OFFSET
1,2
COMMENTS
Conjecture: a(n) > 0 for all n > 0.
By the linked JNT paper, any nonnegative integer can be expressed as the sum of a fourth power and three squares, and each n = 0,1,2,... can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers and z*(y-2*z) = 0. Whether z = 0 or y = 2*z, the number y*(y+240*z) is definitely a square.
LINKS
Zhi-Wei Sun, Table of n, a(n) for n = 1..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(3) = 1 since 3 = 1^4 + 1^2 + 0^2 + 1^2 with 1*(1+240*0) = 1^2.
a(4) = 1 since 4 = 0^4 + 2^2 + 0^2 + 0^2 with 2*(2+240*0) = 2^2.
a(39) = 1 since 39 = 1^4 + 2^2 + 3^2 + 5^2 with 2*(2+240*3) = 38^2.
a(188) = 1 since 188 = 3^4 + 5^2 + 1^2 + 9^2 with 5*(5+240*1) = 35^2.
a(399) = 1 since 399 = 3^4 + 10^2 + 7^2 + 13^2 with 10*(10+240*7) = 130^2.
a(428) = 1 since 428 = 0^4 + 10^2 + 2^2 + 18^2 with 10*(10+240*2) = 70^2.
a(439) = 1 since 439 = 1^4 + 10^2 + 7^2 + 17^2 with 10*(10+240*7) = 130^2.
a(508) = 1 since 508 = 1^4 + 5^2 + 11^2 + 19^2 with 5*(5+240*11) = 115^2.
a(748) = 1 since 748 = 3^4 + 1^2 + 21^2 + 15^2 with 1*(1+240*21) = 71^2.
a(1468) = 1 since 1468 = 2^4 + 10^2 + 26^2 + 26^2 with 10*(10+240*26) = 250^2.
a(2828) = 1 since 2828 = 3^4 + 5^2 + 11^2 + 51^2 with 5*(5+240*11) = 115^2.
MATHEMATICA
SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
Do[r=0; Do[If[SQ[n-x^4-y^2-z^2]&&SQ[y*(y+240*z)], r=r+1], {x, 0, (n-1)^(1/4)}, {y, 1, Sqrt[n-x^4]}, {z, 0, Sqrt[n-x^4-y^2]}]; Print[n, " ", r]; Continue, {n, 1, 80}]
CROSSREFS
KEYWORD
nonn
AUTHOR
Zhi-Wei Sun, Feb 16 2017
STATUS
approved