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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.

A272332 Number of ordered ways to write n as w^2 + x^2 + y^2 + z^2 with 36*x^2*y + 12*y^2*z + z^2*x a square, where w is a positive integer and x,y,z are nonnegative integers.

Original entry on oeis.org

1, 3, 2, 2, 6, 4, 3, 3, 3, 8, 5, 2, 6, 6, 4, 1, 7, 10, 6, 8, 8, 5, 2, 2, 7, 16, 8, 3, 12, 6, 4, 3, 6, 13, 8, 8, 8, 6, 5, 7, 15, 14, 4, 2, 12, 7, 3, 2, 5, 18, 8, 12, 14, 8, 7, 4, 6, 8, 7, 5, 14, 8, 5, 2, 12, 18, 8, 12, 10, 6, 3, 5, 10, 19, 10, 3, 8, 3, 1, 6
Offset: 1

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Author

Zhi-Wei Sun, Apr 26 2016

Keywords

Comments

Conjecture: a(n) > 0 for all n > 0, and a(n) = 1 only for n = 16^k*m (k = 0,1,2,... and m = 1, 79, 591, 599, 1752, 1839, 10264).
We have verified that a(n) > 0 for all n = 1,...,400000.
For more refinements of Lagrange's four-square theorem, see arXiv:1604.06723.

Examples

			a(1) = 1 since 1 = 1^2 + 0^2 + 0^2 + 0^2 with 1 > 0 and 36*0^2*0 + 12*0^2*0 + 0^2*0 = 0^2.
a(79) = 1 since 79 = 7^2 + 1^2 + 5^2 + 2^2 with 7 > 0 and 36*1^2*5 + 12*5^2*2 + 2^2*1 = 28^2.
a(591) = 1 since 591 = 23^2 + 1^2 + 6^2 + 5^2 with 23 > 0 and 36*1^2*6 + 12*6^2*5 + 5^2*1 = 49^2.
a(599) = 1 since 599 = 6^2 + 1^2 + 11^2 + 21^2 with 6 > 0 and 36*1^2*11 + 12*11^2*21 + 21^2*1 = 177^2.
a(1752) = 1 since 1752 = 10^2 + 4^2 + 40^2 + 6^2 with 10 > 0 and 36*4^2*40 + 12*40^2*6 + 6^2*10 = 372^2.
a(1839) = 1 since 1839 = 17^2 + 37^2 + 9^2 + 10^2 with 17 > 0 and 36*37^2*9 + 12*9^2*10 + 10^2*37 = 676^2.
a(10264) = 1 since 10264 = 96^2 + 30^2 + 2^2 + 12^2 with 96 > 0 and 36*30^2*2 + 12*2^2*12 + 12^2*30 = 264^2.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
    Do[r=0;Do[If[SQ[n-x^2-y^2-z^2]&&SQ[36*x^2*y+12*y^2*z+z^2*x],r=r+1],{x,0,Sqrt[n-1]},{y,0,Sqrt[n-1-x^2]},{z,0,Sqrt[n-1-x^2-y^2]}];Print[n," ",r];Continue,{n,1,80}]