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.

LINKS

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

Zhi-Wei Sun, Refining Lagrange's four-square 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[n-x^3-y^2-z^2]&&SQ[x+y-z], r=r+1], {x, 0, n^(1/3)}, {y, 0, Sqrt[(n-x^3)/2]}, {z, y, Sqrt[n-x^3-y^2]}]; Print[n, " ", r]; Continue, {n, 0, 80}]

CROSSREFS

KEYWORD

nonn

AUTHOR

Zhi-Wei Sun, Jul 22 2016

STATUS

approved