Is this sum equal to the Möbius function?

In the wikipedia page Uses of trigonometry under the section Number theory and in the page for the Möbius function there is an explanation for how to calculate the Mobius function from the GCD=1 pattern:

$\mu(n) = \sum\limits_{\stackrel{1\le k \le n }{ \gcd(k,\,n)=1}} e^{2\pi i \tfrac{k}{n}}$

Consider the infinite matrix starting:

$\displaystyle T(n,k) = \begin{bmatrix} +1&+1&+1&+1&+1&+1&+1 \\ +1&-1&+1&-1&+1&-1&+1 \\ +1&+1&-2&+1&+1&-2&+1 \\ +1&-1&+1&-1&+1&-1&+1 \\ +1&+1&+1&+1&-4&+1&+1 \\ +1&-1&-2&-1&+1&+2&+1 \\ +1&+1&+1&+1&+1&+1&-6 \end{bmatrix}$

defined by the recurrence:

$\displaystyle T(n,1)=1, T(1,k)=1, n>=k: -\sum\limits_{i=1}^{k-1} T(n-i,k), n<k: -\sum\limits_{i=1}^{n-1} T(k-i,n)$

Does the Möbius function $\displaystyle \mu(n)$ also equal the following:

$\displaystyle \mu(n) = \frac{1}{n} \sum\limits_{k=1}^{k=n} T(n,k) \cdot e^{i 2 \pi \frac{k}{n}}$

As a Mathematica program this is:

Clear[nn, t, n, k, a, b];
nn = 15;
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}]];
MatrixForm[Table[Table[t[n, k], {k, 1, nn}], {n, 1, nn}]];
MatrixForm[
Table[N[Total[Table[1/n*t[n, k]*Exp[I*2*Pi*k/n], {k, 1, n}]]], {n, 1,
nn}]]


And if needed, the matrix $\displaystyle T(n,k)$ as Excel cell formulas is:

European: