19447Bounds on the (2*N)th Prime
- Jun 23, 2008Thanks David for further explanation of your reasoning and Andrey
Kulsha's graphic. I've done a lot of calculating and believe everything
strongly suggests that the following statement may be true.
Let P represent the Nth prime and Q represent the (2*N)th prime.
Let K represent a rational number between the integers 1 and 2.
Then the equation (Q-2*P)=K*N always has a least one solution. The
maximum number of solutions possible is when K=3/2.
For those in the group that may be interested, here is David's reason for
believing that the equation (4*P+3*N)=2*(the (2*N)th prime), which is
equivalent to (Q-2*P)=K*N with K=(3/2), has a limited number of
solutions. (David calculated 92)
> 1) First, I worked out the second term in the asymptoticThanks folks. Anyone care to comment?
> expansion of
> R(n) = (2*prime(2*n) - 4*prime(n))/n
> as here:
> R(n) ~ 4*log(2)*(1 + 1/log(n))
> 2) Then I proved, by explicit computation,
> that there is no solution to R(n)=3
> with 1500000 > n > 352314 (file attached)
> 3) In the course of this proof, I saw, by outputting
> [n, R(n), 4*log(2)*(1+1/log(n))]
> at intervals of 10,000
> that the actual value, R(n), oscillates only modestly
> about the asymptote 4*log(2)*(1+1/log(n))
> so I believe that I have cause to believe
> (yet cannot prove) that n=352314 is the last solution with
> PS: Feel free to share this, if you might like to...
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