A238134 Number of primes p < n with q = floor((n-p)/4) and prime(q) - q + 1 both prime.

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

Offset: 1

Author

Zhi-Wei Sun, Mar 03 2014

Keywords

• nonn

Comments

Conjecture: Let m > 0 and n > 2*m + 1 be integers. If m = 1 and 2 | n, or m = 3 and n is not congruent to 1 modulo 6, or m = 2, 4, 5, ..., then there is a prime p < n such that q = floor((n-p)/m) and prime(q) - q + 1 are both prime.

In the cases m = 1, 2, this gives refinements of Goldbach's conjecture and Lemoine's conjecture (see also A235189). For m > 2, the conjecture is completely new.

See also A238701 for a similar conjecture involving primes q with q^2 - 2 also prime.

Examples

			 a(29) = 3 since 7, floor((29-7)/4) = 5 and prime(5) - 5 + 1 = 11 - 4 = 7 are all prime; 17, floor((29-17)/4) = 3 and prime(3) - 3 + 1 = 5 - 2 = 3 are all prime; 19, floor((29-19)/4) = 2 and prime(2) - 2 + 1 = 3 - 1 = 2 are all prime.
		

Links

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

Crossrefs

Cf. A000040, A045917, A046927, A234694, A234695, A235189, A237705, A238701.

Programs

Mathematica

PQ[n_]:=PrimeQ[n]&&PrimeQ[Prime[n]-n+1]
a[n_]:=Sum[If[PQ[Floor[(n-Prime[k])/4]],1,0],{k,1,PrimePi[n-1]}]
Table[a[n],{n,1,80}]