cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A350012 Number of ways to write n as 4*x^4 + y^2 + (z^2 + 4^w)/2 with x,y,z,w nonnegative integers.

Original entry on oeis.org

1, 2, 1, 1, 4, 4, 1, 3, 5, 5, 3, 3, 4, 7, 3, 2, 6, 5, 2, 4, 6, 2, 2, 5, 4, 6, 2, 2, 6, 7, 2, 2, 6, 5, 5, 4, 3, 7, 5, 5, 8, 6, 2, 6, 9, 4, 2, 4, 5, 8, 3, 3, 5, 8, 3, 6, 5, 3, 6, 4, 6, 5, 6, 1, 10, 9, 2, 6, 11, 8, 1, 7, 5, 11, 6, 4, 7, 10, 3, 6, 10, 4, 8, 8, 6, 8, 6, 5, 11, 13, 5, 1, 11, 8, 3, 4, 4, 9, 7, 6
Offset: 1

Views

Author

Zhi-Wei Sun, Dec 08 2021

Keywords

Comments

Conjecture: a(n) > 0 for all n > 0.
This is a new refinement of Lagrange's four-square theorem since (x^2 + y^2)/2 = ((x+y)/2)^2 + ((x-y)/2)^2. We have verified the conjecture for n up to 10^6.
See also A349661 for a similar conjecture.
We also have some other conjectures of such a type.

Examples

			a(1) = 4*0^4 + 0^2 + (1^2 + 4^0)/2.
a(3) = 1 with 3 = 4*0^4 + 1^2 + (0^2 + 4)/2.
a(4) = 1 with 4 = 4*0^4 + 0^2 + (2^2 + 4)/2.
a(7) = 1 with 7 = 4*1^4 + 1^2 + (0^2 + 4)/2.
a(71) = 1 with 71 = 4*1^4 + 3^2 + (10^2 + 4^2)/2.
a(92) = 1 with 92 = 4*1^4 + 6^2 + (10^2 + 4)/2.
a(167) = 1 with 167 = 4*1^4 + 9^2 + (10^2 + 4^3)/2.
a(271) = 1 with 271 = 4*1^4 + 11^2 + (6^2 + 4^4)/2.
a(316) = 1 with 316 = 4*1^4 + 4^2 + (24^2 + 4^2)/2.
a(4796) = 1 with 4796 = 4*5^4 + 36^2 + (44^2 + 4^3)/2.
a(14716) = 1 with 14716 = 4*5^4 + 4^2 + (156^2 + 4^3)/2.
a(24316) = 1 with 24316 = 4*3^4 + 84^2 + (184^2 + 4^2)/2.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
    tab={};Do[r=0;Do[If[SQ[2(n-4x^4-y^2)-4^z],r=r+1],{x,0,((n-1)/4)^(1/4)},{y,0,Sqrt[n-1-4x^4]},{z,0,Log[4,2(n-4x^4-y^2)]}];tab=Append[tab,r],{n,1,100}];Print[tab]