A236074 a(n) = |{0 < k < n: p = phi(k) + phi(n-k)/6 + 1, prime(2*p) - 2*prime(p) and prime(p) - 2*prime((p-1)/2) are all prime}|, where phi(.) is Euler's totient function.

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

Offset

1,29

Comments

Conjecture: (i) a(n) > 0 for all n > 116.

(ii) For any integer n > 196, there is a positive integer k < n such that p = phi(k) + phi(n-k)/6 + 1, prime(2*p) - 2*prime(p) and prime(p-1) - 2*prime((p-1)/2) are all prime.

Clearly, part (i) (or part (ii)) implies that there are infinitely many odd primes p with prime(2*p) - 2*prime(p) and prime(p) - 2*prime((p-1)/2) (or prime(p-1) - 2*prime((p-1)/2), resp.) both prime.

Links

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

Example

a(30) = 1 since phi(4) + phi(26)/6 + 1 = 5, prime(2*5) - 2*prime(5) = 29 - 2*11 = 7 and prime(5) - 2*prime((5-1)/2) = 11 - 2*3 = 5 are all prime.
a(204) = 1 since phi(159) + phi(45)/6 + 1 = 109, prime(2*109) - 2*prime(109) = 1361 - 2*599 = 163, and prime(109) - 2*prime((109-1)/2) = 599 - 2*251 = 97 are all prime.

Mathematica

PQ[n_]:=PQ[n]=n>0&&PrimeQ[n]
p[n_]:=PQ[n]&&PQ[Prime[2n]-2Prime[n]]&&PQ[Prime[n]-2*Prime[(n-1)/2]]
f[n_, k_]:=EulerPhi[k]+EulerPhi[n-k]/6+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, A234694, A235924, A236075.

Sequence in context: A051778 A057554 A060575 * A099916 A099917 A137412

Adjacent sequences: A236071 A236072 A236073 * A236075 A236076 A236077

Keyword

nonn

Author

Zhi-Wei Sun, Jan 19 2014

Status

approved