A275344 Number of ordered ways to write n as x^2 + y^2 + z^2 + 2*w^2 with x + 2*y + 3*z a square, where x,y,z,w are nonnegative integers.

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

Offset: 0

Author

Zhi-Wei Sun, Jul 24 2016

Keywords

• nonn

Comments

Conjecture: (i) a(n) > 0 for all n = 0,1,2,..., and a(n) = 1 only for n = 0, 1, 3, 5, 7, 14, 15, 16, 25, 30, 33, 84, 169, 225.

(ii) For each ordered pair (a,b) = (1,2), (1,3), (1,12), (1,23), (2,3), (2,4), (2,6), (2,7), (2,15), (2,16), 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 is a square.

This is similar to the conjecture in A271518. It is known that any natural number can be written as x^2 + y^2 + z^2 + 2*w^2 with x,y,z,w nonnegative integers.

Examples

			a(0) = 1 since 0 = 0^2 + 0^2 + 0^2 + 2*0^2 with 0 + 2*0 + 3*0 = 0^2.
a(1) = 1 since 1 = 1^2 + 0^2 + 0^2 + 2*0^2 with 1 + 2*0 + 3*0 = 1^2.
a(3) = 1 since 3 = 1^2 + 0^2 + 0^2 + 2*1^2 with 1 + 2*0 + 3*0 = 1^2.
a(5) = 1 since 5 = 2^2 + 1^2 + 0^2 + 2*0^2 with 2 + 2*1 + 3*0 = 2^2.
a(7) = 1 since 7 = 2^2 + 1^2 + 0^2 + 2*1^2 with 2 + 2*1 + 3*0 = 2^2.
a(14) = 1 since 14 = 1^2 + 1^2 + 2^2 + 2*2^2 with 1 + 2*1 + 3*2 = 3^2.
a(15) = 1 since 15 = 3^2 + 0^2 + 2^2 + 2*1^2 with 3 + 2*0 + 3*2 = 3^2.
a(16) = 1 since 16 = 4^2 + 0^2 + 0^2 + 2*0^2 with 4 + 2*0 + 3*0 = 2^2.
a(25) = 1 since 25 = 1^2 + 4^2 + 0^2 + 2*2^2 with 1 + 2*4 + 3*0 = 3^2.
a(30) = 1 since 30 = 3^2 + 2^2 + 3^2 + 2*2^2 with 3 + 2*2 + 3*3 = 4^2.
a(33) = 1 since 33 = 1^2 + 0^2 + 0^2 + 2*4^2 with 1 + 2*0 + 3*0 = 1^2.
a(84) = 1 since 84 = 4^2 + 6^2 + 0^2 + 2*4^2 with 4 + 2*6 + 3*0 = 4^2.
a(169) = 1 since 169 = 10^2 + 6^2 + 1^2 + 2*4^2 with 10 + 2*6 + 3*1 = 5^2.
a(225) = 1 since 225 = 10^2 + 6^2 + 9^2 + 2*2^2 with 10 + 2*6 + 3*9 = 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.

Programs

Mathematica

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
Do[r=0;Do[If[SQ[n-2*w^2-x^2-y^2]&&SQ[x+2*y+3*Sqrt[n-2w^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}]