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.

A287616 Number of ways to write n as x(x+1)/2 + y(3y+1)/2 + z(5z+1)/2 with x,y,z nonnegative integers.

Original entry on oeis.org

1, 1, 1, 3, 1, 2, 3, 1, 3, 1, 3, 3, 2, 4, 2, 3, 3, 3, 4, 3, 2, 5, 1, 2, 4, 3, 5, 4, 5, 4, 4, 3, 6, 3, 3, 2, 5, 2, 3, 7, 3, 7, 2, 6, 3, 5, 6, 7, 2, 4, 6, 3, 7, 2, 8, 4, 2, 6, 6, 3, 8, 3, 4, 6, 3, 7, 5, 6, 7, 4, 6, 9, 5, 6, 4, 4, 3, 4, 9, 5, 6
Offset: 0

Views

Author

Zhi-Wei Sun, May 27 2017

Keywords

Comments

Conjecture: a(n) > 0 for all n = 0,1,2,..., and a(n) = 1 only for n = 0, 1, 2, 4, 7, 9, 22.
It was proved in arXiv:1502.03056 that each n = 0,1,2,... can be written as x(x+1)/2 + y(3y+1)/2 + z(5z+1)/2 with x,y,z integers. The author would like to offer 135 US dollars as the prize for the first proof of the conjecture that a(n) is always positive.
See over 400 similar conjectures in the linked a-file.

Examples

			a(4) = 1 since 4 = 1*(1+1)/2 + 0*(3*0+1)/2 + 1*(5*1+1)/2.
a(7) = 1 since 7 = 0*(0+1)/2 + 2*(3*2+1)/2 + 0*(5*0+1)/2.
a(9) = 1 since 9 = 3*(3+1)/2 + 0*(3*0+1)/2 + 1*(5*1+1)/2.
a(22) = 1 since 22 = 5*(5+1)/2 + 2*(3*2+1)/2 + 0*(5*0+1)/2.
		

Crossrefs

Programs

  • Mathematica
    TQ[n_]:=TQ[n]=IntegerQ[Sqrt[8n+1]];
    Do[r=0;Do[If[TQ[n-x(3x+1)/2-y(5y+1)/2],r=r+1],{x,0,(Sqrt[24n+1]-1)/6},{y,0,(Sqrt[40(n-x(3x+1)/2)+1]-1)/10}];Print[n," ",r],{n,0,80}]