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.

A254574 Number of ways to write n = x*(x+1)/2 + y*(3*y+1)/2 + z*(3*z-1)/2 with x,y,z nonnegative integers.

Original entry on oeis.org

1, 2, 2, 3, 2, 2, 3, 3, 5, 2, 3, 3, 3, 5, 2, 6, 3, 5, 5, 2, 4, 3, 8, 4, 3, 4, 4, 6, 6, 6, 7, 3, 4, 5, 3, 6, 5, 8, 5, 4, 6, 8, 5, 8, 5, 5, 4, 6, 10, 1, 7, 6, 10, 5, 4, 7, 6, 7, 9, 6, 6, 6, 8, 10, 4, 7, 5, 9, 7, 7, 4
Offset: 0

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Author

Zhi-Wei Sun, Feb 01 2015

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n. Also, a(n) = 1 only for n = 0, 49.
(ii) Any nonnegative integer not equal to 18 can be written as x*(x+1)/2 + y*(3*y+1) + z*(3*z-1) with x,y,z nonnegative integers.
See also the comments of A254573 for a similar conjecture. We have proved that any nonnegative integer can be written as x*(x+1)/2 + y*(3*y+1)/2 + z*(3*z-1)/2 with x,y,z integers.
Note that Zhi-Wei Sun conjectured in 2009 (cf. Conjecture 1.10 of arXiv:0905.0635) that every n = 0,1,... can be expressed as the sum of a triangular number and two pentagonal numbers.

Examples

			a(14) = 2 since 14 = 0*1/2 + 1*(3*1+1)/2 + 3*(3*3-1)/2 = 3*4/2 + 2*(3*2+1)/2 + 1*(3*1-1)/2.
a(49) = 1 since 49 = 1*2/2 + 4*(3*4+1)/2 + 4*(3*4-1)/2.
		

Crossrefs

Programs

  • Mathematica
    TQ[n_]:=IntegerQ[Sqrt[8n+1]]
    Do[r=0;Do[If[TQ[n-y(3y+1)/2-z(3z-1)/2],r=r+1],{y,0,(Sqrt[24n+1]-1)/6},{z,0,(Sqrt[24(n-y(3y+1)/2)+1]+1)/6}];
    Print[n," ",r];Continue,{n,0,70}]