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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.

A283239 Number of ways to write n as x^2 + y^2 + z*(3*z-1)/2 with x,y,z integers such that x + 2*y is a square.

Original entry on oeis.org

1, 2, 3, 2, 2, 5, 5, 5, 1, 2, 4, 2, 5, 3, 3, 4, 3, 7, 5, 3, 8, 3, 5, 3, 2, 6, 3, 7, 5, 3, 4, 7, 5, 3, 4, 6, 5, 3, 3, 7, 5, 5, 5, 1, 6, 7, 6, 4, 3, 2, 5, 5, 9, 6, 3, 7, 5, 7, 5, 4, 8, 5, 6, 4, 6, 5, 5, 8, 5, 6, 6, 7, 5, 5, 5, 7, 5, 6, 2, 4, 12
Offset: 0

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Author

Zhi-Wei Sun, Mar 03 2017

Keywords

Comments

Conjecture: (i) a(n) > 0 for all n = 0,1,2,..., and a(n) = 1 only for n = 0, 8, 43, 84, 133, 253, 399, 488, 523, 803, 7369.
(ii) Any integer n > 1 can be written as p + x^2 + y^2 with p prime and x + 2*y (or x + 3*y) a square, where x is an integer and y is a nonnegative integer.
Note that those numbers z*(3*z-1)/2 with z integral are called generalized pentagonal numbers (A001318). By Theorem 1.7(ii) of the linked paper in Sci. China Math., each n = 0,1,2,... can be written as the sum of two squares and a generalized pentagonal number.
Ju. V. Linnik proved in 1960 that any sufficiently large integer can be expressed as the sum of a prime and two squares.

Examples

			a(8) = 1 since 8 = 1^2 + 0^2 + (-2)*(3*(-2)-1)/2 with 1 + 2*0 = 1^2.
a(43) = 1 since 43 = 1^2 + 4^2 + (-4)*(3*(-4)-1)/2 with 1 + 2*4 = 3^2.
a(84) = 1 since 84 = 7^2 + (-3)^2 + (-4)*(3*(-4)-1)/2 with 7 + 2*(-3) = 1^2.
a(133) = 1 since 133 = 4^2 + 0^2 + 9*(3*9-1)/2 with 4 + 2*0 = 2^2.
a(253) = 1 since 253 = (-13)^2 + 7^2 + 5*(3*5-1)/2 with (-13) + 2*7 = 1^2.
a(399) = 1 since 399 = 18^2 + (-7)^2 + (-4)*(3*(-4)-1)/2 with 18 + 2*(-7) = 2^2.
a(488) = 1 since 488 = 9^2 + 20^2 + (-2)*(3*(-2)-1)/2 with 9 + 2*20 = 7^2.
a(523) = 1 since 523 = 9^2 + 0^2 + (-17)*(3*(-17)-1)/2 with 9 + 2*0 = 3^2.
a(803) = 1 since 803 = (-17)^2 + 13^2 + (-15)*(3*(-15)-1)/2 with (-17) + 2*13 = 3^2.
a(7369) = 1 since 7369 = 0^2 + 72^2 + (-38)*(3*(-38)-1)/2 with 0 + 2*72 = 12^2.
		

Crossrefs

Programs

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