A273021 Number of ordered ways to write n as x^2 + y^2 + z^2 + w^2 with 2*x*y + y*z - z*w - w*x a square, where w is a positive integer and x,y,z are nonnegative integers with x <= y.

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

Offset: 1

Author

Zhi-Wei Sun, May 13 2016

Keywords

• nonn

Comments

Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 1, 11, 31, 47, 55, 71, 105, 115, 119, 253, 383, 385, 4^k*m (k = 0,1,2,... and m = 2, 22, 23, 30, 330).

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

See arXiv:1604.06723 for more conjectural refinements of Lagrange's four-square theorem.

Examples

			a(1) = 1 since 1 = 0^2 + 0^2 + 0^2 + 1^2 with 0 = 0 and 2*0*0 + 0*0 - 0*1 - 1*0 = 0^2.
a(2) = 1 since 2 = 0^2 + 1^2 + 0^2 + 1^2 with 0 < 1 and 2*0*1 + 1*0 - 0*1 - 1*0 = 0^2.
a(11) = 1 since 11 = 0^2 + 1^2 + 3^2 + 1^2 with 0 < 1 and 2*0*1 + 1*3 - 3*1 - 1*0 = 0^2.
a(22) = 1 since 22 = 0^2 + 3^2 + 2^2 + 3^2 with 0 < 3 and 2*0*3 + 3*2 - 2*3 - 3*0 = 0^2.
a(23) = 1 since 23 = 2^2 + 3^2 + 3^2 + 1^2 with 2 < 3 and 2*2*3 + 3*3 - 3*1 - 1*2 = 4^2.
a(30) = 1 since 30 = 1^2 + 3^2 + 2^2 + 4^2 with 1 < 3 and 2*1*3 + 3*2 - 2*4 - 4*1 = 0^2.
a(31) = 1 since 31 = 3^2 + 3^2 + 2^2 + 3^2 with 3 = 3 and
2*3*3 + 3*2 - 2*3 -3*3 = 3^2.
a(47) = 1 since 47 = 3^2 + 5^2 + 2^2 + 3^2 with 3 < 5 and 2*3*5 + 5*2 - 2*3 - 3*3 = 5^2.
a(55) = 1 since 55 = 1^2 + 7^2 + 2^2 + 1^2 with 1 < 7 and 2*1*7 + 7*2 - 2*1 - 1*1 = 5^2.
a(71) = 1 since 71 = 1^2 + 5^2 + 3^2 + 6^2 with 1 < 5 and 2*1*5 + 5*3 - 3*6 - 6*1 = 1^2.
a(105) = 1 since 105 = 1^2 + 6^2 + 2^2 + 8^2 with 1 < 6 and 2*1*6 + 6*2 - 2*8 - 8*1 = 0^2.
a(115) = 1 since 115 = 1^2 + 8^2 + 7^2 + 1^2 with 1 < 8 and 2*1*8 + 8*7 - 7*1 - 1*1 = 8^2.
a(119) = 1 since 119 = 1^2 + 6^2 + 1^2 + 9^2 with 1 < 6 and 2*1*6 + 6*1 - 1*9 - 9*1 = 0^2.
a(253) = 1 since 253 = 2^2 + 8^2 + 11^2 + 8^2 with 2 < 8 and 2*2*8 + 8*11 - 11*8 - 8*2 = 4^2.
a(330) = 1 since 330 = 4^2 + 13^2 + 8^2 + 9^2 with 4 < 13 and 2*4*13 + 13*8 - 8*9 - 9*4 = 10^2.
a(383) = 1 since 383 = 9^2 + 14^2 + 5^2 + 9^2 with 9 < 14 and 2*9*14 + 14*5 - 5*9 - 9*9 = 14^2.
a(385) = 1 since 385 = 4^2 + 12^2 + 0^2 + 15^2 with 4 < 12 and 2*4*12 + 12*0 - 0*15 - 15*4 = 6^2.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
• Zhi-Wei Sun, Refining Lagrange's four-square theorem, arXiv:1604.06723 [math.GM], 2016.

Crossrefs

Cf. A000118, A000290, A260625, A261876, A262357, A267121, A268197, A268507, A269400, A270073, A271510, A271513, A271518, A271608, A271665, A271714, A271721, A271724, A271775, A271778, A271824, A272084, A272332, A272351, A272620, A272888, A272977.

Programs

Mathematica

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