

A275299


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


4



1, 2, 3, 3, 2, 1, 3, 2, 1, 4, 4, 2, 3, 4, 2, 2, 2, 3, 7, 4, 1, 2, 5, 2, 1, 4, 2, 5, 4, 2, 3, 2, 3, 4, 6, 3, 3, 8, 4, 2, 2, 2, 6, 4, 4, 4, 4, 2, 4, 5, 4, 8, 3, 4, 3, 3, 3, 4, 7, 1, 4, 4, 5, 3, 4, 5, 5, 8, 3, 5, 5, 1, 7, 10, 3, 3, 6, 5, 5, 1, 5
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OFFSET

0,2


COMMENTS

Conjecture: (i) a(n) > 0 for all n = 0,1,2,..., and a(n) = 1 only for n = 0, 5, 8, 20, 24, 59, 71, 79, 119, 184, 575, 743, 764, 1471, 2759.
(ii) For each triple (a,b,c) = (1,2,1), (3,1,1), (4,2,2), (4,3,3), (5,1,1), (6,2,2), (14,2,2), any natural number can be written as x^3 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that a*x + b*y  c*z is a square.
See also A275297 and A275298 for similar conjectures.


LINKS

ZhiWei Sun, Table of n, a(n) for n = 0..10000
ZhiWei Sun, Refining Lagrange's foursquare theorem, arXiv:1604.06723 [math.GM], 2016.


EXAMPLE

a(1) = 2 since 1 = 0^3 + 0^2 + 0^2 + 1^2 with 0 + 0  0 = 0^2 and 0 = 0, and also 1 = 1^3 + 0^2 + 0^2 + 0^2 with 1 + 0  0 = 1^2 and 0 = 0.
a(5) = 1 since 5 = 1^3 + 0^2 + 0^2 + 2^2 with 1 + 0  0 = 1^2 and 0 = 0.
a(8) = 1 since 8 = 0^3 + 2^2 + 2^2 + 0^2 with 0 + 2  2 = 0^2 and 2 = 2.
a(20) = 1 since 20 = 1^3 + 3^2 + 3^2 + 1^2 with 1 + 3  3 = 1^2 and 3 = 3.
a(24) = 1 since 24 = 0^3 + 2^2 + 2^2 + 4^2 with 0 + 2  2 = 0^2 and 2 = 2.
a(59) = 1 since 59 = 0^3 + 5^2 + 5^2 + 3^2 with 0 + 5  5 = 0^2 and 5 = 5.
a(71) = 1 since 71 = 1^3 + 5^2 + 6^2 + 3^2 with 1 + 5  6 = 0^2 and 5 < 6.
a(79) = 1 since 79 = 3^3 + 4^2 + 6^2 + 0^2 with 3 + 4  6 = 1^2 and 4 < 6.
a(119) = 1 since 119 = 1^3 + 3^2 + 3^2 + 10^2 with 1 + 3  3 = 1^2 and 3 = 3.
a(184) = 1 since 184 = 5^3 + 3^2 + 7^2 + 1^2 with 5 + 3  7 = 1^2 and 3 < 7.
a(575) = 1 since 575 = 7^3 + 0^2 + 6^2 + 14^2 with 7 + 0  6 = 1^2 and 0 < 6.
a(743) = 1 since 743 = 1^3 + 2^2 + 3^2 + 27^2 with 1 + 2  3 = 0^2 and 2 < 3.
a(764) = 1 since 764 = 7^3 + 9^2 + 12^2 + 14^2 with 7 + 9  12 = 2^2 and 9 < 12.
a(1471) = 1 since 1471 = 1^3 + 25^2 + 26^2 + 13^2 with 1 + 25  26 = 0^2 and 25 < 26.
a(2759) = 1 since 2759 = 5^3 + 8^2 + 13^2 + 49^2 with 5 + 8  13 = 0^2 and 8 < 13.


MATHEMATICA

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


CROSSREFS

Cf. A000290, A000578, A271518, A275297, A275298.
Sequence in context: A090431 A107336 A282013 * A205100 A156613 A285555
Adjacent sequences: A275296 A275297 A275298 * A275300 A275301 A275302


KEYWORD

nonn


AUTHOR

ZhiWei Sun, Jul 22 2016


STATUS

approved



