A283273 Number of ways to write n as x^2 + y^2 + z^2 with x,y integers and z a nonnegative integer such that x + 2*y + 3*z is a square or twice a square.

1, 2, 4, 3, 2, 6, 3, 0, 4, 6, 4, 5, 3, 5, 7, 0, 2, 3, 4, 6, 6, 7, 5, 0, 3, 4, 9, 5, 0, 15, 4, 0, 4, 3, 6, 9, 6, 4, 7, 0, 4, 7, 7, 4, 5, 9, 3, 0, 3, 2, 6, 9, 5, 11, 12, 0, 7, 5, 4, 13, 0, 9, 6, 0, 2, 9, 11, 2, 3, 6, 5, 0, 4, 5, 12, 6, 6, 11, 5, 0, 6

Offset: 0

Author

Zhi-Wei Sun, Mar 04 2017

Keywords

• nonn

Comments

Conjecture: (i) For any nonnegative integer n not of the form 4^k*(8*m+7) (k,m = 0,1,2,...), we have a(n) > 0.

(ii) Any nonnegative integer not of the form 4^k*(8*m+7) (k,m = 0,1,2,...) can be written as x^2 + y^2 + z^2 with x,y,z integers such that a*x + b*y + c*z is a square or twice a square, whenever (a,b,c) is among the triples (1,2,5), (1,3,4), (1,4,7), (1,6,7), (2,3,4), (2,3,9), (3,4,7), (3,4,9).

The Gauss-Legendre theorem states that a nonnegative integer can be written as the sum of three squares if and only if it is not of the form 4^k*(8*m+7) with k and m nonnegative integers. Thus a(4^k*(8*m+7)) = 0 for all k,m = 0,1,2,..., and part (i) of our conjecture (verified for n up to 1.6*10^6) is stronger than the Gauss-Legendre theorem.

See also A283269 for a similar conjecture.

Examples

			a(82) = 1 since 82 = 0^2 + (-1)^2 + 9^2 with 0 + 2*(-1) + 3*9 = 5^2.
a(328) = 1 since 328 = 0^2 + (-2)^2 + 18^2 with 0 + 2*(-2) + 3*18 = 2*5^2.
a(330) = 1 since 330 = 5^2 + 4^2 + 17^2 with 5 + 2*4 + 3*17 = 8^2.
a(466) = 1 since 466 = 21^2 + 0^2 + 5^2 with 21 + 2*0 + 3*5 = 6^2.
a(1320) = 1 since 1320 = 10^2 + 8^2 + 34^2 with 10 + 2*8 + 3*34 = 2*8^2.
a(1387) = 1 since 1387 = 33^2  + (-17)^2 + 3^2 with 33 + 2*(-17) + 3*3 = 2*2^2.
a(1857) = 1 since 1857 = (-37)^2 + (-2)^2 + 22^2 with (-37) + 2*(-2) + 3*22 = 5^2.
a(1864) = 1 since 1864 = 42^2 + 0^2 + 10^2 with 42 + 2*0 + 3*10 = 2*6^2.
a(2386) = 1 since 2386 = (-44)^2 + 3^2 + 21^2 with (-44) + 2*3 + 3*21 = 5^2.
a(5548) = 1 since 5548 = 66^2 + (-34)^2 + 6^2 with 66 + 2*(-34) + 3*6 = 4^2.
a(7428) = 1 since 7428 = (-74)^2 + (-4)^2 + 44^2 with (-74) + 2*(-4) + 3*44 = 2*5^2.
a(9544) = 1 since 9544 = (-88)^2 + 6^2 + 42^2 with (-88) + 2*6 + 3*42 = 2*5^2.
		

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.

Crossrefs

Cf. A000290, A005875, A271518, A275344, A283170, A283196, A283204, A283205, A283239, A283269.

Programs

Mathematica

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