## 19429Re: Bounds on the (2*N)th Prime

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• Jun 14, 2008
>
> Let P represent an odd prime>3 and N represent the number of primes from
> and including 2 to and including P. Then in the set of positive integers
> beginning with (2*P+N) and ending with 2*(P+N), there exists a prime Q
> where the number of primes from and including 2 to and including Q is
> equal to (2*N). The difference D=(Q-2*P) is always a number between
N and
> 2*N.
> For example, take P=47. N=15. (2*P+N)=109. 2*(P+N)=124. The (2*N)th
prime
> is 113. It lies between 109 and 124. D=(Q-2*P)=19. It lies between N=15
> and 2*N =30.
> Here is a neat consequence of the above statement. Take the expression
> ((4*P+3*N)/2), where P represents a prime>2 and N represents the number
> of primes from and including 2 to and including P. Then for certain
> values of P and its corresponding N, the expression will evaluate to a
> prime Q, where the number of primes from and including 2 to and
including
> Q is equal to (2*N). The Q's are symmetrically located between (2*P+N)
> and 2*(P+N). The D's are symmetrically located between N and 2*N.
> There seems to be an inexhaustible supply of such P's. Here is a partial
> list of the first 10 of such P's. (43, 163, 373, 397, 491, 1997, 2339,
> 4691, 7331, 12149), and the list of corresponding N's. (14, 38, 74, 78,
> 94, 302, 346, 634, 934, 1454).
> The only web reference I was able to find for this, is the
> Bertrand-Chebyshev theorem, which says that there is always at least one
> prime between N and 2*N-2. I can see no connection. I have been
trying to
> prove that it is an obvious consequence of the prime number theorem, but
> can't see my way clear. Can anyone help?
> Thanks folks. Any comments would be appreciated.
> Bill Sindelar

consequence of the prime number theorem (although it certainly doesn't
prove that your result will always hold):

The nth prime occurs around n*log(n), which is close to your p.

The (2*n)th prime occurs around 2*n*log(2*n).

But log(2*n) = log(2) + log(n), so the (2*n)th prime occurs around

2*n*log(n) + 2*n*log(2) =~ 2*n*log(n) + 1.386*n

But n*log(n) is close to your p, so the (2*n)th prime occurs around

2*p + 1.386*n, which is in between 2*p + n and 2*p + 2*n as you have
found.

Mark

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