A235924 a(n) = |{0 < k < n: p = phi(k) + phi(n-k)/3 + 1, q = prime(p) - p + 1 and r = prime(q) - q + 1 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, 1, 0, 0, 0, 1, 0, 2, 0, 3, 0, 1, 0, 2, 3, 4, 3, 3, 1, 2, 3, 1, 6, 2, 9, 2, 5, 3, 4, 3, 8, 1, 4, 3, 9, 2, 3, 5, 6, 6, 7, 3, 8, 7, 6, 4, 4, 5, 7, 3, 6, 5, 1, 4, 6, 6, 2, 3, 4, 5, 4, 11, 4, 5, 4, 7, 2, 5, 5, 5, 2, 6, 2, 5, 5, 7

Offset: 1

Author

Zhi-Wei Sun, Jan 17 2014

Keywords

• nonn

Comments

Conjecture: a(n) > 0 for all n > 37.

This implies that there are infinitely many primes p with q = prime(p) - p + 1 and r = prime(q) - q + 1 both prime.

Examples

			a(20) = 1 since phi(6) + phi(14)/3 + 1 = 5, prime(5) - 4 = 11 - 4 = 7 and prime(7) - 6 = 17 - 6 = 11 are all prime.
a(77) = 1 since phi(59) + phi(18)/3 + 1 = 61, prime(61) - 60 = 283 - 60 = 223 and prime(223) - 222 = 1409 - 222 = 1187 are all prime.
a(1471) = 1 since phi(25) + phi(1446)/3 + 1 = 181, prime(181) - 180 = 1087 - 180 = 907 and prime(907) - 906 = 7057 - 906 = 6151 are all prime.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
• Z.-W. Sun, Problems on combinatorial properties of primes, arXiv:1402.6641, 2014

Crossrefs

Cf. A000010, A000040, A234694, A234695.

Programs

Mathematica

q[n_]:=Prime[n]-n+1
f[n_,k_]:=EulerPhi[k]+EulerPhi[n-k]/3+1
p[n_,k_]:=PrimeQ[f[n,k]]&&PrimeQ[q[f[n,k]]]&&PrimeQ[q[q[f[n,k]]]]
a[n_]:=Sum[If[p[n,k],1,0],{k,1,n-1}]
Table[a[n],{n,1,100}]