A331539 a(n) gives the number of primes of form (2*n+1)*2^m + 1 where m satisfies 2^m <= 2*n+1.

1, 1, 1, 1, 3, 2, 1, 2, 1, 0, 2, 1, 2, 2, 2, 0, 1, 2, 2, 4, 1, 1, 1, 0, 1, 2, 2, 1, 2, 1, 1, 3, 3, 2, 2, 2, 2, 4, 1, 1, 3, 2, 2, 2, 1, 0, 3, 3, 2, 4, 1, 0, 3, 1, 1, 2, 2, 1, 3, 2, 0, 1, 2, 1, 2, 2, 2, 4, 1, 1, 4, 0, 1, 0, 2, 1, 2, 2, 0, 2, 2, 3, 5, 1, 1, 0, 1

Offset: 0

Author

Jeppe Stig Nielsen, Jan 19 2020

Keywords

• nonn

Comments

For each index n, let k = 2*n+1. Then a(n) gives the number of primes of form k*2^m + 1 that are NOT considered Proth primes (A080076) because their m are too small.

In the edge case n=0, so k=1, we count 1*2^0 + 1 = 2 as a non-Proth prime.

Examples

			For n=10, we consider 21*2^m + 1, where m runs from 0 to 4 (the next value m=5 would make 2^m exceed 21). The number of cases where 21*2^m + 1 is prime, is 2, namely m=1 (prime 43) and m=4 (prime 337). So 2 primes means a(10)=2. Compare with the start of A032360, all k=21 primes.
		

Crossrefs

Cf. A080076, A134876.

Programs

Mathematica

a[n_] := Sum[Boole @ PrimeQ[(2n+1)*2^m + 1], {m, 0, Log2[2n+1]}]; Array[a, 100, 0] (* Amiram Eldar, Jan 20 2020 *)

PARI

a(n) = my(k=2*n+1);sum(m=0,logint(k,2),ispseudoprime(k<