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.

A275676 Number of ordered ways to write n as 4^k*(1+5*x^2+y^2) + z^2, where k,x,y,z are nonnegative integers with x <= y.

Original entry on oeis.org

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

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Author

Zhi-Wei Sun, Aug 04 2016

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n > 0.
(ii) Any positive integer can be written as 4^k*(1+5*x^2+y^2) + z^2, where k,x,y,z are nonnegative integers with y <= z.
See also A275656, A275675 and A275678 for similar conjectures.

Examples

			a(4) = 1 since 4 = 4*(1+5*0^2+0^2) + 0^2 with 0 = 0.
a(259) = 1 since 259 = 4^0*(1+5*4^2+13^2) + 3^2 with 4 < 13.
a(333) = 1 since 333 = 4*(1+5*3^2+5^2) + 7^2 with 3 < 5.
a(621) = 1 since 621 = 4*(1+5*0^2+8^2) + 19^2 with 0 < 8.
a(717) = 1 since 717 = 4*(1+5*3^2+11^2) + 7^2 with 3 < 11.
a(1581) = 1 since 1581 = 4*(1+5*1^2+3^2) + 39^2 with 1 < 3.
a(2541) = 1 since 2541 = 4*(1+5*3^2+13^2) + 41^2 with 3 < 13.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
    Do[r=0;Do[If[SQ[n-4^k*(1+5x^2+y^2)],r=r+1],{k,0,Log[4,n]},{x,0,Sqrt[(n/4^k-1)/6]},{y,x,Sqrt[n/4^k-1-5x^2]}];Print[n," ",r];Continue,{n,1,80}]