inverse of the Euler totient function.

This infinite matrix starts:

$$\displaystyle T = \begin{bmatrix} +1&+1&+1&+1&+1&+1&+1&\cdots \\ +1&-1&+1&-1&+1&-1&+1 \\ +1&+1&-2&+1&+1&-2&+1 \\ +1&-1&+1&-1&+1&-1&+1 \\ +1&+1&+1&+1&-4&+1&+1 \\ +1&-1&-2&-1&+1&+2&+1 \\ +1&+1&+1&+1&+1&+1&-6 \\ \vdots&&&&&&&\ddots \end{bmatrix}$$

Prove or disprove that:

$$\sum _{k=1}^n \log \left(\frac{n}{k}\right) T(n,k) = \lim_{s\to 1} \, \zeta (s) \sum _{k=1}^n \frac{T(n,k)}{(k/n)^{s-1}}$$

for $n>1$.

Mathematica:

Clear[nn, t, n, k, i, s];
nn = 12;
t[n_, 1] = 1;
t[1, k_] = 1;
t[n_, k_] :=
t[n, k] =
If[n >= k, -Sum[t[n - i, k], {i, 1, k - 1}], -Sum[
t[k - i, n], {i, 1, n - 1}]];
Exp[Table[
Limit[Zeta[s]*Sum[If[n == 1, 0, t[n, k]]/(k/n)^(s - 1), {k, 1, n}],
s -> 1], {n, 1, nn}]]
Exp[Table[Sum[Log[n/k]*t[n, k], {k, 1, n}], {n, 1, nn}]]