- 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

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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