OFFSET
1,3
COMMENTS
Conjecture: a(n) > 0 for all n > 1. - Dmitry Kamenetsky, Jul 18 2019
LINKS
Alois P. Heinz, Table of n, a(n) for n = 1..10000 (first 1000 terms from Dmitry Kamenetsky)
EXAMPLE
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.
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}]
PROG
(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
CROSSREFS
KEYWORD
nonn
AUTHOR
Jayanta Basu, Apr 21 2013
STATUS
approved