A282161 Number of ways to write n as x^2 + y^2 + z^2 + w^2 with x and (12*x)^2 + (5*y-10*z)^2 both squares, where x,y,z are nonnegative integers and w is a positive integer.

1, 3, 2, 2, 5, 4, 2, 2, 4, 6, 4, 3, 4, 6, 3, 1, 9, 7, 5, 6, 7, 7, 1, 4, 8, 11, 7, 1, 11, 10, 2, 3, 8, 9, 6, 9, 8, 11, 5, 5, 15, 7, 4, 5, 13, 9, 2, 2, 8, 15, 10, 8, 10, 17, 3, 7, 12, 4, 10, 4, 11, 16, 3, 2, 18, 16, 6, 9, 15, 11, 4, 6, 8, 16, 12, 3, 13, 13, 1, 5

Offset

1,2

Comments

Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 16^k*m (k = 0,1,2,... and m = 1, 23, 28, 79, 119, 191, 223, 263, 463, 703, 860, 1052).

(ii) Any positive integer n can be written as x^2 + y^2 + z^2 + w^2 with x and (35*x)^2 + (12*y-24*z)^2 both squares, where x,y,z are nonnegative integers and w is a positive integer.

The author has proved that any nonnegative integer can be written as the sum of a fourth power and three squares.

See also A281976, A281977, A282013 and A282014 for similar conjectures.

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(1) = 1 since 1 = 0^2 + 0^2 + 0^2 + 1^2 with 0 = 0^2 and (12*0)^2 + (5*0-10*0)^2 = 0^2.
a(23) = 1 since 23 = 1^2 + 3^2 + 2^2 + 3^2 with 1 = 1^2 and (12*1)^2 + (5*3-10*2)^2 = 13^2.
a(28) = 1 since 28 = 1^2 + 1^2 + 1^2 + 5^2 with 1 = 1^2 and (12*1)^2 + (5*1-10*1)^2 = 13^2.
a(79) = 1 since 79 = 1^2 + 5^2 + 2^2 + 7^2 with 1 = 1^2 and (12*1)^2 + (5*5-10*2)^2 = 13^2.
a(119) = 1 since 119 = 1^2 + 9^2 + 1^2 + 6^2 with 1 = 1^2 and (12*1)^2 + (5*9-10*1)^2 = 37^2.
a(191) = 1 since 191 = 9^2 + 5^2 + 7^2 + 6^2 with 9 = 3^2 and (12*9)^2 + (5*5-10*7)^2 = 117^2.
a(223) = 1 since 223 = 1^2 + 13^2 + 7^2 + 2^2 with 1 = 1^2 and (12*1)^2 + (5*13-10*7)^2 = 13^2.
a(263) = 1 since 263 = 9^2 + 13^2 + 2^2 + 3^2 with 9 = 3^2 and (12*9)^2 + (5*13-10*2)^2 = 117^2.
a(463) = 1 since 463 = 1^2 + 19^2 + 10^2 + 1^2 with 1 = 1^2 and (12*1)^2 + (5*19-10*10)^2 = 13^2.
a(703) = 1 since 703 = 1^2 + 13^2 + 7^2 + 22^2 with 1 = 1^2 and (12*1)^2 + (5*13-10*7)^2 = 13^2.
a(860) = 1 since 860 = 4^2 + 18^2 + 18^2 + 14^2 with 4 = 2^2 and (12*4)^2 + (5*18-10*18)^2 = 102^2.
a(1052) = 1 since 1052 = 4^2 + 30^2 + 6^2 + 10^2 with 4 = 2^2 and (12*4)^2 + (5*30-10*6)^2 = 102^2.

Mathematica

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
Do[r=0; Do[If[SQ[n-x^4-y^2-z^2]&&SQ[144x^4+(5y-10z)^2], r=r+1], {x, 0, (n-1)^(1/4)}, {y, 0, Sqrt[n-1-x^4]}, {z, 0, Sqrt[n-1-x^4-y^2]}]; Print[n, " ", r]; Continue, {n, 1, 80}]

Crossrefs

Cf. A000118, A000290, A270969, A271714, A273108, A281939, A281941, A281975, A281976, A281977, A282013, A282014.

Sequence in context: A076118 A309045 A210956 * A205675 A342331 A058608

Adjacent sequences: A282158 A282159 A282160 * A282162 A282163 A282164

Keyword

nonn

Author

Zhi-Wei Sun, Feb 07 2017

Status

approved