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

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• ... N and ... prime ... including ... trying to ... Someone s probably privately replied already, but yes your result is a consequence of the prime number
Message 1 of 4 , Jun 14, 2008
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--- In primenumbers@yahoogroups.com, w_sindelar@... wrote:
>
> 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

Someone's probably privately replied already, but yes your result is a
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

.
• Sindelar wrote in primenumbers message 19425: 2 and N represents the number of primes from
Message 2 of 4 , Jun 17, 2008
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Sindelar wrote in primenumbers message 19425:
< 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)...>
David Broadhurst wrote to Sindelar:
<<I believe that there is /no/ solution to
4*prime(n) + 3*n = 2*prime(2*n)
for n > 352314
Explanation:
1) the asymptotic value of
R(n) = (2*prime(2*n) - 4*prime(n))/n
is
R(infinity) = 4*log(2) = 2.7725887...
by the Prime Number Theorem.
2) For small n, there are excursions with R(n) >= 3,
but these soon cease, with
R(352314) = 3
being your last solution
and
R(352316) = 3 + 1/176158
being the last value of R(n) >= 3.
3) Note that you may use
http://primes.utm.edu/nthprime/
to find R(n) for 2*n <= 10^12.
For example:
The 500,000,000,000th prime is 14,638,944,639,703
The 1,000,000,000,000th prime is 29,996,224,275,833
showing that
R(5*10^11) = (2*29996224275833 - 4*14638944639703)/(5*10^11)
= 2.873339985708
is now significantly smaller than 3,
yet still some way above the asymptote 2.7725887...
David
PS: You may quote this argument in "primenumbers", if you wish. >>
Thank you David, for taking the time to respond to my post. It was quite
a surprise. I had not expected to see my paltry list of 10 N's expanded
to 92 so quickly, and to cap it off, be presented with an argument that
makes a case for there being a limit to the number of solutions to the
equation (4*P+3*N)=2*(the (2*N)th prime). Put another way, this means
that for any N>352314, the difference between the (2*N)th prime and twice
the Nth prime can never equal (3*N)/2.
I feel a bit uneasy about this, asymptotics can be tricky. I think the
idea would be of interest to many in the group and I hope it will provoke
a lively discussion. Best regards
Bill Sindelar
____________________________________________________________
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• 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
Message 3 of 4 , Jun 23, 2008
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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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