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%I #14 Aug 29 2018 20:53:03
%S 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,
%T 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,
%U 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
%N Number of primes p such that 2^n <= p <= 2^n + prime(n).
%C 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).
%C a(n)>=1 for n<=2000. But a(1403)=1 is a "near miss". - _Robert Israel_, Aug 29 2018
%H Robert Israel, <a href="/A069923/b069923.txt">Table of n, a(n) for n = 1..2000</a>
%p f:= proc(n) local pn;
%p pn:= ithprime(n);
%p nops(select(isprime, [seq(i,i=2^n+1 .. 2^n+pn, 2)]))
%p end proc:
%p f(1):= 2:
%p map(f, [$1..100]); # _Robert Israel_, Aug 29 2018
%o (PARI) for(n=1,65,print1(sum(i=2^n,2^n+prime(n),isprime(i)),","))
%Y Cf. A014210.
%K easy,nonn
%O 1,1
%A _Benoit Cloitre_, May 05 2002
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