A237531 a(n) = |{0 < k < n/2: phi(k*(n-k)) - 1 and phi(k*(n-k)) + 1 are both prime}|, where phi(.) is Euler's totient function.

0, 0, 0, 0, 0, 2, 2, 2, 3, 2, 3, 1, 3, 1, 4, 1, 1, 1, 1, 4, 3, 2, 1, 2, 1, 1, 3, 2, 5, 4, 3, 3, 6, 2, 3, 1, 5, 4, 5, 2, 3, 5, 5, 3, 7, 6, 3, 7, 3, 8, 7, 4, 4, 5, 6, 4, 3, 9, 6, 8, 9, 8, 9, 9, 10, 7, 6, 3, 5, 4, 8, 4, 8, 5, 10, 2, 7, 9, 5, 7

Offset: 1

Author

Zhi-Wei Sun, Feb 09 2014

Keywords

• nonn

Comments

Conjecture: a(n) > 0 for all n > 5.

Clearly, this implies the twin prime conjecture.

Examples

			a(12) = 1 since 12 = 3 + 9 with phi(3*9) - 1 = 17 and phi(3*9) + 1 = 19 both prime.
a(19) = 1 since 19 = 1 + 18 with phi(1*18) - 1 = 5 and phi(1*18) + 1 = 7 both prime.
a(86) = 1 since 86 = 8 + 78 with phi(8*78) - 1 = 191 and phi(8*78) + 1 = 193 both prime.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..5000
• Z.-W. Sun, Problems on combinatorial properties of primes, arXiv:1402.6641, 2014

Crossrefs

Cf. A000010, A000040, A001359, A006512, A014574, A072281, A233547, A234200, A237127, A237130, A237168, A237523.

Programs

Mathematica

p[n_]:=PrimeQ[EulerPhi[n]-1]&&PrimeQ[EulerPhi[n]+1]
a[n_]:=Sum[If[p[k(n-k)],1,0],{k,1,(n-1)/2}]
Table[a[n],{n,1,80}]