A086668 Number of divisors d of n such that 2d+1 is a prime.

1, 2, 2, 2, 2, 4, 1, 3, 3, 3, 2, 4, 1, 3, 4, 3, 1, 6, 1, 4, 3, 3, 2, 5, 2, 3, 3, 3, 2, 7, 1, 3, 4, 2, 3, 7, 1, 2, 3, 5, 2, 6, 1, 4, 5, 3, 1, 6, 1, 4, 3, 3, 2, 7, 3, 5, 2, 3, 1, 8, 1, 2, 5, 3, 3, 6, 1, 3, 4, 5, 1, 8, 1, 3, 5, 2, 2, 7, 1, 5, 4, 3, 2, 6, 2, 3, 3, 5, 2, 10, 1, 3, 2, 2, 3, 7, 1, 4, 6, 5

Offset: 1

Author

Jon Perry, Jul 27 2003

Keywords

• nonn

Comments

From Antti Karttunen, Jun 15 2018: (Start)

Number of terms of A005097 that divide n.

For all n >= 1, a(n) > A156660(n). Specifically, a(p) = 2 for all p in A005384 (Sophie Germain primes), although 2's occur in other positions as well.

(End)

Examples

			10 has divisors 1,2,5 and 10 of which 2.1+1, 2.2+1 and 2.5+1 are prime, so a(10)=3
		

Links

• Antti Karttunen, Table of n, a(n) for n = 1..65537

Crossrefs

One less than A046886.

Cf. A005097, A005384, A101264, A305818.

Programs

Mathematica

Table[Count[Divisors[n],?(PrimeQ[2#+1]&)],{n,100}] (* _Harvey P. Dale, Apr 29 2015 *)

PARI

for (n=2,100,s=0; fordiv(i=n,i,s+=isprime(2*i+1)); print1(","s))

PARI

A086668(n) = sumdiv(n,d,isprime(d+d+1)); \\ Antti Karttunen, Jun 15 2018

Formula

From Antti Karttunen, Jun 15 2018: (Start)

a(n) = Sum_{d|n} A101264(d).

a(n) = A305818(n) + A101264(n).

(End)

Extensions

Definition modified by Harvey P. Dale, Apr 29 2015