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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.

share|improve this question
    
Do you have a proof of your claim that $\forall n \in \mathbb{N}, \exists p \in \mathbb{P} \; | \; n = \lfloor \sqrt[3]{p^2} \rfloor$? –  Samuel Reid Mar 2 '13 at 20:47
    
unfortunately I do not its actually one of the things I want to proof. –  pbertsch Mar 2 '13 at 20:51
    
Just made some matlab code to check this and you cant make 10 –  Ben Mar 2 '13 at 21:12

1 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
share|improve this answer
    
you are right I checked and made a small Java app. it seems like there is still a interesting pattern. If I correct the equation to $n= \lfloor \sqrt[3]{p^2}+0.5 \rfloor$ it looks like it works. I will re-think my equation and correct it. –  pbertsch Mar 2 '13 at 22:00
    
UPDATE:with +0.5 there are still gaps in the list. –  pbertsch Mar 2 '13 at 22:10
    
I checked the first 25997 Primes and this are the only numbers I cannot generate so far. notIn {24, 36, 45, 47, 93, 109, 122, 221, 265, 416, 996} Any tips how to proceed ? –  pbertsch Mar 2 '13 at 22:32
    
@pbertsch: I don't mean to discourage your attempts, but you may want to pick up a graduate-level analytic number theory textbook and see what machinery is being used to attack problems like the twin primes conjecture. If you are interested in this problem then it will be good motivation for you to learn advanced number theory as such elementary and computational techniques you are suggesting will almost surely not suffice. –  Samuel Reid Mar 3 '13 at 1:14
    
@SamuelReid Thanks for your advice and you definitely not discouraging me. I will bring me up-to-date to attack the problem properly. –  pbertsch Mar 3 '13 at 5:20

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