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Number of ordered ways to write n as x^2 + y^2 + phi(z^2) with 0 <= x <= y, z > 0, 2 | x*y*z, and phi(k^2) < n for all 0 < k < z, where phi(.) is Euler's totient function given by A000010.
5

%I #18 Oct 01 2015 06:33:23

%S 1,2,1,1,1,3,2,1,2,4,2,2,1,2,2,1,2,3,2,3,4,3,1,2,3,4,2,4,2,2,3,1,4,2,

%T 1,2,5,4,1,2,2,6,3,2,4,4,3,3,4,3,3,5,3,3,4,2,6,7,3,4,4,5,2,2,5,6,6,1,

%U 5,4,4,4,6,6,1,4,4,2,4,3,5,6,3,4,5,5,4,2,2,6,5,4,6,3,3,1,5,3,2,3

%N Number of ordered ways to write n as x^2 + y^2 + phi(z^2) with 0 <= x <= y, z > 0, 2 | x*y*z, and phi(k^2) < n for all 0 < k < z, where phi(.) is Euler's totient function given by A000010.

%C Conjecture: a(n) > 0 for all n > 0, and a(n) = 1 only for n = 1, 3, 4, 5, 8, 13, 16, 23, 32, 35, 39, 68, 75, 96, 215, 219, 243, 363, 471, 723, 759, 923, 1443, 1551, 1839, 2739, 2883.

%C It is easy to see that all the numbers phi(n^2) = n*phi(n) (n = 1,2,3,...) are pairwise distinct. We have verified that a(n) > 0 for all n = 1,...,10^7.

%C See also A262311 for a related conjecture.

%H Zhi-Wei Sun, <a href="/A262747/b262747.txt">Table of n, a(n) for n = 1..10000</a>

%e a(5) = 1 since 5 = 0^2 + 2^2 + phi(1^2).

%e a(8) = 1 since 8 = 0^2 + 0^2 + phi(4^2) with 0*0*4 even, and phi(1^2) = 1, phi(2^2) = 2, phi(3^2) = 6 all smaller than 8.

%e a(13) = 1 since 13 = 1^2 + 2^2 + phi(4^2).

%e a(32) = 1 since 32 = 2^2 + 4^2 + phi(6^2).

%e a(68) = 1 since 68 = 0^2 + 6^2 + phi(8^2).

%e a(96) = 1 since 96 = 0^2 + 8^2 + phi(8^2).

%e a(363) = 1 since 363 = 0^2 + 19^2 + phi(2^2).

%e a(471) = 1 since 471 = 0^2 + 19^2 + phi(11^2).

%e a(723) = 1 since 723 = 17^2 + 18^2 + phi(11^2).

%e a(759) = 1 since 759 = 9^2 + 26^2 + phi(2^2).

%e a(923) = 1 since 923 = 16^2 + 25^2 + phi(7^2).

%e a(1443) = 1 since 1443 = 19^2 + 24^2 +phi(23^2).

%e a(1551) = 1 since 1551 = 18^2 + 35^2 + phi(2^2).

%e a(1839) = 1 since 1839 = 3^2 + 30^2 + phi(31^2).

%e a(2739) = 1 since 2739 = 1^2 + 24^2 + phi(47^2).

%e a(2883) = 1 since 2883 = 21^2 + 44^2 + phi(23^2).

%t f[n_]:=EulerPhi[n^2]

%t SQ[n_]:=IntegerQ[Sqrt[n]]

%t Do[r=0;Do[If[f[x]>n,Goto[aa]];Do[If[SQ[n-f[x]-y^2]&&(Mod[x*y,2]==0||Mod[Sqrt[n-f[x]-y^2],2]==0),r=r+1],{y,0,Sqrt[(n-f[x])/2]}];Continue,{x,1,n}];Label[aa];Print[n," ",r];Continue,{n,1,100}]

%Y Cf. A000010, A001481, A002618, A262311, A262746, A262750, A262766.

%K nonn

%O 1,2

%A _Zhi-Wei Sun_, Sep 30 2015