%I #7 Sep 04 2023 12:26:45
%S 3,5,17,17,257,257,257,257,65537,65537,65537,65537,65537,65537,65537,
%T 65537,4398054899713,4398054899713,4398054899713,1125899915231233,
%U 1125899915231233,1125899915231233,1125899915231233,2251799847239681,2251799847239681,1152921513196781569,1152921513196781569
%N a(n) is the first odd prime p such that A000120((2^n-1)*p) = n * A000120(p).
%C a(n) is the first odd prime p such that between any two 1's in the base-2 representation of p there are at least n-1 0's.
%H Robert Israel, <a href="/A365475/b365475.txt">Table of n, a(n) for n = 1..1000</a>
%e a(3) = 17 because A000120((2^3-1) * 17) = A000120(119) = 6 = 3 * A000120(17).
%p f:= proc(m) local S,d,j,r,q;
%p S[0]:= [1];
%p for d from 1 do
%p S[d]:= NULL;
%p for j from 0 to d-m do
%p for r in S[j] do
%p q:= r + 2^d;
%p if isprime(q) then return q fi;
%p S[d]:= S[d],q;
%p od;
%p od;
%p S[d]:= [S[d]];
%p od;
%p end proc:
%p map(f, [$1..30]);
%Y Cf. A000120.
%K nonn,base
%O 1,1
%A _Robert Israel_, Sep 04 2023