

A275298


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


4



1, 2, 1, 1, 2, 2, 1, 1, 3, 4, 2, 1, 2, 2, 2, 1, 3, 5, 2, 3, 4, 3, 1, 1, 5, 5, 4, 2, 3, 6, 3, 3, 3, 6, 3, 4, 6, 3, 3, 1, 6, 7, 3, 2, 3, 5, 1, 2, 3, 5, 6, 7, 7, 5, 4, 2, 5, 4, 2, 4, 6, 7, 4, 3, 6, 8, 5, 5, 7, 7, 1, 3, 6, 4, 5, 6, 6, 4, 3, 5
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

1,2


COMMENTS

Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 1, 3, 4, 7, 8, 12, 16, 23, 24, 40, 47, 71, 167, 311, 599.
(ii) For each triple (a,b,c) = (1,1,1), (2,1,1), (2,1,2), (2,2,2), (3,1,2), any natural number can be written as x^2 + y^2 + z^2 + w^3 with x,y,z,w nonnegative integers such that a*y  b*z  c*w is a square.
See also A275297 and A275299 for similar conjectures.


LINKS

ZhiWei Sun, Table of n, a(n) for n = 1..7000
ZhiWei Sun, Refining Lagrange's foursquare theorem, arXiv:1604.06723 [math.GM], 2016.


EXAMPLE

a(1) = 1 since 1 = 0^3 + 0^2 + 0^2 + 1^2 with 0  0 = 0^2 and 0 < 1 > 0.
a(3) = 1 since 3 = 0^3 + 1^2 + 1^2 + 1^2 with 1  0 = 1^2 and 1 = 1 > 0.
a(4) = 1 since 4 = 0^3 + 0^2 + 0^2 + 2^2 with 0  0 = 0^2 and 0 < 2 > 0.
a(7) = 1 since 7 = 1^3 + 1^2 + 1^2 + 2^2 with 1  1 = 0^2 and 1 < 2 > 1.
a(8) = 1 since 8 = 0^3 + 0^2 + 2^2 + 2^2 with 0  0 = 0^2 and 2 = 2 > 0.
a(12) = 1 since 12 = 1^3 + 1^2 + 1^2 + 3^2 with 1  1 = 0^2 and 1 < 3 > 1.
a(16) = 1 since 16 = 0^3 + 0^2 + 0^2 + 4^2 with 0  0 = 0^2 and 0 < 4 > 0.
a(23) = 1 since 23 = 1^3 + 2^2 + 3^2 + 3^2 with 2  1 = 1^2 and 3 = 3 > 1.
a(24) = 1 since 24 = 0^3 + 4^2 + 2^2 + 2^2 with 4  0 = 2^2 and 2 = 2 > 0.
a(40) = 1 since 40 = 0^3 + 0^2 + 2^2 + 6^2 with 0  0 = 0^2 and 2 < 6 > 0.
a(47) = 1 since 47 = 1^3 + 1^2 + 3^2 + 6^2 with 1  1 = 0^2 and 3 < 6 > 1.
a(71) = 1 since 71 = 1^3 + 5^2 + 3^2 + 6^2 with 5  1 = 2^2 and 3 < 6 > 1.
a(167) = 1 since 167 = 1^3 + 2^2 + 9^2 + 9^2 with 2  1 = 1^2 and 9 = 9 > 1.
a(311) = 1 since 311 = 1^3 + 2^2 + 9^2 + 15^2 with 2  1 = 1^2 and 9 < 15 > 1.
a(599) = 1 since 599 = 5^3 + 5^2 + 7^2 + 20^2 with 5  5 = 0^2 and 7 < 20 > 5.


MATHEMATICA

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
CQ[n_]:=CQ[n]=IntegerQ[n^(1/3)]
Do[r=0; Do[If[CQ[nx^2y^2z^2]&&SQ[x(nx^2y^2z^2)^(1/3)]&&(nx^2y^2z^2)^(1/3)<z, r=r+1], {x, 0, Sqrt[n]}, {y, 0, Sqrt[(nx^2)/2]}, {z, Max[y, Ceiling[(nx^2y^2)^(1/3)]], Sqrt[nx^2y^2]}]; Print[n, " ", r]; Continue, {n, 1, 80}]


CROSSREFS

Cf. A000290, A000578, A271518, A275297, A275299.
Sequence in context: A081206 A156044 A180980 * A048570 A090806 A241926
Adjacent sequences: A275295 A275296 A275297 * A275299 A275300 A275301


KEYWORD

nonn


AUTHOR

ZhiWei Sun, Jul 22 2016


STATUS

approved



