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.

A286885 Number of ways to write 6*n+1 as x^2 + 3*y^2 + 54*z^2 with x,y,z nonnegative integers.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 2, 2, 2, 2, 1, 3, 2, 3, 1, 1, 2, 2, 3, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 1, 2, 4, 4, 3, 2, 2, 4, 2, 3, 3, 3, 3, 3, 2, 2, 4, 3, 4, 1, 3, 2, 3, 4, 3, 3, 3, 3, 2, 3, 3, 2, 4, 3, 2, 3, 2
Offset: 0

Views

Author

Zhi-Wei Sun, Aug 02 2017

Keywords

Comments

Conjecture: a(n) > 0 for all n = 0,1,2,....
In the a-file, we list the tuples (m,r,a,b,c) with 30 >= m > max{2,r} >= 0, 100 >= a >= b >= c > 0, gcd(a,b,c) = 1, and the form a*x^2+b*y^2+c*z^2 irregular, such that all the numbers m*n+r (n = 0,1,2,...) should be representable by a*x^2+b*y^2+c*z^2 with x,y,z integers.

Examples

			a(9) = 1 since 6*9 + 1 = 1^2 + 3*0^2 + 54*1^2.
a(34) = 1 since 6*34 + 1 = 2^2 + 3*7^2 + 54*1^2.
a(125) = 1 since 6*125 + 1 = 26^2 + 3*5^2 + 54*0^2.
a(130) = 1 since 6*130 + 1 = 22^2 + 3*9^2 + 54*1^2.
a(133) = 1 since 6*133 + 1 = 11^2 + 3*8^2 + 54*3^2.
a(203) = 1 since 6*203 + 1 = 25^2 + 3*6^2 + 54*3^2.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
    table={};Do[r=0;Do[If[SQ[6n+1-3y^2-54z^2],r=r+1],{y,0,Sqrt[(6n+1)/3]},{z,0,Sqrt[(6n+1-3y^2)/54]}];table=Append[table,r],{n,0,70}]