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

1,2

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.

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.

Example

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.

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}]

Crossrefs

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

Sequence in context: A096995 A255941 A010264 * A089680 A309507 A306690

Adjacent sequences: A262813 A262814 A262815 * A262817 A262818 A262819

Keyword

nonn

Author

Zhi-Wei Sun, Oct 03 2015

Status

approved