From Wikipedia:
"The prime number theorem is also equivalent to:
$$\lim_{x \rightarrow \infty} \frac{\psi(x)}{x}=1$$
where $$\psi(x) = \sum\limits_{n \leq x} \Lambda(n)$$ is the Chebyshev function. and where: $$\Lambda(n) = \begin{cases} \log p & \text{if }n=p^k \text{ for some prime } p \text{ and integer } k \ge 1, \\ 0 & \text{otherwise.} \end{cases}$$ is the von Mangoldt function.
The von Mangoldt function can be calculated as.
Edit 27.1.2018, I rewrote the whole question from here on, trying to use more conventional notation. $n$ stands for row index, and $k$ stands for column index.
Mathematica knows that: $$\log(n)=\lim\limits_{s \rightarrow 1}\zeta(s)\left(1-\frac{1}{n^{s-1}}\right) \; \; \; \; \; \; (1)$$ and it has been proven that for $n>1$ the von Mangoldt function is: $$\Lambda(n)=\lim\limits_{s \rightarrow 1}\zeta(s)\sum\limits_{d|n} \frac{\mu(d)}{d^{s-1}} \; \; \; \; \; \; (2)$$ The Dirichlet series associated with the von Mangoldt function defined in such away as above, is the infinite symmetric square matrix $T_1$:
$$T_1=a(GCD(n,k)) \; \; \; \; \; \; (3)$$
which starts:
$$T_1 = \left( \begin{array}{ccccccc} +1&+1&+1&+1&+1&+1&+1&\cdots \\ +1&-1&+1&-1&+1&-1&+1 \\ +1&+1&-2&+1&+1&-2&+1 \\ +1&-1&+1&-1&+1&-1&+1 \\ +1&+1&+1&+1&-4&+1&+1 \\ +1&-1&-2&-1&+1&+2&+1 \\ +1&+1&+1&+1&+1&+1&-6 \\ \vdots&&&&&&&\ddots \end{array} \right)$$
where:
$$a(n) = \sum\limits_{d|n} d \cdot \mu(d) \; \; \; \; \; \; (4)$$
(better known as the Dirichlet inverse of the Euler totient.)
Now by periodicity of the entries in the columns for $m=0,1,2,3,4,5,...$ or $m \in \mathcal ℕ_0$, the column sums satisfy:
$$\sum\limits_{n=1+m}^{n=k+m} T_1(n,k)=\begin{cases}1 & \mbox{ if } k=1\\ 0&\mbox{ if } k>1.\end{cases} \; \; \; \; \; \; (5)$$
and by periodicity of the entries in the rows, the row sums satisfy:
$$\sum\limits_{k=1+m}^{k=n+m} T_1(n,k)=\begin{cases}1 & \mbox{ if } n=1\\ 0&\mbox{ if } n>1.\end{cases} \; \; \; \; \; \; (6)$$ by symmetry.
Since from $(2)$ for $n>1$:
$$\Lambda(n)=\sum\limits_{k=1}^{k=\infty}\frac{T_1(n,k)}{k} \; \; \; \; \; \; (7)$$
and:
$$\sum\limits_{n=1+m}^{n=k+m} \frac{T_1(n,k)}{k} =\begin{cases}1 & \mbox{ if } k=1\\ 0&\mbox{ if } k>1.\end{cases} \; \; \; \; \; \; (8)$$
and since the denominator is $k$ in $(8)$ and the length of the period in the $k$-the column is also $k$, we can say that the average contribution to $\Lambda(n)$ from an entry $T_1(n,k)$ for $k>1$ in matrix $T_1$ must be:
$$\sum\limits_{n=1+m}^{n=k+m} \frac{T_1(n,k)}{k} =\begin{cases}1 & \mbox{ if } k=1\\ 0&\mbox{ if } k>1.\end{cases} \; \; \; \; \; \; (9)$$
Writing: $$\sum_{n \leq x} \Lambda(n)=\sum_{n \leq x} \sum\limits_{k=1}^{k=\infty}\frac{T_1(n,k)}{k} \; \; \; \; \; \; (10)$$
Combining $(9)$ with the right hand side of $(10)$ and multiplying the denominators $k \cdot k = k^2$ we get:
$$\sum_{n \leq x} \Lambda(n)=\sum_{n \leq x} \sum\limits_{k=1}^{k=\infty}\sum\limits_{n=1+m}^{n=k+m} \frac{T_1(n,k)}{k^2} =\begin{cases} (1+o(1))x & \mbox{ if } k=1\\ (0+o(1))x &\mbox{ if } k>1.\end{cases} \; \; \; \; \; \; (11)$$
and thereby at least heuristically:
$$\sum_{n \leq x} \Lambda(n) = (1+o(1)) x \; \; \; \; \; \; (12)$$
$\square$
Associated Mathematica 8 code:
nn = 7
a[n_] := If[n < 1, 0, Sum[d MoebiusMu@d, {d, Divisors[n]}]]
MatrixForm[
Table[Table[If[k == 1, 0, a[GCD[n, k]]], {k, 1, nn}], {n, 1, nn}]]
MatrixForm[
Table[Table[
Sum[If[k == 1, 0, a[GCD[n, k]]], {k, 1, x}], {x, 1, nn}], {n, 1,
nn}]]
MatrixForm[
Table[Table[
Sum[If[k == 1, 0, a[GCD[n, k]]], {k, 1, x}]/x, {x, 1, nn}], {n, 1,
nn}]]
Edit 18.2.2018: I will try to add some details.
Starting with matrix $T_1$ defined above, we form the matrix $T_2$, but before that we state again the property of matrix $T_1$, namely:
$$\Lambda(k) = \sum\limits_{n=1}^{n=\infty}\frac{T_1(n,k)}{n}$$
and then matrix $T_2$:
$$T_2(n,k)=\sum\limits_{k=1}^{k=g}T_1(n,k)$$ where: $$g=1,2,3,4,5,...$$ and: $$n=1,2,3,4,5,...$$
Matrix $T_2$ in turn has the property:
$$\psi(x)=\sum\limits_{k \leq x} \Lambda(k) = \sum\limits_{n=1}^{n=\infty}\frac{T_2(n,k)}{n}$$
We then form matrix $T_3$:
$$T_3(n,k)=\frac{T_2(n,k)}{n \cdot k}$$
Since for $n>1$:
$$\lim\limits_{k \rightarrow \infty} T_3(n,k) = 0$$
and since:
$$\lim\limits_{k \rightarrow \infty} T_3(1,k) = 1$$
the prime number theorem is true/follows.
$\square$
In case I did not get it entirely right I attach this second program in Mathematica as a verification:
(*start*)
nn = 12;
TableForm[
A = Table[Table[If[Mod[n, k] == 0, 1, 0], {k, 1, nn}], {n, 1, nn}]];
TableForm[
B = Table[
Table[If[Mod[k, n] == 0, MoebiusMu[n]*n, 0], {k, 1, nn}], {n, 1,
nn}]];
TableForm[T1 = (A.B)];
TableForm[
T2 = Table[Table[Sum[T1[[n, k]], {k, 1, g}], {g, 1, nn}], {n, 1, nn}]]
TableForm[T3 = Table[Table[T2[[n, k]]/n/k, {k, 1, nn}], {n, 1, nn}]]
(*end*)
