A237819 Number of primes p < n such that floor(sqrt(n-p)) is a Sophie Germain prime.

0, 0, 0, 0, 0, 1, 2, 2, 3, 3, 4, 4, 4, 4, 5, 5, 6, 5, 4, 4, 4, 4, 4, 4, 4, 4, 5, 6, 5, 6, 6, 7, 7, 7, 7, 8, 8, 8, 6, 6, 6, 7, 6, 7, 7, 7, 6, 7, 6, 6, 7, 7, 5, 6, 5, 6, 6, 6, 4, 4, 4, 5, 5, 5, 5, 6, 5, 6, 5, 5, 6, 7, 6, 6, 6, 6, 5, 6, 5, 5

Offset: 1

Author

Zhi-Wei Sun, Feb 13 2014

Keywords

• nonn

Comments

Conjecture: (i) a(n) > 0 for all n > 5.

(ii) For any integer n > 10, there is a prime p < n such that q = floor(sqrt(n-p)) and q + 2 are both prime.

Examples

			a(6) = 1 since 2, floor(sqrt(6-2)) = 2 and 2*2 + 1 = 5 are all prime.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000040, A001359, A005384, A006512, A237720, A237721, A237815, A237817.

Programs

Mathematica

f[n_]:=Floor[Sqrt[n]]
q[n_]:=PrimeQ[f[n]]&&PrimeQ[2*f[n]+1]
a[n_]:=Sum[If[q[n-Prime[k]],1,0],{k,1,PrimePi[n-1]}]
Table[a[n],{n,1,80}]