

A236192


a(n) = {0 < k < n: p = phi(k) + phi(nk)/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.


3



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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(nk)/3  1, prime(p)^2 + (p1)^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 + (p1)^2 and p^2 + (prime(p)1)^2 both prime.


LINKS

ZhiWei 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(124183)/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[nk]/4+1
a[n_]:=Sum[If[p[f[n, k]], 1, 0], {k, 1, n1}]
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

ZhiWei Sun, Jan 20 2014


STATUS

approved



