A262816 Number of ordered ways to write n as x^3 + y^2 + z*(3*z-1)/2, where x and y are nonnegative integers, and z is a nonzero integer.

1, 3, 3, 1, 2, 4, 3, 2, 3, 4, 4, 3, 4, 4, 3, 5, 5, 3, 4, 2, 3, 4, 4, 6, 2, 4, 6, 4, 5, 4, 6, 6, 3, 4, 5, 5, 4, 8, 6, 5, 5, 4, 7, 5, 5, 3, 2, 6, 5, 5, 8, 8, 4, 3, 4, 4, 6, 6, 8, 3, 4, 6, 3, 5, 7, 9, 6, 5, 6, 6, 8, 6, 4, 6, 6, 6, 7, 9, 9, 5, 4, 6, 7, 6, 6, 6, 11, 5, 4, 7, 5, 5, 7, 11, 4, 6, 4, 5, 3, 6

Offset: 1

Author

Zhi-Wei Sun, Oct 03 2015

Keywords

• nonn

Comments

Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 1, 4, 216.

(ii) Any positive integer can be written as x^3 + y*(3*y-1)/2 + z*(3*z-1)/2, where x and y are nonnegative integers, and z is a nonzero integer.

Conjectures (i) and (ii) verified for n up to 10^9. - Mauro Fiorentini, Jul 21 2023

See also A262813 and A262815 for similar conjectures.

By Theorem 1.7(ii) in the linked paper, any nonnegative integer can be written as x^2 + y^2 + z*(3*z-1)/2, where x, y and z are integers.

Examples

			a(1) = 1 since 1 = 0^3 + 0^2 + 1*(3*1-1)/2.
a(4) = 1 since 4 = 1^3 + 1^2 + (-1)*(3*(-1)-1)/2.
a(8) = 2 since 8 = 0^3 + 1^2 + (-2)*(3*(-2)-1)/2 = 1^3 + 0^2 + (-2)*(3*(-2)-1)/2.
a(216) = 1 since 216 = 2^3 + 14^2 + 3*(3*3-1)/2.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
• Zhi-Wei Sun, On universal sums of polygonal numbers, Sci. China Math. 58(2015), no. 7, 1367-1396.

Crossrefs

Cf. A000290, A000326, A000578, A001318, A160325, A160326, A262813, A262815.

Programs

Mathematica

PenQ[n_]:=n>0&&IntegerQ[Sqrt[24n+1]]
Do[r=0;Do[If[PenQ[n-x^3-y^2],r=r+1],{x,0,n^(1/3)},{y,0,Sqrt[n-x^3]}];Print[n," ",r];Continue,{n,1,100}]