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.

A255191 Number of solutions to the equation n = x^2 + y*(y+1)/2 + z*(z+1)/2 (x,y,z=0,1,...) with x == Floor[sqrt(n)] (mod 2) and y <= z.

Original entry on oeis.org

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

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Author

Zhi-Wei Sun, Feb 16 2015

Keywords

Comments

By Dickson's book in the reference, E. Lionnet observed in 1872 that any nonnegative integer can be written as the sum of a square and two triangular numbers.
We have a(n) > 0 for all n. This follows from Theorem 1(i) and Lemma 2 of Sun's reference in 2007. Note that if n = m*(m+1) with m a nonnegative integer then floor(sqrt(n)) = m.

Examples

			a(14) = 1 since 14 = 1^2 + 2*3/2 + 4*5/2 with floor(sqrt(14)) == 1 (mod 2).
a(44) = 1 since 44 = 4^2 + 0*1/2 + 7*8/2 with floor(sqrt(44)) == 4 (mod 2).
a(63) = 1 since 63 = 5^2 + 4*5/2 + 7*8/2 with floor(sqrt(63)) == 5 (mod 2).
		

References

  • L. E. Dickson, History of the Theory of Numbers, vol. II, AMS Chelsea Publ., 1999, p. 24.

Crossrefs

Programs

  • Mathematica
    TQ[n_]:=IntegerQ[Sqrt[8n+1]]
    Do[r=0;Do[If[Mod[Floor[Sqrt[n]]-x,2]==0&&TQ[n-x^2-y*(y+1)/2], r=r+1],{x,0,Sqrt[n]},{y,0,(Sqrt[4(n-x^2)+1]-1)/2}];
    Print[n," ",r];Label[aa];Continue,{n,0,100}]