A273108 Number of ordered ways to write n as x^2 + y^2 + z^2 + w^2 with (x + y)^2 + (4z)^2 a square, where x,y,z,w are nonnegative integers with x <= y > z.

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

Offset: 1

Author

Zhi-Wei Sun, May 15 2016

Keywords

• nonn

Comments

Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 39, 47, 95, 543, 4^k*m (k = 0,1,2,... and m = 1, 3, 7, 15, 23, 135, 183).

(ii) Any natural number can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that (a*x+b*y)^2 + (c*z)^2 is a square, whenever (a,b,c) is among the triples (1,2,4), (1,2,12), (1,4,8), (1,4,12), (1,10,20), (1,15,12), (2,7,20), (2,7,60), (2,21,60), (3,3,4), (3,3,40), (3,4,12), (3,5,60), (3,6,20), (3,9,20), (3,11,24), (3,12,8), (3,27,20), (3,27,56), (3,29,60), (3,30,28), (3,45,20), (4,4,3), (4,4,5), (4,4,9), (4,4,15), (4,8,5), (4,12,15), (4,12,21), (4,12,45), (4,16,45), (4,19,40), (4,20,21), (4,36,21), (4,36,33), (4,52,63), (5,5,8), (5,5,12), (5,5,24), (5,6,12), (5,8,24), (5,10,4), (5,15,24), (5,25,16), (5,30,12), (5,35,48), (5,40,24), (6,10,15), (6,15,28), (6,45,28), (7,7,20), (7,7,24), (7,21,12), (7,63,36), (8,8,15), (8,12,45), (8,16,35), (8,16,45), (8,32,15), (8,32,21), (8,48,45), (9,9,40), (9,18,28), (9,27,16), (9,45,20), (10,15,12), (10,25,28), (11,11,60), (12,12,5), (12,12,35), (12,20,63), (12,60,55).

See also A271714, A273107, A273110 and A273134 for similar conjectures related to Pythagorean triples. For more conjectural refinements of Lagrange's four-square theorem, one may consult arXiv:1604.06723.

Examples

			a(1) = 1 since 1 = 0^2 + 1^2 + 0^2 + 0^2 with 0 < 1 > 0 and (0+1)^2 + (4*0)^2 = 1^2.
a(3) = 1 since 3 = 1^2 + 1^2 + 0^2 + 1^2 with 1 = 1 > 0 and (1+1)^2 + (4*0)^2 = 2^2.
a(7) = 1 since 7 = 1^2 + 2^2 + 1^2 + 1^2 with 1 < 2 > 1 and (1+2)^2 + (4*1)^2 = 5^2.
a(15) = 1 since 15 = 1^2 + 2^2 + 1^2 + 3^2 with 1 < 2 > 1 and (1+2)^2 + (4*1)^2 = 5^2.
a(23) = 1 since 23 = 3^2 + 3^2 + 2^2 + 1^2 with 3 = 3 > 2 and (3+3)^2 + (4*2)^2 = 10^2.
a(39) = 1 since 39 = 1^2 + 5^2 + 2^2 + 3^2 with 1 < 5 > 2 and (1+5)^2 + (4*2)^2 = 10^2.
a(47) = 1 since 47 = 3^2 + 3^2 + 2^2 + 5^2 with 3 = 3 > 2 and (3+3)^2 + (4*2)^2 = 10^2.
a(95) = 1 since 95 = 3^2 + 7^2 + 6^2 + 1^2 with 3 < 7 > 6 and (3+7)^2 + (4*6)^2 = 26^2.
a(135) = 1 since 135 = 3^2 + 6^2 + 3^2 + 9^2 with 3 < 6 > 3 and (3+6)^2 + (4*3)^2 = 15^2.
a(183) = 1 since 183 = 2^2 + 7^2 + 3^2 + 11^2 with 2 < 7 > 3 and (2+7)^2 + (4*3)^2 = 15^2.
a(543) = 1 since 543 = 2^2 + 13^2 + 9^2 + 17^2 with 2 < 13 > 9 and (2+13)^2 + (4*9)^2 = 39^2.
		

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, A273021, A273107, A273110, A273134.

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+16*z^2],r=r+1],{x,0,Sqrt[n/2]},{y,x,Sqrt[n-x^2]},{z,0,Min[y-1,Sqrt[n-x^2-y^2]]}];Print[n," ",r];Continue,{n,1,80}]

Extensions

All statements in examples checked by Rick L. Shepherd, May 29 2016