A272620 Number of ordered ways to write n as w^2 + x^2 + y^2 + z^2 with w + x + y - z a square, where w is an integer and x,y,z are nonnegative integers with |w| <= x >= y <= z < x + y.

1, 1, 1, 1, 2, 1, 1, 2, 1, 4, 1, 1, 3, 3, 2, 3, 1, 7, 1, 2, 3, 2, 1, 3, 3, 7, 2, 3, 1, 7, 1, 1, 4, 5, 3, 2, 1, 9, 2, 5, 3, 6, 5, 3, 3, 7, 2, 2, 5, 6, 3, 3, 5, 9, 4, 4, 4, 9, 4, 4, 5, 6, 6, 1, 6, 12, 2, 2, 7, 4, 4, 6, 5, 11, 7, 3, 5, 9, 4, 5

Offset: 1

Author

Zhi-Wei Sun, May 03 2016

Keywords

• nonn

Comments

Conjecture: a(n) > 0 for all n > 0.

In contrast, the author has proved that any natural number can be written as w^2 + x^2 + y^2 + z^2 with w,x,y,z integers such that x + y + z is a square. See arXiv:1604.06723.

Yu-Chen Sun and the author proved in arXiv:1605.03074 that any nonnegative integer can be written as w^2 + x^2 + y^2 + z^2 with w,x,y,z integers such that w + x + y + z is a square. - Zhi-Wei Sun, May 10 2016

Examples

			a(1) = 1 since 1 = 0^2 + 1^2 + 0^2 + 0^2 with 0 < 1 > 0 = 0 < 1 + 0 and 0 + 1 + 0 - 0 = 1^2.
a(2) = 1 since 2 = (-1)^2 + 1^2 + 0^2 + 0^2 with 1 = 1 > 0 = 0 < 1 + 0 and -1 + 1 + 0 - 0 = 0^2.
a(3) = 1 since 3 = 0^2 + 1^2 + 1^2 + 1^2 with 0 < 1 = 1 = 1 < 1 + 1 and 0 + 1 + 1 - 1 = 1^2.
a(4) = 1 since 4 = (-1)^2 + 1^2 + 1^2 + 1^2 with 1 = 1 = 1 = 1 < 1 + 1 and -1 + 1 + 1 - 1 = 0^2.
a(6) = 1 since 6 = (-1)^2 + 2^2 + 0^2 + 1^2 with 1 < 2 > 0 < 1 < 2 + 0 and -1 + 2 + 0 - 1 = 0^2.
a(7) = 1 since 7 = (-1)^2 + 2^2 + 1^2 + 1^2 with 1 < 2 > 1 = 1 < 2 + 1 and -1 + 2 + 1 - 1 = 1^2.
a(9) = 1 since 9 = 0^2 + 2^2 + 1^2 + 2^2 with 0 < 2 > 1 < 2 < 2 + 1 and 0 + 2 + 1 - 2 = 1^2.
a(11) = 1 since 11 = (-1)^2 + 3^2 + 0^2 + 1^2 with 1 < 3 > 0 < 1 < 3 + 0 and -1 + 3 + 0 - 1 = 1^2.
a(12) = 1 since 12 = 1^2 + 3^2 + 1^2 + 1^2 with 1 < 3 > 1 = 1 < 3 + 1 and 1 + 3 + 1 - 1 = 2^2.
a(17) = 1 since 17 = 0^2 + 2^2 + 2^2 + 3^2 with 0 < 2 = 2 < 3 < 2 + 2 and 0 + 2 + 2 - 3 = 1^2.
a(19) = 1 since 19 = 0^2 + 3^2 + 1^2 + 3^2 with 0 < 3 > 1 < 3 < 3 + 1 and 0 + 3 + 1 - 3 = 1^2.
a(23) = 1 since 23 = (-1)^2 + 3^2 + 2^2 + 3^2 with 1 < 3 > 2 < 3 < 3 + 2 and -1 + 3 + 2 - 3 = 1^2.
a(29) = 1 since 29 = 0^2 + 3^2 + 2^2 + 4^2 with 0 < 3 > 2 < 4 < 3 + 2 and 0 + 3 + 2 - 4 = 1^2.
a(31) = 1 since 31 = (-2)^2 + 3^2 + 3^2 + 3^2 with 2 < 3 = 3 = 3 < 3 + 3 and -2 + 3 + 3 - 3 = 1^2.
a(37) = 1 since 37 = (-1)^2 + 4^2 + 2^2 + 4^2 with 1 < 4 > 2 < 4 < 4 + 2 and -1 + 4 + 2 - 4 = 1^2.
a(92) = 1 since 92 = 3^2 + 5^2 + 3^2 + 7^2 with 3 < 5 > 3 < 7 < 5 + 3 and 3 + 5 + 3 - 7 = 2^2.
a(284) = 1 since 284 = 3^2 + 9^2 + 5^2 + 13^2 with 3 < 9 > 5 < 13 < 9 + 5 and 3 + 9 + 5 - 13 = 2^2.
a(572) = 1 since 572 = 3^2 + 11^2 + 9^2 + 19^2 with 3 < 11 > 9 < 19 < 11 + 9 and 3 + 11 + 9 - 19 = 2^2.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..10000
• Yu-Chen Sun and Zhi-Wei Sun, Two refinements of Lagrange's four-square theorem, arXiv:1605.03074 [math.NT], 2016.
• 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, A268507, A269400, A271510, A271513, A271518, A271608, A271665, A271714, A271721, A271724, A271775, A271778, A271824, A272084, A272332, A272351.

Programs

Mathematica

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
Do[r=0;Do[If[Sqrt[n-x^2-y^2-z^2]<=x&&SQ[n-x^2-y^2-z^2]&&SQ[x+y-z+(-1)^k*Sqrt[n-x^2-y^2-z^2]],r=r+1],{y,0,Sqrt[n/3]},{x,y,Sqrt[n-y^2]},{z,y,Min[x+y-1,Sqrt[n-x^2-y^2]]},{k,0,Min[1,Sqrt[n-x^2-y^2-z^2]]}];Print[n," ",r];Continue,{n,1,80}]

Extensions

Rick L. Shepherd, May 27 2016: I checked all the statements in each example.