A229974 Number of ways to write n = x + y + z (x, y, z > 0) with the six numbers 2*x-1, 2*x+1, 2*x*y-1, 2*x*y+1, 2*x*y*z-1, 2*x*y*z+1 all prime.

0, 0, 0, 1, 1, 4, 2, 1, 2, 4, 5, 3, 3, 8, 1, 9, 4, 6, 3, 8, 16, 8, 4, 8, 7, 3, 10, 7, 3, 14, 4, 6, 8, 13, 12, 14, 6, 8, 13, 7, 13, 15, 13, 9, 9, 10, 7, 13, 14, 7, 16, 15, 12, 8, 16, 31, 11, 6, 16, 13, 16, 15, 26, 8, 10, 17, 10, 12, 11, 17, 9, 9, 13, 18, 17, 23, 14, 10, 7, 13, 29, 13, 18, 14, 9, 19, 21, 14, 19, 14, 25, 11, 14, 18, 13, 21, 15, 26, 14, 8

Offset: 1

Author

Zhi-Wei Sun, Oct 05 2013

Keywords

• nonn

Comments

Conjecture: (i) a(n) > 0 for all n > 3. Moreover, any integer n > 3 can be written as x + y + z with x among 2, 3, 6 such that {2*x*y-1, 2*x*y+1} and {2*x*y*z-1, 2*x*y*z+1} are twin prime pairs.

(ii) Each integer n > 11 can be written as x + y + z (x, y, z > 0) with x-1, x+1, x*y-1, x*y+1, x*y*z-1, x*y*z+1 all prime, moreover we may require that x is among 4, 6, 12.

(iii) Any integer n > 3 not equal to 10 can be written as x + y + z (x, y, z > 0) such that the three numbers 2*x-1, 2*x*y-1 and 2*x*y*z-1 are Sophie Germain primes, moreover we may require that x is among 2, 3, 6.

Note that part (i) or (ii) of the above conjecture implies the twin prime conjecture, while part (iii) implies that there are infinitely many Sophie Germain primes.

See also the comments of A229969 for other similar conjectures.

Examples

			a(4) = 1 since 4 = 2+1+1 with 2*2-1 and 2*2+1 both prime.
a(5) = 1 since 5 = 3+1+1 with 2*3-1 and 2*3+1 both prime.
a(15) = 1 since 15 = 6+5+4 with 2*6-1, 2*6+1, 2*6*5-1, 2*6*5+1, 2*6*5*4-1, 2*6*5*4+1 all prime.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..4000
• Zhi-Wei Sun, Conjectures involving primes and quadratic forms, preprint, arXiv:1211.1588.

Crossrefs

Cf. A001359, A006512, A005384, A229969, A219842, A219864.

Programs

Mathematica

a[n_]:=Sum[If[PrimeQ[2i-1]&&PrimeQ[2i+1]&&PrimeQ[2*i*j-1]&&PrimeQ[2i*j+1]&&PrimeQ[2i*j*(n-i-j)-1]&&PrimeQ[2i*j*(n-i-j)+1],1,0],{i,1,n-2},{j,1,n-1-i}]
Table[a[n],{n,1,100}]