A236192 a(n) = |{0 < k < n: p = phi(k) + phi(n-k)/4 + 1, prime(p)^2 + (2*p)^2 and p^2 + (2*prime(p))^2 are all prime}|, where phi(.) is Euler's totient function.

0, 0, 0, 0, 0, 1, 1, 0, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0

Offset

1

Comments

Conjecture: (i) a(n) > 0 for all n > 370.

(ii) For any integer n >= 700, there is a positive integer k < n such that p = phi(k) + phi(n-k)/3 - 1, prime(p)^2 + (p-1)^2 and p^2 + (prime(p)-1)^2 are all prime.

Clearly, part (i) of this conjecture implies that there are infinitely many primes p with prime(p)^2 + (2*p)^2 and p^2 + (2*prime(p))^2 both prime, and part (ii) implies that there are infinitely many primes p with prime(p)^2 + (p-1)^2 and p^2 + (prime(p)-1)^2 both prime.

Links

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

Example

a(10) = 1 since phi(2) + phi(8)/4 + 1 = 3, prime(3)^2 + (2*3)^2 = 5^2 + 6^2 = 61 and 3^2 + (2*prime(3))^2 = 3^2 + 10^2 = 109 are all prime.
a(1241) = 1 since phi(83) + phi(1241-83)/4 + 1 = 82 + 96 + 1 = 179, prime(179)^2 + (2*179)^2 = 1063^2 + 358^2 = 1258133 and 179^2 + (2*prime(179))^2 = 179^2 + 2126^2 = 4551917 are all prime.

Mathematica

p[n_]:=PrimeQ[n]&&PrimeQ[Prime[n]^2+(2*n)^2]&&PrimeQ[n^2+(2*Prime[n])^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, A236097, A236138, A236193.

Sequence in context: A316345 A118952 A011668 * A011670 A011666 A011669

Adjacent sequences: A236189 A236190 A236191 * A236193 A236194 A236195

Keyword

nonn

Author

Zhi-Wei Sun, Jan 20 2014

Status

approved