A233549 Number of ways to write n = p + q (q > 0) with p and (phi(p)*phi(q))^4 + 1 prime, where phi(.) is Euler's totient function (A000010).

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

Offset: 1

Author

Zhi-Wei Sun, Dec 12 2013

Keywords

• nonn

Comments

Conjecture: (i) a(n) > 0 for all n > 2.

(ii) If n > 2 is not equal to 26, then there is a prime p < n with (phi(p)*phi(n-p))^2 + 1 prime.

(iii) If n > 3 is different from 9 and 16, then there is a prime p < n with ((p+1)*phi(n-p))^2 + 1 prime.

Part (i) of the conjecture implies that there are infinitely many primes of the form x^4 + 1. We have verified it for n up to 10^7.

Examples

			a(11) = 1 since 11 = 2 + 9 with 2 and (phi(2)*phi(9))^4 + 1 = 6^4 + 1 = 1297 both prime.
a(13) = 1 since 13 = 5 + 8 with 5 and (phi(5)*phi(8))^4 + 1 = 16^4 + 1 = 65537 both prime.
a(258) = 1 since 258 = 167 + 91 with 167 and (phi(167)*phi(91))^4 + 1 = (166*72)^4 + 1 = 20406209352892417 both prime.
		

Links

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

Crossrefs

Cf. A000010, A000040, A000068, A037896, A233542, A233544, A233547.

Programs

Mathematica

a[n_]:=Sum[If[PrimeQ[((Prime[k]-1)*EulerPhi[n-Prime[k]])^4+1],1,0],{k,1,PrimePi[n-1]}]
Table[a[n],{n,1,100}]