A299825 Number of ways to write n as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that x <= y, x == y (mod 2), and |x+y-z| is a power of 4 (including 4^0 = 1).

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

Offset: 1

Author

Zhi-Wei Sun, Feb 19 2018

Keywords

• nonn

Comments

Conjecture: a(n) > 0 for all n > 0, and a(n) = 1 only for n = 1, 2, 6, 13, 15, 18, 27, 43, 53, 63, 93, 107, 109, 123, 141, 159, 165, 173, 477, 493, 653, 1005, 16^k*m (k = 0,1,2,... and m = 3, 4, 7, 8).

We have verified that a(n) > 0 for all n = 1..5*10^6.

A weaker version of the conjecture was proved by the author in arXiv:1701.05868.

Examples

			a(8) = 1 since 8 = 2^2 + 2^2 + 0^2 + 0^2 with 2 == 2 (mod 2) and 2 + 2 - 0 = 4.
a(13) = 1 since 13 = 0^2 + 2^2 + 3^2 + 0^2 with 0 == 2 (mod 2) and 0 + 2 - 3 = -4^0.
a(109) = 1 since 109 = 2^2 + 4^2 + 5^2 + 8^2 with 2 == 4 (mod 2) and 2 + 4 - 5 = 4^0.
a(123) = 1 since 123 = 1^2 + 3^2 + 8^2 + 7^2 with 1 == 3 (mod 2) and 1 + 3 - 8 = -4.
a(477) = 1 since 477 = 0^2 + 10^2 + 11^2 + 16^2 with 0 == 10 (mod 2) and 0 + 10 - 11 = -4^0.
a(653) = 1 since 653 = 8^2 + 12^2 + 21^2 + 2^2 with 8 == 12 (mod 2) and 8 + 12 - 21 = -4^0.
a(1005) = 1 since 1005 = 0^2 + 10^2 + 11^2 + 28^2 with 0 == 10 (mod 2) and 0 + 10 - 11 = -4^0.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 1..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.

Crossrefs

Cf. A000118, A271518, A279612, A281945, A281976, A282463.

Programs

Mathematica

SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
Pow[n_]:=Pow[n]=IntegerQ[Log[4,n]];
Do[r=0;Do[If[Mod[x-y,2]==0&&Pow[Abs[x+y-z]]&&SQ[n-x^2-y^2-z^2],r=r+1],{x,0,Sqrt[n/2]},{y,x,Sqrt[n-x^2]},{z,0,Sqrt[n-x^2-y^2]}];Print[n," ",r],{n,1,80}]