A270073 Number of ordered ways to write n as x^2 + y^2 + z^2 + w^2 with x*y + 2*z*w a square, where x,y,z are nonnegative integers and w is an integer with x <= y and z >= |w|.

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

Offset: 0

Author

Zhi-Wei Sun, May 07 2016

Keywords

• nonn

Comments

Conjecture: (i) a(n) > 0 for all n = 0,1,2,..., and a(n) = 1 only for n = 0, 3, 11, 15, 23, 43, 67, 79, 155, 211, 331, 347, 403, 427, 659, 899, 1443, 1955, 2^k*m (k = 0,1,2,... and m = 14, 35, 62, 158, 382).

(ii) Each n = 0,1,2,... can be written as x^2 + y^2 + z^2 + w^2 with x*y + z*w/2 a square, where x,y,z are nonnegative integers and w is an integer with 2 | z*w and x <= y >= |w| <= z.

We have verified that a(n) > 0 for all n = 0,...,10^5.

For more refinements of Lagrange's four-square theorem, see arXiv:1604.06723.

Examples

			a(3) = 1 since 3 = 1^2 + 1^2 + 1^2 + 0^2 with 1 = 1, 1 > 0 and 1*1 + 2*1*0 = 1^2.
a(11) = 1 since 11 = 1^2 + 1^2 + 3^2 + 0^2 with 1 = 1, 3 > 0 and 1*1 + 2*3*0 = 1^2.
a(14) = 1 since 14 = 0^2 + 3^2 + 2^2 + 1^2 with 0 < 3, 2 > 1 and 0*3 + 2*2*1 = 2^2.
a(15) = 1 since 15 = 2^2 + 3^2 + 1^2 + (-1)^2 with 2 < 3, 1 = |-1| and 2*3 + 2*1*(-1) = 2^2.
a(23) = 1 since 23 = 2^2 + 3^2 + 3^2 + (-1)^2 with 2 < 3, 3 > |-1| and 2*3 + 2*3*(-1) = 0^2.
a(35) = 1 since 35 = 3^2 + 3^2 + 4^2 + (-1)^2 with 3 = 3, 4 > |-1| and 3*3 + 2*4*(-1) = 1^2.
a(43) = 1 since 43 = 3^2 + 3^2 + 5^2 + 0^2 with 3 = 3, 5 > 0 and 3*3 + 2*5*0 = 3^2.
a(62) = 1 since 62 = 3^2 + 4^2 + 6^2 + (-1)^2 with 3 < 4, 6 > |-1| and 3*4 + 2*6*(-1) = 0^2.
a(67) = 1 since 67 = 3^2 + 3^2 + 7^2 + 0^2 with 3 = 3, 7 > 0 and 3*3 + 2*7*0 = 3^2.
a(79) = 1 since 79 = 2^2 + 7^2 + 5^2 + (-1)^2 with 2 < 7, 5 > |-1| and 2*7 + 2*5*(-1) = 2^2.
a(155) = 1 since 155 = 3^2 + 11^2 + 4^2 + (-3)^2 with 3 < 11, 4 > |-3| and 3*11 + 2*4*(-3) = 3^2.
a(158) = 1 since 158 = 1^2 + 12^2 + 3^2 + (-2)^2 with 1 < 12, 3 > |-2| and 1*12 + 2*3*(-2) = 0^2.
a(211) = 1 since 211 = 9^2 + 9^2 + 7^2 + 0^2 with 9 = 9, 7 > 0 and 9*9 + 2*7*0 = 9^2.
a(331) = 1 since 331 = 9^2 + 9^2 + 13^2 + 0^2 with 9 = 9, 13 > 0 and 9*9 + 2*13*0 = 9^2.
a(347) = 1 since 347 = 13^2 + 13^2 + 3^2 + 0^2 with 13 = 13, 3 > 0 and 13*13 + 2*3*0 = 13^2.
a(382) = 1 since 382 = 5^2 + 16^2 + 10^2 + 1^2 with 5 < 16, 10 > 1 and 5*16 + 2*10*1 = 10^2.
a(403) = 1 since 403 = 13^2 + 13^2 + 7^2 + 4^2 with 13 = 13, 7 > 4 and 13*13 + 2*7*4 = 15^2.
a(427) = 1 since 427 = 11^2 + 11^2 + 13^2 + 4^2 with 11 = 11, 13 > 4 and 11*11 + 2*13*4 = 15^2.
a(659) = 1 since 659 = 17^2 + 17^2 + 9^2 + 0^2 with 17 = 17, 9 > 0 and 17*17 + 2*9*0 = 17^2.
a(899) = 1 since 899 = 7^2 + 15^2 + 24^2 + 7^2 with 7 < 15, 24 > 7 and 7*15 + 2*24*7 = 21^2.
a(1443) = 1 since 1443 = 7^2 + 31^2 + 17^2 + 12^2 with 7 < 31, 17 > 12 and 7*31 + 2*17*12 = 25^2.
a(1955) = 1 since 1955 = 19^2 + 27^2 + 28^2 + (-9)^2 with 19 < 27, 28 > |-9| and 19*27 + 2*28*(-9) = 3^2.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
• Zhi-Wei Sun, Refining Lagrange's four-square theorem, arXiv:1604.06723 [math.GM], 2016.
• Zhi-Wei Sun, Refine Lagrange's four-square theorem, a message to Number Theory List, April 26, 2016.

Crossrefs

Cf. A000118, A000290, A260625, A261876, A262357, A267121, A268197, A268507, A269400, A271510, A271513, A271518, A271608, A271665, A271714, A271721, A271724, A271775, A271778, A271824, A272084, A272332, A272351, A272620.

Programs

Mathematica

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

Extensions

All statements in the examples checked by Rick L. Shepherd, May 27 2016