OFFSET
1,4
COMMENTS
Conjecture: a(n) > 0 for all n > 0, and a(n) = 1 only for n = 1, 3, 6, 7, 11, 14, 15, 23, 38, 47, 71, 77, 143, 152, 191, 608, 2^(2k+1) (k = 0,1,2,...).
This is stronger than Lagrange's four-square theorem.
See A275656 for a stronger conjecture.
LINKS
Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
Zhi-Wei Sun, Refining Lagrange's four-square theorem, arXiv:1604.06723 [math.GM], 2016.
EXAMPLE
a(1) = 1 since 1 = 1^2*(1+0^2+0^2) + 0^2 with 0 = 0 = 0.
a(2) = 1 since 2 = 1^2*(1+0^2+0^2) + 1^2 with 0 = 0 < 1.
a(3) = 1 since 3 = 1^2*(1+0^2+1^2) + 1^2 with 0 < 1 = 1.
a(6) = 1 since 6 = 1^2*(1+0^2+1^2) + 2^2 with 0 < 1 < 2.
a(7) = 1 since 7 = 1^2*(1+1^2+1^2) + 2^2 with 1 = 1 < 2.
a(11) = 1 since 11 = 1^2*(1+0^2+1^2) + 3^2 with 0 < 1 < 3.
a(14) = 1 since 14 = 1^2*(1+0^2+2^2) + 3^2 with 0 < 2 < 3.
a(15) = 1 since 15 = 1^2*(1+1^2+2^2) + 3^2 with 1 < 2 < 3.
a(23) = 1 since 23 = 1^2*(1+2^2+3^2) + 3^2 with 2 < 3 = 3.
a(38) = 1 since 38 = 1^2*(1+0^2+1^2) + 6^2 with 0 < 1 < 6.
a(47) = 1 since 47 = 1^2*(1+1^2+3^2) + 6^2 with 1 < 3 < 6.
a(71) = 1 since 71 = 1^2*(1+3^2+5^2) + 6^2 with 3 < 5 < 6.
a(77) = 1 since 77 = 1^2*(1+2^2+6^2) + 6^2 with 2 < 6 = 6.
a(143) = 1 since 143 = 1^2*(1+5^2+6^2) + 9^2 with 5 < 6 < 9.
a(152) = 1 since 152 = 2^2*(1+0^2+1^2) + 12^2 with 0 < 1 < 12.
a(191) = 1 since 191 = 1^2*(1+3^2+9^2) + 10^2 with 3 < 9 < 10.
a(608) = 1 since 608 = 4^2*(1+0^2+1^2) + 24^2 with 0 < 1 < 24.
MATHEMATICA
SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
Do[r=0; Do[If[SQ[n-x^2*(1+y^2+z^2)], r=r+1], {x, 1, Sqrt[n]}, {y, 0, Sqrt[(n-x^2)/(2x^2+1)]}, {z, y, Sqrt[(n-x^2*(1+y^2))/(x^2+1)]}]; Print[n, " ", r]; Continue, {n, 1, 80}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Aug 04 2016
STATUS
approved