A237837 Number of primes p < n such that the number of Sophie Germain primes among 1, ..., n-p is a cube.

0, 0, 1, 2, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0, 1, 1, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 1, 1, 0, 0, 1, 1, 1, 1, 0, 0, 1, 1, 0, 0, 0, 0, 1, 2, 2, 2, 3, 3, 5, 5, 5, 5, 5, 5, 6, 6, 7, 7, 7, 7, 9, 9, 9, 9, 9, 9, 9, 9, 10

Offset: 1

Author

Zhi-Wei Sun, Feb 13 2014

Keywords

• nonn

Comments

Conjecture: a(n) > 0 for all n > 53.

Examples

			a(55) = 2 since 53 is prime and there is exactly 1^3 = 1 Sophie Germain prime not exceeding 55 - 53 = 2, and 2 is prime and there are exactly 2^3 = 8 Sophie Germain primes not exceeding 55 - 2 = 53 (namely, they are 2, 3, 5, 11, 23, 29, 41, 53).
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
• Z.-W. Sun, Problems on combinatorial properties of primes, arXiv:1402.6641, 2014

Crossrefs

Cf. A000040, A000578, A005384, A237815.

Programs

Mathematica

sg[n_]:=Sum[If[PrimeQ[2*Prime[k]+1],1,0],{k,1,PrimePi[n]}]
CQ[n_]:=IntegerQ[n^(1/3)]
a[n_]:=Sum[If[CQ[sg[n-Prime[k]]],1,0],{k,1,PrimePi[n-1]}]
Table[a[n],{n,1,80}]