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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).
4
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
1,8
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.
EXAMPLE
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.
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}]
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Dec 13 2013
STATUS
approved