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.

A301375 Number of ways to write n as x^2 + y^2 + z^2 + w^2 with x,y,z nonnegative integers and w a positive integer such that x*(y+3*z) is a cube or half a cube and y <= z <= w if x = 0.

Original entry on oeis.org

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

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Author

Zhi-Wei Sun, Mar 19 2018

Keywords

Comments

Conjecture: a(n) > 0 for all n > 0.
We have verified this for all n = 1..3*10^6.

Examples

			a(14) = 1 since 14 = 0^2 + 1^2 + 2^2 + 3^2 with 0*(1+3*2) = 0^3.
a(15) = 1 since 15 = 2^2 + 1^2 + 1^2 + 3^2 with 2*(1+3*1) = 2^3.
a(16) = 1 since 16 = 0^2 + 0^2 + 0^2 + 4^2 with 0*(0+3*0) = 0^3.
a(60) = 1 since 60 = 4^2 + 2^2 + 2^2 + 6^2 with 4*(2+3*2) = 4^3/2.
a(92) = 1 since 92 = 6^2 + 6^2 + 4^2 + 2^2 with 6*(6+3*4) = 6^3/2.
a(240) = 1 since 240 = 2^2 + 14^2 + 6^2 + 2^2 with 2*(14+3*6) = 4^3.
a(807) = 1 since 807 = 1^2 + 21^2 + 2^2 + 19^2 with 1*(21+3*2) = 3^3.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
    CQ[n_]:=CQ[n]=IntegerQ[n^(1/3)];
    QQ[n_]:=QQ[n]=CQ[n]||CQ[2n];
    tab={};Do[r=0;Do[If[QQ[x(y+3z)]&&SQ[n-x^2-y^2-z^2],r=r+1],{x,0,Sqrt[n-1]},{y,0,Sqrt[If[x==0,n/3,n-1-x^2]]},{z,If[x==0,y,0],Sqrt[If[x==0,(n-x^2-y^2)/2,n-1-x^2-y^2]]}]; tab=Append[tab,r],{n,1,80}];Print[tab]