|
%I #14 Jul 02 2013 15:46:06
%S 5,6,15,21,25,26,28,29,30,33,35,38,39,44,47,50,62,63,65,67,74,78,80,
%T 86,94,105,108,109,112,114,153,155,164,165,170,171,172,173,174,207,
%U 208,209,215,276,279,282,283,285,287,293,294,297,298,299,504,704,712,714,1308,1316,1322,1328
%N Number of primes between n and 2n inclusive equals the number of primes between 2n and 3n inclusive.
%C There are almost certainly no more terms. Can anyone prove this conclusively?
%C The analogous sequence where the number of primes between 0 and n equals the number of primes from n to 2n is finite, consisting of only 2, 4, and 10. See Ehrhart. - _Charles R Greathouse IV_, Jul 02 2013
%H Eugene Ehrhart, <a href="http://www.fq.math.ca/Scanned/26-3/ehrhart.pdf">On prime numbers</a>, Fibonacci Quarterly 26:3 (1988), pp. 271-274.
%F Numbers n such that pi(3n) + pi(n-1) = 2pi(2n).
%e 4 is not in the sequence since the interval [n,2n] = [4,8] contains two primes (5 and 7), while the interval [2n,3n] = [8,12] contains only one prime (11).
%e 6 is in the sequence since the intervals [6,12] contains two primes (7 and 11), and the interval [12,18] also contains two primes (13 and 17).
%t Do[If[PrimePi[3n]+PrimePi[n-1]==2*PrimePi[2n],Print[n]],{n,4,10^6}]
%o (PARI) is(n)=my(t=primepi(3*n)+primepi(n-1));t%2==0 && t==2*primepi(2*n) \\ _Charles R Greathouse IV_, Jul 02 2013
%K nonn
%O 1,1
%A _Zak Seidov_, Jun 30 2013
|
