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

1,13

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).

Links

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

Example

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.

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}]

Crossrefs

Cf. A000010, A000040, A007530, A236508.

Sequence in context: A281772 A082886 A287179 * A235924 A097304 A136745

Adjacent sequences: A236508 A236509 A236510 * A236512 A236513 A236514

Keyword

nonn

Author

Zhi-Wei Sun, Jan 27 2014

Status

approved