19433Re: Bounds on the (2*N)th Prime
- Jun 17, 2008Sindelar 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
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
1) the asymptotic value of
R(n) = (2*prime(2*n) - 4*prime(n))/n
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
R(352316) = 3 + 1/176158
being the last value of R(n) >= 3.
3) Note that you may use
to find R(n) for 2*n <= 10^12.
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
R(5*10^11) = (2*29996224275833 - 4*14638944639703)/(5*10^11)
is now significantly smaller than 3,
yet still some way above the asymptote 2.7725887...
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
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