A192390 Numbers n such that number of primes in the range (2^n-sqrt(2^n), 2^n] is equal to number of primes in the range (2^n, 2^n+sqrt(2^n)].

1, 2, 6, 10, 21

Offset

1,2

Comments

a(6) > 60. Probably a finite sequence. [Charles R Greathouse IV, Jun 30 2011]

Links

Table of n, a(n) for n=1..5.

Example

a(1)=1 because 2 in range (2^1-sqrt(1), 2^1]=(1, 2] and 3 in range (2^1, 2^1+sqrt(1)]=(2, 3].
a(3)=6 because 59, 61 in range (2^6-sqrt(2^6), 2^6]=(56, 64] and 67, 71 in range (2^6, 2^6+sqrt(2^6)]=(64, 72].
a(4)=10 because 997, 1009, 1013, 1019, 1021 in range (2^10-sqrt(2^10), 2^10]=(992, 1024] and 1031, 1033, 1039, 1049, 1051 in range (2^10, 2^10+sqrt(2^10)]=(1024, 1056].

Prog

(PARI) ct(a, b)=sum(k=floor(a)+1, b, isprime(k))
is(n)=ct(2^n-sqrt(2^n-.5), 2^n)==ct(2^n, 2^n+sqrt(2^n+.5)) \\ Charles R Greathouse IV, Jul 19 2011

Crossrefs

Sequence in context: A025588 A009325 A308346 * A272967 A067716 A125518

Adjacent sequences: A192387 A192388 A192389 * A192391 A192392 A192393

Keyword

nonn,more

Author

Juri-Stepan Gerasimov, Jun 29 2011

Status

approved