cp's OEIS Frontend

A262781 Number of ordered ways to write n as x^2 + phi(y^2) + phi(z^2) (x >= 0 and 0 < y <= z) with y or z prime, where phi(.) is Euler's totient function given by A000010.

%I A262781 #11 Oct 02 2015 03:38:23
%S A262781 0,0,1,2,1,0,2,3,1,1,2,3,2,3,2,2,1,3,3,1,2,3,4,1,1,3,2,3,2,4,1,3,2,2,
%T A262781 3,1,3,3,4,2,2,3,5,5,1,4,4,4,2,6,4,4,4,6,3,4,5,4,5,4,4,3,6,4,2,3,3,5,
%U A262781 4,4,4,3,1,4,5,4,3,6,3,1,2,3,4,4,5,5,3,3,2,8,5,3,4,2,4,4,2,3,7,2
%N A262781 Number of ordered ways to write n as x^2 + phi(y^2) + phi(z^2) (x >= 0 and 0 < y <= z) with y or z prime, where phi(.) is Euler's totient function given by A000010.
%C A262781 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.
%C A262781 (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.
%C A262781 See also A262311 for a similar conjecture.
%H A262781 Zhi-Wei Sun, <a href="/A262781/b262781.txt">Table of n, a(n) for n = 1..10000</a>
%F A262781 a(3) = 1 since 3 = 0^2 + phi(1^2) + phi(2^2) with 2 prime.
%F A262781 a(5) = 1 since 5 = 1^2 + phi(2^2) + phi(2^2) with 2 prime.
%F A262781 a(9) = 1 since 9 = 1^2 + phi(2^2) + phi(3^2) with 2 and 3 both prime.
%F A262781 a(10) = 1 since 10 = 0^2 + phi(2^2) + phi(4^2) with 2 prime.
%F A262781 a(17) = 1 since 17 = 3^2 + phi(2^2) + phi(3^2) with 2 and 3 both prime.
%F A262781 a(20) = 1 since 20 = 4^2 + phi(2^2) + phi(2^2) with 2 prime.
%F A262781 a(24) = 1 since 24 = 4^2 + phi(2^2) + phi(3^2) with 2 and 3 both prime.
%F A262781 a(25) = 1 since 25 = 2^2 + phi(1^2) + phi(5^2) with 5 prime.
%F A262781 a(31) = 1 since 31 = 3^2 + phi(2^2) + phi(5^2) with 2 and 5 both prime.
%F A262781 a(36) = 1 since 36 = 2^2 + phi(5^2) + phi(6^2) with 5 prime.
%F A262781 a(45) = 1 since 45 = 1^2 + phi(2^2) + phi(7^2) with 2 and 7 both prime.
%F A262781 a(73) = 1 since 73 = 5^2 + phi(3^2) + phi(7^2) with 3 and 7 both prime.
%F A262781 a(80) = 1 since 80 = 6^2 + phi(2^2) + phi(7^2) with 2 and 7 both prime.
%F A262781 a(101) = 1 since 101 = 7^2 + phi(5^2) + phi(8^2) with 5 prime.
%F A262781 a(136) = 1 since 136 = 5^2 + phi(1^2) + phi(11^2) with 11 prime.
%F A262781 a(145) = 1 since 145 = 7^2 + phi(7^2) + phi(9^2) with 7 prime.
%F A262781 a(388) = 1 since 388 = 2^2 + phi(7^2) + phi(19^2) with 7 and 19 both prime.
%F A262781 a(649) = 1 since 649 = 11^2 + phi(7^2) + phi(27^2) with 7 prime.
%t A262781 SQ[n_]:=IntegerQ[Sqrt[n]]
%t A262781 f[n_]:=EulerPhi[n^2]
%t A262781 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}]
%Y A262781 Cf. A000010, A000040, A000290, A002618, A262311, A262746, A262747.
%K A262781 nonn
%O A262781 1,4
%A A262781 _Zhi-Wei Sun_, Oct 01 2015