Here is what I want to do, I have this matrix:

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

that I have posted many times before, and which has the definition:


where $a$ is the Dirichlet inverse of the Euler totient function and $GCD(n,k)$ is the Greatest Common Divisiors of $n$ and $k$.

I know that the sequence of primes is:

$$-\Lambda (n) (\mu (n)) = 0,\log (2),\log (3),0,\log (5),0,\log (7),0,0,0,\log(11),0...$$

where $(\mu (n))$ is the Möbius function, and $\Lambda (n)$ is the von Mangoldt function.

The von Mangoldt function can be calculated as a limit of the matrix $T$:

$$\Lambda (n) = \sum\limits_{k=1}^{\infty} \frac{T(n,k)}{k}$$

So going about this problem this way I intend therefore to multiply elementwise the Mobius function with the elements in $T(n,k)$:

$$\mu(n) \sum\limits_{k=1}^{\infty} \frac{T(n,k)}{k}$$

After an OEIS search I found, by combining two sequences, that:

$$\left(\sum\limits_{n=1}^{\infty} \frac{(-1)^{(n+1)}\mu(n)}{n^s}\right)^{-1}$$ $$ =\lim_{S\to \text{s}} \, \left(\sum _{b=0}^{\infty } \left(\sum _{a=0}^{\infty } \left(\frac{1}{\left((2\cdot 2)^a (2 b+1)\right)^S}-\frac{1}{\left(2 (2\cdot 2)^a (2 b+1)\right)^S}\right)\right)\right)$$

and after another OEIS search numerical evidence suggests that this is equal to:

$$=\frac{\left(2^s-1\right) \zeta (s)}{2^s+1}$$

where $\zeta (s)$ is the Riemann zeta function.

Well and fine, but how about the rest? I guess it will blow up when I reach $$n=6$$ since:

$$\Lambda(6)^{-1} = \infty$$

but I would like to try anyways for $n=3$. $$\mu(n) \sum\limits_{k=1}^{\infty} \frac{T(n,k)}{k}$$ This is my guess:

$$\mu(n) \sum\limits_{n=1}^{\infty} \frac{T(n,3)}{n} = \frac{\left(3^s-2\right) \zeta (s)}{3^s+1}$$

but it doesn't fit numerically very well. Also the switch between $n$ and $k$ confuses me. Can you help me find the sum for any value of $k$ and $n$?


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