A224989 Let p = prime(n). a(n) = number of primes q less than p, such that both p-q+1 and p-q-1 are primes.

0, 0, 0, 1, 2, 1, 3, 2, 4, 3, 2, 3, 4, 3, 5, 4, 5, 3, 3, 6, 5, 6, 6, 6, 3, 7, 5, 5, 6, 9, 4, 8, 3, 7, 7, 5, 6, 5, 7, 6, 8, 8, 9, 5, 10, 9, 12, 8, 6, 8, 9, 13, 10, 12, 9, 8, 12, 9, 7, 14, 9, 10, 8, 13, 9, 9, 11, 10, 6, 13, 12, 11, 8, 9, 17, 9, 12, 6, 11, 14

Offset: 1

Author

Jayanta Basu, Apr 22 2013

Keywords

• nonn

Examples

			For n=3, p=5, there are no primes q(<5) such that both 5-q+1 and 5-q-1 are primes and hence a(3)=0. Also for n=5, p=11, there are a(5)=2 solutions 5,7 since 11-5+1=7, 11-5-1=5 and 11-7+1=5, 11-7-1=3.
		

Links

• Harvey P. Dale, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A224748, A224908, A224979, A224980.

Programs

Mathematica

Table[p = Prime[n]; c = 0; i = 1; While[i < n, p1 = p - Prime[i]; If[PrimeQ[p1 + 1] && PrimeQ[p1 - 1], c = c + 1]; i++]; c, {n, 80}]
Table[Count[Prime[p]-Prime[Range[p-1]],?(AllTrue[#+{1,-1},PrimeQ]&)],{p,80}] (* _Harvey P. Dale, Aug 12 2025 *)