# Are the elements in the n-th row of the first matrix a permutation of the elements in the n-th row of the second matrix?

From my previous questions here and here the following two matrices arise for twin primes and cousin primes from Dirichlet convolution:

For $h=2$ twin primes:

$$T_2(n,m)=\sum\limits_{\substack{k=1 \\ k|n}}^n a(\gcd (k,m+2)) a\left(\gcd \left(\frac{n}{k},m\right)\right)\ \ \ \ \ \ \ \ \ \ (1)$$

For $h=4$ cousin primes:

$$T_4(n,m)=\sum\limits_{\substack{k=1 \\ k|n}}^n a(\gcd (k,m+4)) a\left(\gcd \left(\frac{n}{k},m\right)\right)\ \ \ \ \ \ \ \ \ \ (2)$$

where $$a(n) = \sum\limits_{d|n}\mu(d)d\ \ \ \ \ \ \ \ \ \ (3)$$, $\text{gcd}$ is the greatest common divisor and $\mu(n)$ is the Möbius function.

The general form for any $h$ is: $$T_h(n,m)=\sum\limits_{\substack{k=1 \\ k|n}}^n a(\gcd (k,m+h)) a\left(\gcd \left(\frac{n}{k},m\right)\right)\ \ \ \ \ \ \ \ \ \ (4)$$

Question:

Are the elements in the $n$-th row of matrix $T_2$ a permutation of the elements in the $n$-th row of matrix $T_4$?

If they are then the asymptotic density of twin primes should be equal to the asymptotic density of cousin primes because the answer by Ofir Gorodetsky here:

$$\lim_{x \to \infty} \frac{\sum_{1 \le n\le x} \Lambda(n)\Lambda(n+h)}{x} =C(h) \ \ \ \ \ \ \ \ \ \ (5)$$

can be substituted with:

$$\Lambda(n)=\lim\limits_{s \rightarrow 1} \zeta(s)\sum\limits_{d|n} \frac{\mu(d)}{d^{(s-1)}} \ \ \ \ \ \ \ \ \ \ (6)$$

$$\Lambda(n)\Lambda(n+h) = \left(\lim\limits_{s \rightarrow 1} \zeta(s)\sum\limits_{d|n} \frac{\mu(d)}{d^{(s-1)}}\right)\left( \lim\limits_{s \rightarrow 1} \zeta(s)\sum\limits_{d|n+h} \frac{\mu(d)}{d^{(s-1)}}\right) \ \ \ \ \ \ \ \ \ \ (7)$$

and $(6)$ is equivalent to:

$$\Lambda(n)=\sum\limits_{k=1}^{\infty} \frac{a(gcd(n,k))}{k}\ \ \ \ \ \ \ \ \ \ (8)$$

so that from the combination of $(8)$ with $(7)$ we get $(4)$ through Dirichlet convolution.
From $(5)$ and if the question about the elements in the $n$-th rows in $T_2$ and $T_4$ is answered in the positive and if the rows are periodic then:

$$\lim\limits_{x \rightarrow \infty} \frac{\sum\limits_{k=1}^{k=x} T_2(n,k)}{x}=\lim\limits_{x \rightarrow \infty} \frac{\sum\limits_{k=1}^{k=x} T_4(n,k)}{x} \ \ \ \ \ \ \ \ \ (9)$$ for all $n$, which fulfils $(5)$ so that $$\lim_{x \to \infty} \frac{\sum_{1 \le n\le x} \Lambda(n)\Lambda(n+2)}{x}=\lim_{x \to \infty} \frac{\sum_{1 \le n\le x} \Lambda(n)\Lambda(n+4)}{x} \ \ \ \ \ \ \ \ \ \ (10)$$.

Associated Mathematica program:

Clear[nn, a, n, d, h, T2, T4, k, m];
nn = 25;
a[n_] := If[n < 1, 0, Sum[d MoebiusMu@d, {d, Divisors[n]}]];
h = 2;
TableForm[
T2 = Table[
Table[Sum[
If[Mod[n, k] == 0, a[GCD[n/k, m]]*a[GCD[k, m + h]], 0], {k, 1,
n}], {m, 1, nn}], {n, 1, nn}]]
h = 4;
TableForm[
T4 = Table[
Table[Sum[
If[Mod[n, k] == 0, a[GCD[n/k, m]]*a[GCD[k, m + h]], 0], {k, 1,
n}], {m, 1, nn}], {n, 1, nn}]]
N[Table[Sum[T2[[3, n]], {n, 1, X}]/X, {X, 1, nn}]]
N[Table[Sum[T4[[3, n]], {n, 1, X}]/X, {X, 1, nn}]]
N[Table[Sum[T2[[16, n]], {n, 1, X}]/X, {X, 1, nn}]]
N[Table[Sum[T4[[16, n]], {n, 1, X}]/X, {X, 1, nn}]]


Matrix $T_2$ is an infinite (square) matrix whose lower triangular part starts:

$$T_2=\begin{array}{llllllllllll} 1 & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} \\ 2 & -2 & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} \\ -1 & 2 & -1 & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} \\ 3 & -1 & 3 & -1 & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} \\ 2 & 2 & -3 & 2 & -3 & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} \\ -2 & -4 & -2 & 2 & 4 & 2 & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} \\ 2 & 2 & 2 & 2 & -5 & 2 & -5 & \text{} & \text{} & \text{} & \text{} & \text{} \\ 4 & 0 & 4 & 0 & 4 & 0 & 4 & 0 & \text{} & \text{} & \text{} & \text{} \\ -3 & 3 & -3 & -3 & 3 & -3 & -3 & 3 & -3 & \text{} & \text{} & \text{} \\ 4 & -4 & -6 & -4 & -6 & -4 & 4 & 6 & 4 & 6 & \text{} & \text{} \\ 2 & 2 & 2 & 2 & 2 & 2 & 2 & 2 & -9 & 2 & -9 & \text{} \\ -3 & -2 & -3 & 1 & 6 & 1 & -3 & -2 & -3 & 1 & 6 & 1 \end{array}$$

Matrix $T_4$ is also an infinite (square) matrix whose lower triangular part starts:

$$T_4=\begin{array}{llllllllllll} 1 & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} \\ 2 & -2 & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} \\ 2 & -1 & -1 & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} \\ 3 & -1 & 3 & -1 & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} \\ -3 & 2 & 2 & 2 & -3 & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} \\ 4 & 2 & -2 & -4 & -2 & 2 & \text{} & \text{} & \text{} & \text{} & \text{} & \text{} \\ 2 & 2 & -5 & 2 & 2 & 2 & -5 & \text{} & \text{} & \text{} & \text{} & \text{} \\ 4 & 0 & 4 & 0 & 4 & 0 & 4 & 0 & \text{} & \text{} & \text{} & \text{} \\ 3 & -3 & -3 & 3 & -3 & -3 & 3 & -3 & -3 & \text{} & \text{} & \text{} \\ -6 & -4 & 4 & -4 & -6 & 6 & 4 & -4 & 4 & 6 & \text{} & \text{} \\ 2 & 2 & 2 & 2 & 2 & 2 & -9 & 2 & 2 & 2 & -9 & \text{} \\ 6 & 1 & -3 & -2 & -3 & 1 & 6 & 1 & -3 & -2 & -3 & 1 \end{array}$$