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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*)
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Such proofs are heuristic. They can be used to demonstrate that a certain result is probably true. Similar arguments can be used to "prove" the twin prime conjecture or show that there are infinitely many Mersenne primes. Such arguments are never rigorous as they do not actually prove anything, they can simply serve as an indication that a particular result could be true. I believe Terence Tao is interested in these so-called heuristic justifications, so you might want read his blog. Also, I remember watching a talk he gave about this area of research on YouTube.

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  • 1
    $\begingroup$ I unselected this answer because of the changes I made to the question. I made an attempt at a proof. $\endgroup$ – Mats Granvik Mar 21 '14 at 23:12

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