A300139 Number of ways to write n as x^2 + y^2 + z^2 + w^2 with 4*x - 3*y a square, where x,y,z,w are nonnegative integers with z <= w such that 10*x or y is a square.

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

Offset: 0

Author

Zhi-Wei Sun, Mar 12 2018

Keywords

• nonn

Comments

Conjecture 1: a(n) > 0 for all n >= 0, and a(n) = 1 only for n = 16^k*m with k = 0,1,2,... and m = 0, 7, 8, 13, 22, 23, 24, 31, 44, 56, 71, 79, 88, 109, 120, 152, 184, 472, 1912, 6008, 9080.

Conjecture 2: Each n = 0,1,2,... can be written as x^2 + y^2 + z^2 + w^2 with 3*x - y twice a square, where x,y,z,w are nonnegative integers such that 5*x or y is a square.

By the author's 2017 JNT paper, any nonnegative integer can be written as the sum of a fourth power and three squares.

See also A281976, A300666, A300667, A300708 and A300712 for similar conjectures.

a(n) > 0 for all n = 0..10^8. Also, Conjecture 2 holds for all n = 0..10^8. - Zhi-Wei Sun, Oct 05 2020

Examples

			a(22) = 1 since 22 = 1^2 + 1^2 + 2^2 + 4^2 with 1 = 1^2 and 4*1 - 3*1 = 1^2.
a(23) = 1 since 23 = 3^2 + 1^2 + 2^2 + 3^2 with 1 = 1^2 and 4*3 - 3*1 = 3^2.
a(109) = 1 since 109 = 0^2 + 0^2 + 3^2 + 10^2 with 0 = 0^2 and 4*0 - 3*0 = 0^2.
a(184) = 1 since 184 = 10^2 + 8^2 + 2^2 + 4^2 with 10*10 = 10^2 and 4*10 - 3*8 = 4^2.
a(6008) = 1 since 6008 = 12^2 + 16^2 + 42^2 + 62^2 with 16 = 4^2 and 4*12 - 3*16 = 0^2.
a(9080) = 1 since 9080 = 10^2 + 12^2 + 0^2 + 94^2 with 10*10 = 10^2 and 4*10 - 3*12 = 2^2.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
• Zhi-Wei Sun, Refining Lagrange's four-square theorem, J. Number Theory 175(2017), 167-190.
• Zhi-Wei Sun, Restricted sums of four squares, arXiv:1701.05868 [math.NT], 2017-2018.

Crossrefs

Cf. A000118, A000290, A271518, A281976, A282542, A300666, A300667, A300219, A300708, A300712, A300751, A300752.

Programs

Mathematica

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
tab={};Do[r=0;Do[If[Mod[m^2+3y,4]==0&&(SQ[10(m^2+3y)/4]||SQ[y]), Do[If[SQ[n-((m^2+3y)/4)^2-y^2-z^2],r=r+1],{z,0,Sqrt[Max[0,(n-((m^2+3y)/4)^2-y^2)/2]]}]],{m,0,2n^(1/4)},{y,0,4/5*Sqrt[n-m^4/16]}];tab=Append[tab,r],{n,0,80}];Print[tab]