A237130 Number of ordered ways to write n = k + m with k > 0 and m > 0 such that both {3*k - 1, 3*k + 1} and {phi(m) - 1, phi(m) + 1} are twin prime pairs, where phi(.) is Euler's totient function.

0, 0, 0, 0, 0, 0, 1, 0, 2, 1, 3, 2, 2, 3, 3, 3, 2, 3, 3, 3, 2, 3, 4, 3, 3, 1, 3, 4, 4, 3, 3, 5, 5, 3, 2, 2, 3, 4, 2, 3, 5, 5, 3, 4, 4, 5, 3, 5, 2, 3, 4, 4, 4, 2, 6, 4, 3, 4, 3, 5, 1, 5, 5, 5, 4, 2, 5, 4, 4, 2, 4, 6, 5, 6, 3, 5, 5, 6, 5, 1, 5, 3, 5, 3, 6, 4, 5, 7, 3, 5, 3, 5, 5, 3, 7, 3, 9, 4, 6, 5

Offset: 1

Author

Zhi-Wei Sun, Feb 04 2014

Keywords

• nonn

Comments

Conjecture: (i) a(n) > 0 for all n > 8.

(ii) Any integer n > 6 can be written as k + m with k > 0 and m > 0 such that both {prime(k), prime(k) + 2} and {phi(m) - 1, phi(m) + 1} are twin prime pairs.

(iii) Each n = 12, 13, ... can be written as p + q (q > 0) with p, p + 6, phi(q) - 1 and phi(q) + 1 all prime.

(iv) If n > 2 is neither 10 nor 430, then n can be written as k + m with k > 0 and m > 0 such that both {3k - 1, 3*k + 1} and {6*m - 1, 6*m + 1} are twin prime pairs.

Note that each part of the above conjecture implies the twin prime conjecture.

Examples

			a(7) = 1 since 7 = 2 + 5 with 3*2 - 1 = 5, 3*2 + 1 =7, phi(5) - 1 = 3 and phi(5) + 1 = 5 all prime.
a(140) = 1 since 140 = 104 + 36 with 3*104 - 1 = 311, 3*104 + 1 = 313, phi(36) - 1 = 11 and phi(36) + 1 = 13 all prime.
		

Links

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

Crossrefs

Cf. A000010, A000040, A001359, A006512, A014574, A072281, A182662, A236531, A236566, A236831, A236968, A237127.

Programs

Mathematica

PQ[n_]:=PrimeQ[EulerPhi[n]-1]&&PrimeQ[EulerPhi[n]+1]
a[n_]:=Sum[If[PrimeQ[3k-1]&&PrimeQ[3k+1]&&PQ[n-k],1,0],{k,1,n-1}]
Table[a[n],{n,1,100}]