Thanks 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 asymptotic

> 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

> R(n)=3.

>

> David

>

> PS: Feel free to share this, if you might like to...

Thanks folks. Anyone care to comment?

Bill Sindelar

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