A282562 Number of ways to write n as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that both 2*x - y and 9*z^2 + 666*z*w + w^2 are squares.

1, 2, 2, 2, 4, 4, 5, 2, 3, 5, 2, 2, 1, 5, 6, 2, 2, 5, 5, 1, 5, 7, 8, 1, 3, 4, 2, 4, 3, 9, 7, 5, 2, 5, 5, 4, 8, 3, 9, 3, 4, 4, 5, 3, 1, 7, 6, 3, 2, 13, 7, 5, 5, 5, 8, 3, 2, 3, 5, 2, 1, 6, 10, 6, 4, 8, 9, 1, 7, 8, 11, 3, 3, 6, 6, 5, 3, 11, 4, 4, 4

Offset

0,2

Comments

Conjecture: a(n) > 0 for all n = 0,1,2,..., and a(n) = 1 only for n = 0, 16^k*m (k = 0,1,2,... and m = 12, 19, 23, 44, 60, 67, 139, 140, 248, 264, 347, 427, 499, 636, 1388, 1867, 1964, 4843).

By the linked JNT paper, any nonnegative integer can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers and z*(z-w) = 0. Whether z = 0 or z = w, the number 9*z^2 + 666*z*w + w^2 is definitely a square.

See also A282561 for a similar conjecture.

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 = 2^2 + 0^2 + 2^2 + 2^2 with 2*2 - 0 = 2^2 and 9*2^2 + 666*2*2 + 2^2 = 52^2.
a(19) = 1 since 19 = 1^2 + 1^2 + 4^2 + 1^2 with 2*1 - 1 = 1^2 and 9*4^2 + 666*4*1 + 1^2 = 53^2.
a(44) = 1 since 44 = 5^2 + 1^2 + 3^2 + 3^2 with 2*5 - 1 = 3^2 and 9*3^2 + 666*3*3 + 3^2 = 78^2.
a(60) = 1 since 60 = 3^2 + 5^2 + 1^2 + 5^2 with 2*3 - 5 = 1^2 and 9*1^2 + 666*1*5 + 5^2 = 58^2.
a(67) = 1 since 67 = 4^2 + 7^2 + 1^2 + 1^2 with 2*4 - 7 = 1^2 and 9*1^2 + 666*1*1 + 1^2 = 26^2.
a(139) = 1 since 139 = 8^2 + 7^2 + 1^2 + 5^2 with 2*8 - 7 = 3^2 and 9*1^2 + 666*1*5 + 5^2 = 58^2.
a(140) = 1 since 140 = 3^2 + 5^2 + 5^2 + 9^2 with 2*3 - 5 = 1^2 and 9*5^2 + 666*5*9 + 9^2 = 174^2.
a(264) = 1 since 264 = 8^2 + 0^2 + 10^2 + 10^2 with 2*8 - 0 = 4^2 and 9*10^2 + 666*10*10 + 10^2 = 260^2.
a(499) = 1 since 499 = 7^2 + 5^2 + 20^2 + 5^2 with 2*7 - 5 = 3^2 and 9*20^2 + 666*20*5 + 5^2 = 265^2.
a(1388) = 1 since 1388 = 15^2 + 21^2 + 19^2 + 19^2 with 2*15 - 21 = 3^2 and 9*19^2 + 666*19*19 + 19^2 = 494^2.
a(1867) = 1 since 1867 = 16^2 + 31^2 + 5^2 + 25^2 with 2*16 - 31 = 1^2 and 9*5^2 + 666*5*25 + 25^2 = 290^2.
a(4843) = 1 since 4843 = 11^2 + 13^2 + 52^2 + 43^2 with 2*11 - 13 = 3^2 and 9*52^2 + 666*52*43 + 43^2 = 1231^2.

Mathematica

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
Do[r=0; Do[If[SQ[2x-y], Do[If[SQ[n-x^2-y^2-z^2]&&SQ[9z^2+666z*Sqrt[n-x^2-y^2-z^2]+(n-x^2-y^2-z^2)], r=r+1], {z, 0, Sqrt[n-x^2-y^2]}]], {y, 0, Sqrt[4n/5]}, {x, Ceiling[y/2], Sqrt[n-y^2]}]; Print[n, " ", r]; Continue, {n, 0, 80}]

Crossrefs

Cf. A000118, A000290, A271518, A281939, A281976, A282463, A282494, A282495, A282545, A282561.

Sequence in context: A089873 A275433 A096323 * A035682 A054543 A240861

Adjacent sequences: A282559 A282560 A282561 * A282563 A282564 A282565

Keyword

nonn

Author

Zhi-Wei Sun, Feb 18 2017

Status

approved