

A262816


Number of ordered ways to write n as x^3 + y^2 + z*(3*z1)/2, where x and y are nonnegative integers, and z is a nonzero integer.


24



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
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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*y1)/2 + z*(3*z1)/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
By Theorem 1.7(ii) in the linked paper, any nonnegative integer can be written as x^2 + y^2 + z*(3*z1)/2, where x, y and z are integers.


LINKS



EXAMPLE

a(1) = 1 since 1 = 0^3 + 0^2 + 1*(3*11)/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*31)/2.


MATHEMATICA

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


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



