

A282091


Number of ways to write n as x^2 + y^2 + z^2 + w^2 with x + y  z a cube of an integer, where x,y,z,w are nonnegative integers with x >= y <= z and x == y (mod 2).


1



1, 2, 1, 1, 2, 2, 2, 2, 1, 3, 2, 1, 3, 1, 2, 2, 1, 4, 1, 2, 2, 2, 2, 1, 2, 3, 4, 2, 3, 2, 2, 1, 1, 5, 2, 3, 4, 2, 1, 2, 1, 4, 5, 1, 4, 2, 1, 2, 1, 5, 3, 3, 3, 1, 3, 4, 1, 4, 2, 1, 5, 3, 4, 2, 3, 5, 3, 3, 6, 3, 5, 3, 4, 6, 1, 3, 5, 3, 2, 3, 2
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OFFSET

0,2


COMMENTS

Conjecture: (i) a(n) > 0 for all n = 0,1,2,.... Also, any nonnegative integer can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers and x <= y <= z such that x + y  z is a cube of an integer.
(ii) Any nonnegative integer can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that P(x,y,z,w) is a cube of an integer, whenever P(x,y,z,w) is among the following polynomials: 2xy, 4(2xy), 4(x+yz), 2x+yz, 2*(2x+yz), 4(2x+yz), x+2y2z, 4(x+2y2z), x+3y3z, 4(x+3y3z), 2x+3y3z, 2(2x+3y3z), 4(2x+3y3z), x+5y5z, 4(x+5y5z), 2x+4y10z, 4x+8y20z, 2x+yzw, 4(2x+yzw), 4x+y2zw, 2(4x+y2zw), 4(4x+y2zw).
The author has proved that each n = 0,1,2,... can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that x (or 4x) is a cube.


LINKS

ZhiWei Sun, Table of n, a(n) for n = 0..10000
ZhiWei Sun, Refining Lagrange's foursquare theorem, J. Number Theory 175(2017), 167190.


EXAMPLE

a(2) = 1 since 2 = 0^2 + 0^2 + 1^2 + 1^2 with 0 = 0 < 1, 0 == 0 (mod 2), and 0 + 0  1 = (1)^3.
a(13) = 1 since 13 = 2^2 + 0^2 + 3^2 + 0^2 with 2 > 0 < 3, 2 == 0 (mod 2), and 2 + 0  3 = (1)^3.
a(18) = 1 since 18 = 2^2 + 2^2 + 3^2 + 1^2 with 2 = 2 < 3, 2 == 2 (mod 2), and 2 + 2  3 = 1^3.
a(31) = 1 since 31 = 1^2 + 1^2 + 2^2 + 5^2 with 1 = 1 < 2, 1 == 1 (mod 2), and 1 + 1  2 = 0^3.
a(95) = 1 since 95 = 9^2 + 1^2 + 2^2 + 3^2 with 9 > 1 < 2, 9 == 1 (mod 2), and 9 + 1  2 = 2^3.
a(479) = 1 since 479 = 15^2 + 7^2 + 14^2 + 3^2 with 15 > 7 < 14, 15 == 7 (mod 2), and 15 + 7  14 = 2^3.
a(653) = 1 since 653 = 12^2 + 8^2 + 21^2 + 2^2 with 12 > 8 < 21, 12 == 8 (mod 2), and 12 + 8  21 = (1)^3.
a(1424) = 1 since 1424 = 8^2 + 0^2 + 8^2 + 36^2 with 8 > 0 < 8, 8 == 0 (mod 2), and 8 + 0  8 = 0^3.
a(2576) = 0 since 2576 = 24^2 + 16^2 + 40^2 + 12^2 with 24 > 16 < 40, 24 == 16 (mod 2), and 24 + 16  40 = 0^3.
a(2960) = 1 since 2960 = 24^2 + 8^2 + 32^2 + 36^2 with 24 > 8 < 32, 24 == 8 (mod 2), and 24 + 8  32 = 0^3.


MATHEMATICA

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
CQ[n_]:=CQ[n]=IntegerQ[CubeRoot[n]];
Do[r=0; Do[If[SQ[nx^2y^2z^2]&&CQ[x+yz]&&Mod[xy, 2]==0, r=r+1], {y, 0, Sqrt[n/3]}, {x, y, Sqrt[ny^2]}, {z, y, Sqrt[nx^2y^2]}]; Print[n, " ", r]; Continue, {n, 0, 80}]


CROSSREFS

Cf. A000118, A000290, A000578, A271518, A273429, A273432, A273458.
Sequence in context: A183018 A067390 A256945 * A015718 A008350 A019556
Adjacent sequences: A282088 A282089 A282090 * A282092 A282093 A282094


KEYWORD

nonn


AUTHOR

ZhiWei Sun, Feb 06 2017


STATUS

approved



