A236511 a(n) = |{0 < k < n: p = 3*phi(k) + phi(n-k) - 1, p + 2, p + 6 and p + 8 are all prime}|, where phi(.) is Euler's totient function.

0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 2, 0, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 1, 1, 1, 1, 2, 0, 0, 0, 0, 1, 1, 2, 1, 1, 1, 1, 0, 2, 2, 2, 2, 2, 2, 0, 2, 0, 4, 4, 2, 1, 3, 4, 2, 2, 3, 0, 1, 3, 2, 3, 1, 4, 4, 3, 1

Offset: 1

Author

Zhi-Wei Sun, Jan 27 2014

Keywords

• nonn

Comments

Conjecture: a(n) > 0 for all n > 1075.

We have verified this for n up to 50000.

The above conjecture implies the well-known conjecture that there are infinitely many prime quadruplets (p, p + 2, p + 6, p + 8).

Examples

			a(10) = 1 since 3*phi(3) + phi(7) - 1 = 6 + 6 - 1 = 11, 11 + 2 = 13, 11 + 6 = 17 and 11 + 8 = 19 are all prime.
a(57) = 1 since 3*phi(31) + phi(26) - 1 = 90 + 12 - 1 = 101, 101 + 2 = 103, 101 + 6 = 107 and 101 + 8 = 109 are all prime.
		

Links

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

Crossrefs

Cf. A000010, A000040, A007530, A236508.

Programs

Mathematica

p[n_]:=PrimeQ[n]&&PrimeQ[n+2]&&PrimeQ[n+6]&&PrimeQ[n+8]
f[n_,k_]:=3*EulerPhi[k]+EulerPhi[n-k]-1
a[n_]:=Sum[If[p[f[n,k]],1,0],{k,1,n-1}]
Table[a[n],{n,1,100}]