A271510 Number of ordered ways to write n as x^2 + y^2 + z^2 + w^2 with x >= y >= 0, z >= 0 and w >= 0 such that x^2 + 8*y^2 + 16*z^2 is a square.

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

Offset

0,2

Comments

Conjecture: (i) a(n) > 0 for all n = 0,1,2,..., and a(n) = 1 only for n = 0, 7, 23, 71, 77, 105, 191, 215, 311, 335, 2903, 4^k*q (k = 0,1,2,... and q = 6, 15, 47, 138).

(ii) Any natural number can be written as x^2 + y^2 + z^2 + w^2 with x >= y >= 0, z >=0 and w >= 0 such that 4*x^2 + 21*y^2 + 24*z^2 (or 5*x^2 + 40*y^2 + 4*z^2, 20*x^2 + 85*y^2 +16*z^2, 25*x^2 + 480*y^2 + 96*z^2, 36*x^2 + 45*y^2 + 40*z^2, 40*x^2 + 72*y^2 + 9*z^2) is a square.

(iii) For any ordered pair (b, c) = (48, 112), (63, 7), (112, 1008), (136, 24), (136, 216), (360, 40), (840, 280), (1008, 112), each natural number can be written as x^2 + y^2 + z^2 + w^2 with x >= y >= 0, z >=0 and w >= 0 such that 9*x^2 + b*y^2 + c*z^2 is a square.

(iv) For any ordered pair (b, c) = (80, 25), (81, 48), (144, 9), (144, 153), (177, 48), each natural number can be written as x^2 + y^2 + z^2 + w^2 with x >= y >= 0, z >=0 and w >= 0 such that 16*x^2 + b*y^2 + c*z^2 is a square.

This conjecture is much stronger than Lagrange's four-square theorem. It is apparent that a(m^2*n) >= a(n) for all m,n = 1,2,3,....

See also A271513 and A271518 for related conjectures.

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. Also available from arXiv:1604.06723 [math.NT], 2016-2017.

Example

a(6) = 1 since 6 = 1^2 + 1^2 + 0^2 + 2^2 with 1 = 1 and 1^2 + 8*1^2 + 16*0^2 = 3^2.
a(7) = 1 since 7 = 1^2 + 1^2 + 1^2 + 2^2 with 1 = 1 and 1^2 + 8*1^2 + 16*1^2 = 5^2.
a(15) = 1 since 15 = 3^2 + 1^2 + 2^2 + 1^2 with 3 > 1 and 3^2 + 8*1^2 + 16*2^2 = 9^2.
a(23) = 1 since 23 = 3^2 + 1^2 + 2^2 + 3^2 with 3 > 1 and 3^2 + 8*1^2 + 16*2^2 = 9^2.
a(47) = 1 since 47 = 3^2 + 2^2 + 5^2 + 3^2 with 3 > 2 and 3^2 + 8*2^2 + 16*5^2 = 21^2.
a(71) = 1 since 71 = 7^2 + 2^2 + 3^2 + 3^2 with 7 > 2 and 7^2 + 8*2^2 + 16*3^2 = 15^2.
a(77) = 1 since 77 = 5^2 + 4^2 + 6^2 + 0^2 with 5 > 4 and 5^2 + 8*4^2 + 16*6^2 = 27^2.
a(105) = 1 since 105 = 6^2 + 2^2 + 4^2 + 7^2 with 6 > 2 and 6^2 + 8*2^2 + 16*4^2 = 18^2.
a(138) = 1 since 138 = 3^2 + 2^2 + 5^2 + 10^2 with 3 > 2 and 3^2 + 8*2^2 + 16*5^2 = 21^2.
a(191) = 1 since 191 = 9^2 + 3^2 + 1^2 + 10^2 with 9 > 3 and 9^2 + 8*3^2 + 16*1^2 = 13^2.
a(215) = 1 since 215 = 11^2 + 7^2 + 6^2 + 3^2 with 11 > 7 and 11^2 + 8*7^2 + 16*6^2 = 33^2.
a(311) = 1 since 311 = 15^2 + 6^2 + 1^2 + 7^2 with 15 > 6 and 15^2 + 8*6^2 + 16*1^2 = 23^2.
a(335) = 1 since 335 = 17^2 + 1^2 + 3^2 + 6^2 with 17 > 1 and 17^2 + 8*1^2 + 16*3^2 = 21^2.
a(2903) = 1 since 2903 = 49^2 + 14^2 + 15^2 + 9^2 with 49 > 14 and 49^2 + 8*14^2 + 16*15^2 = 87^2.

Mathematica

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
Do[r=0; Do[If[SQ[n-x^2-y^2-z^2]&&SQ[x^2+8y^2+16z^2], r=r+1], {y, 0, Sqrt[n/2]}, {x, y, Sqrt[n-y^2]}, {z, 0, Sqrt[n-x^2-y^2]}]; Print[n, " ", r]; Continue, {n, 0, 80}]

Crossrefs

Cf. A000118, A000290, A270969, A271513, A271518.

Sequence in context: A106686 A106702 A303338 * A349661 A282545 A306471

Adjacent sequences: A271507 A271508 A271509 * A271511 A271512 A271513

Keyword

nonn

Author

Zhi-Wei Sun, Apr 09 2016

Status

approved