A182662 Number of ordered ways to write n = p + q with q > 0 such that p, 3*(p + prime(q)) - 1 and 3*(p + prime(q)) + 1 are all prime.

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

Offset: 1

Author

Zhi-Wei Sun, Jan 31 2014

Keywords

• nonn

Comments

Conjecture: a(n) > 0 if n is not a divisor of 12.

Clearly, this implies the twin prime conjecture.

Examples

			a(11) = 1 since 11 = 7 + 4 with 7, 3*(7 + prime(4)) - 1 = 3*14 - 1 = 41 and 3*(7 + prime(4)) + 1 = 3*14 + 1 = 43 all prime.
a(210) = 1 since 210 = 97 + 113 with 97, 3*(97 + prime(113)) - 1 = 3*(97 + 617) - 1 = 2141 and 3*(97 + prime(113)) + 1 = 3*(97 + 617) + 1 =  2143 all prime.
		

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, A001359, A006512, A236531, A236831.

Programs

Mathematica

p[n_,m_]:=PrimeQ[3(m+Prime[n-m])-1]&&PrimeQ[3(m+Prime[n-m])+1]
a[n_]:=Sum[If[p[n,Prime[k]],1,0],{k,1,PrimePi[n-1]}]
Table[a[n],{n,1,100}]