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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.

A286944 Number of ways to write n as x^2 + 15*y^2 + z*(3z+1)/2, where x and y are nonnegative integers and z is an integer.

Original entry on oeis.org

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

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Author

Zhi-Wei Sun, May 16 2017

Keywords

Comments

Conjecture: a(n) > 0 for any nonnegative integer n, and a(n) = 1 only for n = 0, 3, 4, 7, 8, 10, 12, 13, 14, 29, 33, 34, 48, 68, 113, 129, 220.
Let a,b,c,d,e,f be nonnegative integers with a > b, c > d, e > f, a == b (mod 2), c == d (mod 2), e == f (mod 2), a >= c >= e >= 2, b >= d if a = c, and d >= f if c = e. We have shown in arXiv:1502.03056 that if the ordered tuple (a,b,c,d,e,f) is universal, i.e., each n = 0,1,2,... can be written as x(ax+b)/2 + y(cy+d)/2 + z(ez+f)/2 with x,y,z integers, then (a,b,c,d,e,f) must be among the 12082 tuples listed in the linked a-file. We also conjecture that all the listed tuples not yet proved to be universal are indeed universal. Note that those numbers x(4x+2)/2 with x integral coincide with triangular numbers.

Examples

			a(33) = 1 since 33 = 4^2 + 15*1^2 + 1*(3*1+1)/2.
a(34) = 1 since 34 = 2^2 + 15*1^2 + 3*(3*3+1)/2.
a(48) = 1 since 48 = 6^2 + 15*0^2 + (-3)*(3*(-3)+1)/2.
a(68) = 1 since 68 = 1^2 + 15*2^2 + 2*(2*3+1)/2.
a(113) = 1 since 113 = 6^2 + 15*0^2 + 7*(3*7+1)/2.
a(129) = 1 since 129 = 8^2 + 15*2^2 + (-2)*(3*(-2)+1)/2.
a(220) = 1 since 220 = 13^2 + 15*0^2 + 6*(3*6+1)/2.
		

Crossrefs

Programs

  • Mathematica
    SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];Do[r=0;Do[If[SQ[24(n-x^2-15y^2)+1],r=r+1],{x,0,Sqrt[n]},{y,0,Sqrt[(n-x^2)/15]}];Print[n," ",r],{n,0,80}]