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A236456 Number of ordered ways to write n = k + m with k > 0 and m > 0 such that p = phi(k) + phi(n-k)/4 - 1, q = p + 2 and r = prime(q) + 2 are all prime, where phi(.) is Euler's totient function. 9
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 1, 2, 3, 3, 2, 4, 2, 2, 4, 5, 4, 2, 3, 1, 3, 2, 3, 4, 3, 4, 5, 0, 2, 2, 3, 2, 4, 2, 4, 3, 2, 2, 1, 2, 5, 2, 3, 1, 4, 2, 2, 4, 1, 4, 1, 5, 4, 2, 2, 1, 2, 1, 5, 3, 3, 1, 2, 2, 4, 1, 3, 4, 2, 2, 1, 0, 2, 4, 2, 1, 3, 1, 4, 3, 5, 3, 2, 1, 3, 2, 3, 2, 0 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,19

COMMENTS

Conjecture: a(n) > 0 for all n > 357.

This is much stronger than the twin prime conjecture. Actually it implies that there are infinitely many primes p such that {p, p + 2} and {prime(p+2), prime(p+2) + 2} are both twin prime pairs. See A236457 for such primes p.

LINKS

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

EXAMPLE

a(18) = 1 since 18 = 3 + 15 with phi(3) + phi(15)/4 - 1  = 3,  3 + 2 = 5 and prime(5) + 2 = 13 all prime.

a(50) = 1 since 50 = 16 + 34 with phi(16) + phi(34)/4 - 1 = 11, 11 + 2 = 13 and prime(13) + 2 = 43 all prime.

a(929) = 1 since 929 = 441 + 488 with phi(441) + phi(488)/4 - 1 = 252 + 60 - 1 = 311, 311 + 2 = 313 and prime(313) + 2 = 2083 all prime.

MATHEMATICA

  p[n_]:=PrimeQ[n]&&PrimeQ[n+2]&&PrimeQ[Prime[n+2]+2]

f[n_, k_]:=EulerPhi[k]+EulerPhi[n-k]/4-1

a[n_]:=Sum[If[p[f[n, k]], 1, 0], {k, 1, n-1}]

Table[a[n], {n, 1, 100}]

CROSSREFS

Cf. A000010, A000040, A001359, A006512, A236097, A236457, A236458.

Sequence in context: A090414 A068227 A235343 * A046824 A130156 A139169

Adjacent sequences:  A236453 A236454 A236455 * A236457 A236458 A236459

KEYWORD

nonn

AUTHOR

Zhi-Wei Sun, Jan 26 2014

STATUS

approved

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Last modified September 24 17:29 EDT 2021. Contains 347651 sequences. (Running on oeis4.)