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%I #4 Apr 21 2013 16:03:36
%S 0,0,2,3,1,2,2,2,1,2,1,1,3,2,1,1,1,3,2,2,1,3,1,0,4,0,1,2,5,0,2,2,1,2,
%T 2,2,1,2,2,1,1,4,2,1,0,2,5,1,1,3,1,3,3,3,0,1,2,4,1,4,4,2,2,2,6,2,5,2,
%U 3,3,2,4,5,3,2,1,3,1,3,3,3,2,2,3,2
%N Let p = prime(n). a(n) = number of primes q less than p, such that both p*q+p+q and p*q-p-q are primes.
%e For n=3, p=5, there are a(3)=2 solutions 2,3 since 5*2+5+2=17, 5*2-5-2=3 and 5*3+5+3=23, 5*3-5-3=7. Also for n=5, p=11, there is a(5)=1 solution in the form of 11*3+11+3=47, 11*3-11-3=19.
%t Table[p = Prime[n]; c = 0; i = 1; While[i < n, q1 = p*Prime[i]; q2 = p + Prime[i]; If[PrimeQ[q1 + q2] && PrimeQ[q1 - q2], c = c + 1]; i++]; c, {n, 85}]
%Y Cf. A224748, A224908, A224925, A224962.
%K nonn
%O 1,3
%A _Jayanta Basu_, Apr 21 2013
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