A275409 Number of ordered ways to write n as 2*w^2 + x^2 + y^2 + z^2 with w + x + 2*y + 4*z a square, where w,x,y,z are nonnegative integers.

1, 2, 1, 0, 2, 2, 2, 1, 1, 1, 0, 3, 1, 2, 1, 1, 3, 2, 5, 3, 4, 3, 1, 1, 1, 1, 2, 2, 2, 4, 2, 2, 4, 2, 7, 3, 1, 6, 2, 1, 2, 3, 4, 5, 1, 1, 3, 5, 3, 3, 4, 3, 7, 3, 2, 4, 3, 4, 4, 3, 1, 4, 5, 3, 6, 4, 4, 4, 5, 7, 7, 3, 6, 5, 5, 4, 3, 11, 2, 2, 4

Offset: 0

Author

Zhi-Wei Sun, Jul 26 2016

Keywords

• nonn

Comments

Conjecture: (i) a(n) > 0 except for n = 3, 10, and a(n) = 1 only for n = 0, 2, 7, 8, 9, 12, 14, 15, 22, 23, 24, 25, 36, 39, 44, 45, 60, 87, 98, 106, 110, 111, 183.

(ii) Any natural number can be written as x^2 + y^2 + z^2 + 2*w^2 with x,y,z,w nonnegative integers such that x + 2*y + 3*z - 3*w is a square.

(iii) For each triple (a,b,c) = (1,2,1), (1,2,3), (1,3,1), (2,4,1), (2,4,2), (2,4,3), (2,4,4), (2,4,8), (8,9,5), any natural number can be written as x^2 + y^2 + z^2 + 2*w^2 with x,y,z,w nonnegative integers such that a*x + b*y - c*z is a square.

(iv) Any natural number can be written as x^2 + y^2 + z^2 + 2*w^2 with x,y,z,w nonnegative integers such that x + 2*y - 2*z is twice a nonnegative cube. Also, each natural number can be written as x^2 + y^2 + z^2 + 2*w^3 with x,y,z,w nonnegative integers such that x + 3*y - z is a square.

See also A275344 and A275301 for related conjectures. We are able to show that each natural number can be written as x^2 + y^2 + z^2 + 2*w^2 with x,y,z,w integers such that x + y + z = t^2 for some t = 0, 1, 2.

Examples

			a(2) = 1 since 2 = 2*1^2 + 0^2 + 0^2 + 0^2 with 1 + 0 + 2*0 + 4*0 = 1^2.
a(7) = 1 since 7 = 2*1^2 + 0^2 + 2^2 + 1^2 with 1 + 0 + 2*2 + 4*1 = 3^2.
a(8) = 1 since 8 = 2*1^2 + 2^2 + 1^2 + 1^2 with 1 + 2 + 2*1 + 4*1 = 3^2.
a(9) = 1 since 9 = 2*2^2 + 0^2 + 1^2 + 0^2 with 2 + 0 + 2*1 + 4*0 = 2^2.
a(12) = 1 since 12 = 2*2^2 + 2^2 + 0^2 + 0^2 with 2 + 2 + 2*0 + 4*0 = 2^2.
a(14) = 1 since 14 = 2*0^2 + 2^2 + 1^2 + 3^2 with 0 + 2 + 2*1 + 4*3 = 4^2.
a(15) = 1 since 15 = 2*1^2 + 2^2 + 3^2 + 0^2 with 1 + 2 + 2*3 + 4*0 = 3^2.
a(22) = 1 since 22 = 2*1^2 + 4^2 + 2^2 + 0^2 with 1 + 4 + 2*2 + 4*0 = 3^2.
a(23) = 1 since 23 = 2*3^2 + 2^2 + 0^2 + 1^2 with 3 + 2 + 2*0 + 4*1 = 3^2.
a(24) = 1 since 24 = 2*0^2 + 4^2 + 2^2 + 2^2 with 0 + 4 + 2*2 + 4*2 = 4^2.
a(25) = 1 since 25 = 2*0^2 + 4^2 + 0^2 + 3^2 with 0 + 4 + 2*0 + 4*3 = 4^2.
a(36) = 1 since 36 = 2*3^2 + 1^2 + 4^2 + 1^2 with 3 + 1 + 2*4 + 4*1 = 4^2.
a(39) = 1 since 39 = 2*1^2 + 6^2 + 1^2 + 0^2 with 1 + 6 + 2*1 + 4*0 = 3^2.
a(44) = 1 since 44 = 2*3^2 + 0^2 + 1^2 + 5^2 with 3 + 0 + 2*1 + 4*5 = 5^2.
a(45) = 1 since 45 = 2*0^2 + 5^2 + 2^2 + 4^2 with 0 + 5 + 2*2 + 4*4 = 5^2.
a(60) = 1 since 60 = 2*2^2 + 6^2 + 4^2 + 0^2 with 2 + 6 + 2*4 + 4*0 = 4^2.
a(87) = 1 since 87 = 2*3^2 + 2^2 + 8^2 + 1^2 with 3 + 2 + 2*8 + 4*1 = 5^2.
a(98) = 1 since 98 = 2*4^2 + 1^2 + 8^2 + 1^2 with 4 + 1 + 2*8 + 4*1 = 5^2.
a(106) = 1 since 106 = 2*2^2 + 8^2 + 3^2 + 5^2 with 2 + 8 + 2*3 + 4*5 = 6^2.
a(110) = 1 since 110 = 2*6^2 + 5^2 + 3^2 + 2^2 with 6 + 5 + 2*3 + 4*2 = 5^2.
a(111) = 1 since 111 = 2*5^2 + 3^2 + 6^2 + 4^2 with 5 + 3 + 2*6 + 4*4 = 6^2.
a(183) = 1 since 183 = 2*3^2 + 10^2 + 4^2 + 7^2 with 3 + 10 + 2*4 + 4*7 = 7^2.
		

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.

Crossrefs

Cf. A000290, A271518, A275297, A275301, A275344.

Programs

Mathematica

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
Do[r=0;Do[If[SQ[n-2*w^2-x^2-y^2]&&SQ[w+x+2y+4*Sqrt[n-2*w^2-x^2-y^2]],r=r+1],{w,0,Sqrt[n/2]},{x,0,Sqrt[n-2*w^2]},{y,0,Sqrt[n-2*w^2-x^2]}];Print[n," ",r];Continue,{n,0,80}]