Idea about Twin-Primes and the generation of natural numbers

Question

Years ago (6 years to be exact) I was fascinate by prime-twins, and still I am, but the years went by and I almost forgot about it until yesterday.

I found my notes again and I don't know if I am on something. Maybe someone could show me the right way proving it or even tell me if it is worth going on with this. So...

Let $ \mathbb{P} $ are all the prime numbers und $ \mathbb{P_2} $ all the twin-primes (I'm not assuming there are infinite).

Be $p, q \space \epsilon \space \mathbb{P} $.

$ \exists p: $ $ q = \lfloor \sqrt[3]{p^2} \rfloor $ and $ q+2 \space \epsilon \space \mathbb{P} \Rightarrow q, q+2 \space \epsilon \space \mathbb{P_2} \\ $
Every natural number n can be formed by $ n=\lfloor \sqrt[3]{p^2} \rfloor $

Let $ n \space \epsilon \space \mathbb{N} \\ n=\lfloor \sqrt[3]{p^2} \rfloor \Leftrightarrow n\leq \sqrt[3]{p^2} < n+1 \Leftrightarrow n^3 \leq p^2 < (n+1)^3 \Leftrightarrow \sqrt{n^3} \leq p < \sqrt{(n+1)^3} $

$ \forall \space n \space \epsilon\space \mathbb{N}\space \exists\space p \space\epsilon \space \mathbb{P}:$ $ \sqrt{n^3} \leq p < \sqrt{(n+1)^3} $

By looking at my approach I came across the idea to sum the reciprocal $\sqrt{ n^3} $

After looking closer I realized that the sum that I got is a geometric series. $ \sum n^{-\alpha} \; (\forall \alpha > 1) $ I think I could show the convergence with Cauchy.

This series even looks like a Zeta-Function which converges near $e \space(2.6149)$. I'm not sure if this is already it or if it could converge to $e$

I do this in my spare time and I always wanted to ask someone about my idea. I lost all my computations but if someone thinks it could be worth the effort I would start again.

Thanks.

Answer 1

This is not the case as the array notIn in the following code is $\lbrace 10, 20 , 24 , 27 , 32 , 65 , 121 , 139 , 141 , 187 , 306 , 321 , 348 , 1006 , 1051\rbrace$ and so the claim you can make any natural number is false.

The code:

 clc
 clear
 A= primes(1000000);

 for i = 1:length(A(1,:))
 n(i) = floor(A(1,i)^(2/3)); %numbers made as floor(p^(2/3)
 end

 naturals = (1:max(n));      list of naturals from 1 to max made in above 
                             for loop
 index = 1;
 for i = 1 : max(n)
 check = 0;
 for j = i : length(n)       just check if this natural occurs in 
                             the list n
    if naturals(i) == n(j)
        check = check +1;
        break
    end
end
if check == 0                if we have made it through without a  
                             match then n cant be made as claimed
    notIn(index) = naturals(i);
    index = index +1;
end 
end

notIn