Consider the bipartite graphs defined here: Why is this bipartite graph a partial cube, if it is?
We do random walks on them with equal propability and since the graphs are finite and connected the markov chain has a stationary distribution, which I suspect to be :
$$p_n = ( \deg(1) , \cdots, \deg(n) ) \cdot \frac{1}{2 |E_n|}$$
where if it is of interest, we have:
$$\deg(v) = \pi(n/v)+\omega(v)$$
Here $\pi$ counts the prime numbers and $\omega$ counts the prime divisors of $n$ without multipliclity.
Let the spin $\sigma_v$ of a number $v$ be defined as:
$$\sigma_v = (-1)^{\Omega(v)}$$
Consider the vector $X_n$:
$$X_n = (\sigma_1,\cdots,\sigma_n)$$
Now, I have checked empirically that, the expected value $E(\sigma_v)$ is zero:
$$0 = \left < p_n, X_n \right > = E(\sigma_v)$$
Then the variance must be $ \operatorname{Var}(\sigma_v) = 1$.
Now if we allow the not so random numbers $\sigma_1,\cdots,\sigma_n$ to be treaten as random numbers with expected value $\mu=0$ and variance $\sigma=1$:
$$\sigma_v,v=1,\cdots,n$$
we could apply the central limit theorem to the "random variable":
$$S_n :=\frac{\sum_{v=1}^n\sigma_v}{n}$$
and get:
$$\frac{\sqrt{n}}{\sigma}(S_n - \mu) = \frac{\sum_{v=1}^n\sigma_v}{\sqrt{n}}$$
converges in probability to the normal distribution $\mathbf{N}(0,1)$ with mean $0$.
Suppose that the stationary distribution could be proven. Is there anything wrong with this approach to proving the prime number theorem? Thanks for your help!
Edit: Here is a proof that the expected value equals zero and as a consequence we get a formula for the prime counting function $\pi$:
$$\pi(n) = - \sum_{v=2}^n ( \pi(n/v)+\omega(v))(-1)^{\Omega(v)}$$
"The expected value is zero" is true in all finite bipartite graphs:
$$0 = \sum_{v \in V_n} \deg(v) \sigma_v $$
where $\sigma_v = (-1)^{d(1,v)}$ where $d$ is the shortest path distance in the finite bipartite undirected graph $G_n = (V_n,E_n)$.
Proof: We have
$$\sigma_v \deg(v) = \sigma_v \sum_{u\approx v} +1 = -\sigma_v \sum_{u \approx v} (-1)^{d(u,v)} = \ldots$$
Since $G_n$ is a bipartite graph we have only even length closed walks, so :
$$d(u,v) \equiv d(1,v)+d(1,u) \mod (2)$$
and we get:
$$ \ldots = -\sigma_v \sum_{u \approx v}(-1)^{d(1,u)}(-1)^{d(1,v)} = -\sigma_v \sum_{u \approx v} \sigma_u \sigma_v$$
$$ = -\sum_{u \approx v} \sigma_u$$
But since $V_n = A_n \cup B_n$ is a bipartition of $G_n$, where $A_n = \{ v| \sigma_v = +1\}$ and $B_n = \{ v | \sigma_v = -1 \}$, we get:
$$\sum_{v \in V_n} \deg(v) \sigma_v = -\sum_{v \in V_n}\sum_{u \approx v} \sigma_u$$
$$ = -\sum_{v \in A_n}\sum_{u \approx v} \sigma_u -\sum_{v \in B_n}\sum_{u \approx v} \sigma_u$$
$$ = -\sum_{v \in A_n} (-1)\deg(v) - \sum_{v \in B_n} (+1)\deg(v)$$ $$ = \sum_{v \in A_n} \deg(v) - \sum_{v \in B_n} \deg(v)$$ $$ = |E_n|-|E_n| = 0$$
which proves the assertion.
The stationary vector can be proved directly I think by showing that it is an eigenvector to the eigenvalue $1$ and using the Perron-Frobenius theorem on stochastic matrices.
Second edit:
From this it follows with:
$$P_n(v) = \frac{\deg(v)}{2|E_n|}$$
that:
$$P_n(\sigma_v = +1) = P_n(\sigma_v = -1) = 1/2$$
This is because
$$P_n(\sigma_v = +1) = P_n(A_n) = \sum_{v \in A_n} P_n(v)$$
and
$$P_n(\sigma_v = -1) = P_n(B_n) = \sum_{v \in B_n} P_n(v)$$
But $P_n(B_n) = 1-P_n(A_n)$ since $V_n = A_n \cup B_n$ and so it follows that:
$$P_n(\sigma_v = +1) = P_n(\sigma_v = -1) = 1/2$$
So the random variables $\sigma_v, v \in V_n$ are Rademacher variables which take the values $\pm 1$ with probability $1/2$.
Now in the Riemann hypothesis, we are interested if the limit:
$$\lim_{n \rightarrow \infty} \left | \frac{\sum_{v=1}^n \sigma_v }{\sqrt{n}} \right | $$
exists, to conclude that $\forall \epsilon>0$:
$$\lim_{n \rightarrow \infty} \frac{\sum_{v=1}^n \sigma_v }{\sqrt{n} n^{\epsilon}} =0 $$
But for independent random variables $x_v$ with Rademacher distribution, we have ( https://mathworld.wolfram.com/RandomWalk1-Dimensional.html ) the expected value is given by:
$$\lim_{n \rightarrow \infty} E \left ( \left | \frac{\sum_{v=1}^n x_v }{\sqrt{n}} \right | \right ) = \sqrt{\frac{2}{\pi}} $$
So if we think about the $\sigma_v$ heuristically as random variables, motivated by the $P_n( \sigma_v = +1) = 1/2 = P_n( \sigma_v = -1)$, then we should expect that:
$$\lim_{n \rightarrow \infty} \left | \frac{\sum_{v=1}^n \sigma_v }{\sqrt{n}} \right | = \sqrt{\frac{2}{\pi}}$$
Here is some python code for simulation, with the following picture:
