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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.

A275648 Number of ordered ways to write n as x^2*(1+y^2+z^2)+w^2, where x is a positive integer and y,z,w are nonnegative integers with y <= z <= w.

Original entry on oeis.org

1, 1, 1, 2, 2, 1, 1, 1, 3, 3, 1, 2, 4, 1, 1, 2, 3, 3, 3, 3, 2, 2, 1, 2, 4, 3, 3, 5, 3, 2, 2, 1, 4, 5, 2, 5, 4, 1, 2, 4, 4, 3, 3, 2, 5, 2, 1, 2, 6, 4, 4, 7, 4, 5, 3, 2, 4, 5, 2, 4, 5, 2, 4, 2, 6, 4, 4, 4, 4, 4, 1, 4, 7, 4, 4, 7, 1, 2, 3, 3
Offset: 1

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Author

Zhi-Wei Sun, Aug 04 2016

Keywords

Comments

Conjecture: a(n) > 0 for all n > 0, and a(n) = 1 only for n = 1, 3, 6, 7, 11, 14, 15, 23, 38, 47, 71, 77, 143, 152, 191, 608, 2^(2k+1) (k = 0,1,2,...).
This is stronger than Lagrange's four-square theorem.
See A275656 for a stronger conjecture.

Examples

			a(1) = 1 since 1 = 1^2*(1+0^2+0^2) + 0^2 with 0 = 0 = 0.
a(2) = 1 since 2 = 1^2*(1+0^2+0^2) + 1^2 with 0 = 0 < 1.
a(3) = 1 since 3 = 1^2*(1+0^2+1^2) + 1^2 with 0 < 1 = 1.
a(6) = 1 since 6 = 1^2*(1+0^2+1^2) + 2^2 with 0 < 1 < 2.
a(7) = 1 since 7 = 1^2*(1+1^2+1^2) + 2^2 with 1 = 1 < 2.
a(11) = 1 since 11 = 1^2*(1+0^2+1^2) + 3^2 with 0 < 1 < 3.
a(14) = 1 since 14 = 1^2*(1+0^2+2^2) + 3^2 with 0 < 2 < 3.
a(15) = 1 since 15 = 1^2*(1+1^2+2^2) + 3^2 with 1 < 2 < 3.
a(23) = 1 since 23 = 1^2*(1+2^2+3^2) + 3^2 with 2 < 3 = 3.
a(38) = 1 since 38 = 1^2*(1+0^2+1^2) + 6^2 with 0 < 1 < 6.
a(47) = 1 since 47 = 1^2*(1+1^2+3^2) + 6^2 with 1 < 3 < 6.
a(71) = 1 since 71 = 1^2*(1+3^2+5^2) + 6^2 with 3 < 5 < 6.
a(77) = 1 since 77 = 1^2*(1+2^2+6^2) + 6^2 with 2 < 6 = 6.
a(143) = 1 since 143 = 1^2*(1+5^2+6^2) + 9^2 with 5 < 6 < 9.
a(152) = 1 since 152 = 2^2*(1+0^2+1^2) + 12^2 with 0 < 1 < 12.
a(191) = 1 since 191 = 1^2*(1+3^2+9^2) + 10^2 with 3 < 9 < 10.
a(608) = 1 since 608 = 4^2*(1+0^2+1^2) + 24^2 with 0 < 1 < 24.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
    Do[r=0;Do[If[SQ[n-x^2*(1+y^2+z^2)],r=r+1],{x,1,Sqrt[n]},{y,0,Sqrt[(n-x^2)/(2x^2+1)]},{z,y,Sqrt[(n-x^2*(1+y^2))/(x^2+1)]}];Print[n," ",r];Continue,{n,1,80}]