OFFSET
0,2
COMMENTS
Conjecture: a(n) > 0 for all n = 0,1,2,....
This is stronger than the 1-3-5 conjecture (cf. A271518).
By the linked JNT paper, any nonnegative integer can be expressed as the sum of a fourth power and three squares.
LINKS
Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
Zhi-Wei Sun, Refining Lagrange's four-square theorem, J. Number Theory 175(2017), 167-190.
Zhi-Wei Sun, Restricted sums of four squares, arXiv:1701.05868 [math.NT], 2017.
EXAMPLE
a(12) = 1 since 12 = 1^2 + 1^2 + 1^2 + 3^2 with 1 + 3*1 + 5*1 = 3^2 and 1 = 1^2.
a(28) = 1 since 28 = 1^2 + 1^2 + 1^2 + 5^2 with 1 + 3*1 + 5*1 = 3^2 and 1 = 1^2.
a(47) = 1 since 47 = 3^2 + 1^2 + 6^2 + 1^2 with 3 + 3*1 + 5*6 = 6^2 and 1 = 1^2.
a(92) = 1 since 92 = 1^2 + 1^2 + 9^2 + 3^2 with 1 + 3*1 + 5*1 = 3^2 and 9 = 3^2.
a(188) = 1 since 188 = 7^2 + 9^2 + 3^2 + 7^2 with 7 + 3*9 + 5*3 = 7^2 and 9 = 3^2.
a(248) = 1 since 248 = 10^2 + 2^2 + 0^2 + 12^2 with 10 + 3*2 + 5*0 = 4^2 and 0 = 0^2.
a(388) = 1 since 388 = 13^2 + 1^2 + 13^2 + 7^2 with 13 + 3*1 + 5*13 = 9^2 and 1 = 1^2.
MATHEMATICA
SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
Do[r=0; Do[If[SQ[n-x^2-y^2-z^2]&&(SQ[y]||SQ[z]||SQ[Sqrt[n-x^2-y^2-z^2]])&&SQ[x+3y+5z], r=r+1], {x, 0, n^(1/2)}, {y, 0, Sqrt[n-x^2]}, {z, 0, Sqrt[n-x^2-y^2]}]; Print[n, " ", r]; Continue, {n, 0, 80}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Feb 17 2017
STATUS
approved