A069923 Number of primes p such that 2^n <= p <= 2^n + prime(n).

2, 2, 2, 3, 3, 3, 3, 4, 2, 5, 3, 5, 5, 4, 7, 9, 4, 5, 5, 7, 3, 4, 7, 3, 7, 6, 8, 6, 5, 8, 4, 6, 10, 3, 5, 3, 7, 6, 7, 7, 8, 6, 7, 5, 7, 5, 8, 4, 2, 7, 6, 6, 7, 3, 6, 6, 11, 6, 6, 9, 8, 8, 7, 7, 6, 6, 10, 8, 7, 10, 9, 7, 5, 5, 9, 6, 8, 11, 9, 5, 8, 6, 10, 9, 5, 9, 12, 6, 7, 4, 7, 6, 9, 8, 5, 7, 6, 7, 3, 4, 8

Offset

1,1

Comments

For any n>0, is there always at least one prime p such that 2^n <= p <= 2^n + prime(n)? (checked up to n=250) In this case, that would be stronger than the Schinzel conjecture: "for m > 1 there's at least one prime p such that m <= p <= m + log(m)^2" since, for n > 2, prime(n) < log(2^n)^2 = n^2*log(2).

a(n)>=1 for n<=2000. But a(1403)=1 is a "near miss". - Robert Israel, Aug 29 2018

Links

Robert Israel, Table of n, a(n) for n = 1..2000

Maple

f:= proc(n) local pn;
  pn:= ithprime(n);
  nops(select(isprime, [seq(i, i=2^n+1 .. 2^n+pn, 2)]))
end proc:
f(1):= 2:
map(f, [$1..100]); # Robert Israel, Aug 29 2018

Prog

(PARI) for(n=1, 65, print1(sum(i=2^n, 2^n+prime(n), isprime(i)), ", "))

Crossrefs

Cf. A014210.

Sequence in context: A097561 A162345 A048689 * A095840 A131343 A089051

Adjacent sequences: A069920 A069921 A069922 * A069924 A069925 A069926

Keyword

easy,nonn

Author

Benoit Cloitre, May 05 2002

Status

approved