A262982 - OEIS

%I #20 Oct 06 2015 18:35:43

%S 0,1,2,2,2,1,2,3,3,2,2,4,3,2,2,3,3,4,3,1,3,4,7,4,2,1,5,4,3,5,3,2,3,5,

%T 3,3,4,5,5,1,3,5,6,3,4,5,4,5,6,3,5,4,4,5,3,5,8,7,3,3,5,4,5,7,3,2,4,6,

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

%N Number of ordered ways to write n as x^4 + phi(y^2) + z*(z+1)/2 with x >= 0, y > 0 and z > 0, where phi(.) is Euler's totient function given by A000010.

%C Conjecture: a(n) > 0 for all n > 1, and a(n) = 1 only for n = 2, 6, 20, 26, 40, 82, 89, 105, 305, 416, 470, 725, 6135, 25430, 90285.

%C Compare this with the conjectures in A262311, A262785 and A262813.

%H Zhi-Wei Sun, <a href="/A262982/b262982.txt">Table of n, a(n) for n = 1..10000</a>

%e a(2) = 1 since 1 = 0^4 + phi(1^2) + 1*2/2.

%e a(6) = 1 since 6 = 1^4 + phi(2^2) + 2*3/2.

%e a(20) = 1 since 20 = 2^4 + phi(1^2) + 2*3/2.

%e a(26) = 1 since 26 = 0^4 + phi(5^2) + 3*4/2.

%e a(40) = 1 since 40 = 0^4 + phi(6^2) + 7*8/2.

%e a(82) = 1 since 82 = 0^4 + phi(9^2) + 7*8/2.

%e a(89) = 1 since 89 = 3^4 + phi(2^2) + 3*4/2.

%e a(105) = 1 since 105 = 0^4 + phi(14^2) + 6*7/2.

%e a(305) = 1 since 305 = 4^4 + phi(12^2) + 1*2/2.

%e a(416) = 1 since 416 = 4^4 + phi(10^2) + 15*16/2.

%e a(470) = 1 since 470 = 2^4 + phi(12^2) + 28*29/2.

%e a(725) = 1 since 725 = 2^4 + phi(3^2) + 37*38/2.

%e a(6135) = 1 since 6135 = 6^4 + phi(81^2) + 30*31/2.

%e a(25430) = 1 since 25430 = 5^4 + phi(152^2) + 166*167/2.

%e a(90285) = 1 since 90285 = 16^4 + phi(212^2) + 73*74/2.

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

%t TQ[n_]:=n>0&&IntegerQ[Sqrt[8n+1]]

%t Do[r=0;Do[If[f[x]>n,Goto[aa]];Do[If[TQ[n-f[x]-y^4],r=r+1],{y,0,(n-f[x])^(1/4)}];Label[aa];Continue,{x,1,n}];Print[n," ",r];Continue,{n,1,100}]

%Y Cf. A000010, A000217, A000290, A000583, A002618, A262311, A262781, A262785, A262813, A262941, A262979.

%K nonn

%O 1,3

%A _Zhi-Wei Sun_, Oct 06 2015