A233567 Number of ways to write n = p + q (q > 0) with p and p^4 + phi(q)^4 both prime, where phi(.) is Euler's totient function (A000010).

0, 0, 1, 1, 0, 1, 1, 2, 2, 2, 3, 1, 3, 2, 4, 2, 3, 4, 3, 4, 5, 3, 5, 2, 6, 4, 3, 4, 5, 2, 1, 2, 3, 5, 5, 1, 3, 3, 4, 3, 3, 7, 6, 4, 7, 2, 5, 5, 5, 5, 3, 7, 4, 7, 4, 6, 5, 3, 5, 6, 6, 5, 5, 8, 9, 6, 7, 5, 6, 5, 7, 7, 5, 8, 7, 6, 6, 6, 8, 8, 5, 8, 11, 3, 7, 6, 7, 8, 7, 1, 8, 5, 6, 9, 10, 8, 9, 12, 8, 6

Offset: 1

Author

Zhi-Wei Sun, Dec 13 2013

Keywords

• nonn

Comments

Conjecture: If n > 2 is not equal to 5, then we have a(n) > 0, also there is a prime p < n with p^2 + phi(n-p)^2 prime.

We have verified this for n up to 10^7. The first assertion in the conjecture implies that there are infinitely many primes of the form p^4 + q^4, where p is a prime and q is a positive integer.

Examples

			a(7) = 1 since 7 = 3 + 4 with 3 and 3^4 + phi(4)^4 = 81 + 16 = 97 both prime.
a(12) = 1 since 12 = 7 + 5 with 7 and 7^4 + phi(5)^4 = 7^4 + 4^4 = 2657 both prime.
a(31) = 1 since 31 = 23 + 8 with 23 and 23^4 + phi(8)^4 = 23^4 + 4^4 = 280097 both prime.
a(36) = 1 since 36 = 3 + 33 with 3 and 3^4 + phi(33)^4 = 3^4 + 20^4 = 160081 both prime.
a(90) = 1 since 90 = 79 + 11 with 79 and 79^4 + phi(11)^4 = 79^4 + 10^4 = 38960081 both prime.
		

Links

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

Crossrefs

Cf. A000010, A000040, A002645, A233542, A233544, A233547, A233549, A233566.

Programs

Mathematica

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