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
KEYWORD
easy,nonn
AUTHOR
Benoit Cloitre, May 05 2002
STATUS
approved