OFFSET
1,4
COMMENTS
Conjecture: (i) a(n) > 0 for all n > 6, and a(n) = 1 only for n = 3, 5, 9, 10, 17, 20, 24, 25, 31, 36, 45, 73, 80, 101, 136, 145, 388, 649.
(ii) For any integer n > 4, we can write 2*n as phi(p^2) + phi(x^2) + phi(y^2) with p prime and p <= x <= y.
See also A262311 for a similar conjecture.
LINKS
Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
FORMULA
a(3) = 1 since 3 = 0^2 + phi(1^2) + phi(2^2) with 2 prime.
a(5) = 1 since 5 = 1^2 + phi(2^2) + phi(2^2) with 2 prime.
a(9) = 1 since 9 = 1^2 + phi(2^2) + phi(3^2) with 2 and 3 both prime.
a(10) = 1 since 10 = 0^2 + phi(2^2) + phi(4^2) with 2 prime.
a(17) = 1 since 17 = 3^2 + phi(2^2) + phi(3^2) with 2 and 3 both prime.
a(20) = 1 since 20 = 4^2 + phi(2^2) + phi(2^2) with 2 prime.
a(24) = 1 since 24 = 4^2 + phi(2^2) + phi(3^2) with 2 and 3 both prime.
a(25) = 1 since 25 = 2^2 + phi(1^2) + phi(5^2) with 5 prime.
a(31) = 1 since 31 = 3^2 + phi(2^2) + phi(5^2) with 2 and 5 both prime.
a(36) = 1 since 36 = 2^2 + phi(5^2) + phi(6^2) with 5 prime.
a(45) = 1 since 45 = 1^2 + phi(2^2) + phi(7^2) with 2 and 7 both prime.
a(73) = 1 since 73 = 5^2 + phi(3^2) + phi(7^2) with 3 and 7 both prime.
a(80) = 1 since 80 = 6^2 + phi(2^2) + phi(7^2) with 2 and 7 both prime.
a(101) = 1 since 101 = 7^2 + phi(5^2) + phi(8^2) with 5 prime.
a(136) = 1 since 136 = 5^2 + phi(1^2) + phi(11^2) with 11 prime.
a(145) = 1 since 145 = 7^2 + phi(7^2) + phi(9^2) with 7 prime.
a(388) = 1 since 388 = 2^2 + phi(7^2) + phi(19^2) with 7 and 19 both prime.
a(649) = 1 since 649 = 11^2 + phi(7^2) + phi(27^2) with 7 prime.
MATHEMATICA
SQ[n_]:=IntegerQ[Sqrt[n]]
f[n_]:=EulerPhi[n^2]
Do[r=0; Do[If[f[z]>n, Goto[aa]]; Do[If[SQ[n-f[z]-f[y]]&&(PrimeQ[y]||PrimeQ[z]), r=r+1], {y, 1, z}]; Label[aa]; Continue, {z, 1, n}]; Print[n, " ", r]; Continue, {n, 1, 100}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Oct 01 2015
STATUS
approved