A224962 a(n) = number of primes of the form p*q+(p+q) where p = prime(n) and q is any prime < p.

0, 1, 2, 3, 2, 3, 4, 4, 4, 7, 3, 5, 6, 4, 7, 6, 8, 4, 5, 6, 2, 6, 10, 11, 8, 8, 5, 7, 10, 8, 5, 11, 8, 9, 14, 6, 6, 7, 11, 11, 14, 9, 12, 6, 13, 9, 10, 7, 16, 11, 11, 22, 9, 16, 17, 17, 21, 9, 4, 11, 6, 21, 10, 14, 13, 22, 10, 12, 21, 15, 20, 22, 13, 11, 12

Offset: 1

Author

Jayanta Basu, Apr 21 2013

Keywords

• nonn

Comments

Conjecture: a(n) > 0 for all n > 1. - Dmitry Kamenetsky, Jul 18 2019

Examples

			For n=3, p=5, there are a(3)=2 solutions from 5*2+(5+2)=17 and 5*3+(5+3)=23.
For n=4, p=7, there are a(4)=3 solutions in the form of 7*2+(7+2)=23, 7*3+(7+3)=31 and 7*5+(7+5)=47.
		

Links

• Alois P. Heinz, Table of n, a(n) for n = 1..10000 (first 1000 terms from Dmitry Kamenetsky)

Crossrefs

Cf. A224748, A224908, A224925.

Programs

Maple

a:= n-> (p-> add((q-> `if`(isprime((p+1)*(q+1)-1),
       1, 0))(ithprime(j)), j=1..n-1))(ithprime(n)):
seq(a(n), n=1..100);  # Alois P. Heinz, Jul 18 2019

Mathematica

Table[p = Prime[n]; c = 0; i = 1; While[i < n, If[PrimeQ[p*Prime[i] + (p + Prime[i])], c = c + 1]; i++]; c, {n, 75}]

PARI

a(n) = my(p=prime(n), q); sum(k=1, n-1, q=prime(k); isprime(p*q+(p+q))); \\ Michel Marcus, Jul 18 2019