A274274 Number of ordered ways to write n as x^3 + y^2 + z^2, where x,y,z are nonnegative integers with y <= z.

1, 2, 2, 1, 1, 2, 1, 0, 2, 3, 3, 1, 1, 2, 1, 0, 2, 3, 3, 1, 1, 2, 0, 0, 1, 3, 4, 2, 2, 2, 1, 1, 2, 3, 2, 2, 2, 4, 1, 0, 3, 2, 2, 1, 2, 3, 1, 1, 1, 2, 3, 2, 3, 4, 1, 0, 1, 1, 3, 2, 1, 3, 1, 1, 3, 4, 4, 1, 3, 3, 0, 0, 4, 5, 3, 1, 2, 3, 0, 1, 4

Offset

0,2

Comments

Conjecture: Let n be any nonnegative integer.

(i) Either a(n) > 0 or a(n-2) > 0. Also, a(n) > 0 or a(n-6) > 0. Moreover, if n has the form 2^k*(4m+1) with k and m nonnegative integers, then a(n) > 0 except for n = 813, 4404, 6420, 28804.

(ii) Either n or n-3 can be written as x^3 + y^2 + 3*z^2 with x,y,z nonnegative integers.

(iii) For each d = 4, 5, 11, 12, either n or n-d can be written as x^3 + y^2 + 2*z^2 with x,y,z nonnegative integers.

We have verified that a(n) or a(n-2) is positive for every n = 0..2*10^6. Note that for each n = 0,1,2,... either n or n-2 can be written as x^2 + y^2 + z^2 with x,y,z nonnegative integers, which follows immediately from the Gauss-Legendre theorem on sums of three squares.

Links

Zhi-Wei Sun, Table of n, a(n) for n = 0..10000

Example

a(6) = 1 since 6 = 1^3 + 1^2 + 2^2.
a(14) = 1 since 14 = 1^3 + 2^2 + 3^2.
a(31) = 1 since 31 = 3^3 + 0^2 + 2^2.

Mathematica

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
Do[r=0; Do[If[SQ[n-x^3-y^2], r=r+1], {x, 0, n^(1/3)}, {y, 0, Sqrt[(n-x^3)/2]}]; Print[n, " ", r]; Continue, {n, 0, 80}]

Crossrefs

Cf. A000290, A000578, A022551, A022552, A262857, A272979.

Sequence in context: A343334 A157865 A362423 * A072550 A037810 A335211

Adjacent sequences: A274271 A274272 A274273 * A274275 A274276 A274277

Keyword

nonn

Author

Zhi-Wei Sun, Jul 14 2016

Status

approved