A262982 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.

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, 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, 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

Offset: 1

Author

Zhi-Wei Sun, Oct 06 2015

Keywords

• nonn

Comments

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.

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

Examples

			a(2) = 1 since 1 = 0^4 + phi(1^2) + 1*2/2.
a(6) = 1 since 6 = 1^4 + phi(2^2) + 2*3/2.
a(20) = 1 since 20 = 2^4 + phi(1^2) + 2*3/2.
a(26) = 1 since 26 = 0^4 + phi(5^2) + 3*4/2.
a(40) = 1 since 40 = 0^4 + phi(6^2) + 7*8/2.
a(82) = 1 since 82 = 0^4 + phi(9^2) + 7*8/2.
a(89) = 1 since 89 = 3^4 + phi(2^2) + 3*4/2.
a(105) = 1 since 105 = 0^4 + phi(14^2) + 6*7/2.
a(305) = 1 since 305 = 4^4 + phi(12^2) + 1*2/2.
a(416) = 1 since 416 = 4^4 + phi(10^2) + 15*16/2.
a(470) = 1 since 470 = 2^4 + phi(12^2) + 28*29/2.
a(725) = 1 since 725 = 2^4 + phi(3^2) + 37*38/2.
a(6135) = 1 since 6135 = 6^4 + phi(81^2) + 30*31/2.
a(25430) = 1 since 25430 = 5^4 + phi(152^2) + 166*167/2.
a(90285) = 1 since 90285 = 16^4 + phi(212^2) + 73*74/2.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..10000

Crossrefs

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

Programs

Mathematica

f[n_]:=EulerPhi[n^2]
TQ[n_]:=n>0&&IntegerQ[Sqrt[8n+1]]
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}]