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a(n) = number of primes of the form p*q+(p+q) where p = prime(n) and q is any prime < p.
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%I #21 Jul 18 2019 11:30:18

%S 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,

%T 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,

%U 11,6,21,10,14,13,22,10,12,21,15,20,22,13,11,12

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

%C Conjecture: a(n) > 0 for all n > 1. - _Dmitry Kamenetsky_, Jul 18 2019

%H Alois P. Heinz, <a href="/A224962/b224962.txt">Table of n, a(n) for n = 1..10000</a> (first 1000 terms from Dmitry Kamenetsky)

%e For n=3, p=5, there are a(3)=2 solutions from 5*2+(5+2)=17 and 5*3+(5+3)=23.

%e 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.

%p a:= n-> (p-> add((q-> `if`(isprime((p+1)*(q+1)-1),

%p 1, 0))(ithprime(j)), j=1..n-1))(ithprime(n)):

%p seq(a(n), n=1..100); # _Alois P. Heinz_, Jul 18 2019

%t 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}]

%o (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

%Y Cf. A224748, A224908, A224925.

%K nonn

%O 1,3

%A _Jayanta Basu_, Apr 21 2013