%I #24 Dec 06 2016 03:18:45

%S 1,3,3,2,5,5,2,2,4,9,5,3,7,4,3,1,7,13,6,7,9,4,2,4,10,13,10,4,9,6,3,3,

%T 9,15,7,10,8,6,5,6,14,14,7,3,14,7,2,3,5,14,12,11,12,9,5,5,9,12,6,6,10,

%U 5,4,2,11,20,10,10,12,4,2,6,13,14,10,4,7,5,1,5

%N Number of ways to write n as w^2 + x^2 + y^2 + z^2 with w a positive integer and x,y,z nonnegative integers such that x^3 + 4*y*z*(y-z) is a square.

%C Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 16^k*q (k = 0,1,2,... and q = 1, 79, 143, 184, 575).

%C (ii) Any positive integer n can be written as w^2 + x^2 + y^2 + z^2 with w a positive integer and x,y,z nonnegative integers such that x^3 + 8*y*z*(2y-z) is a square.

%C We have verified a(n) > 0 and part (ii) of the conjecture for n up to 3*10^5.

%C For more conjectural refinements of Lagrange's four-square theorem, see Section 4 of arXiv:1604.06723.

%H Zhi-Wei Sun, <a href="/A279056/b279056.txt">Table of n, a(n) for n = 1..10000</a>

%H Zhi-Wei Sun, <a href="http://arxiv.org/abs/1604.06723">Refining Lagrange's four-square theorem</a>, arXiv:1604.06723 [math.GM], 2016.

%e a(1) = 1 since 1 = 1^2 + 0^2 + 0^2 + 0^2 with 0^3 + 4*0*0*(0-0) = 0^2.

%e a(79) = 1 since 79 = 7^2 + 1^2 + 5^2 + 2^2 with 1^3 + 4*5*2*(5-2) = 11^2.

%e a(143) = 1 since 143 = 9^2 + 1^2 + 6^2 + 5^2 with 1^3 + 4*6*5*(6-5) = 11^2.

%e a(184) = 1 since 184 = 10^2 + 8^2 + 4^2 + 2^2 with 8^3 + 4*4*2*(4-2) = 24^2.

%e a(575) = 1 since 575 = 1^2 + 22^2 + 3^2 + 9^2 with 22^3 + 4*3*9*(3-9) = 100^2.

%t SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];

%t table={};Do[r=0;Do[If[SQ[n-x^2-y^2-z^2]&&SQ[x^3+4y*z(y-z)],r=r+1],{x,0,Sqrt[n-1]},{y,0,Sqrt[n-1-x^2]},{z,0,Sqrt[n-1-x^2-y^2]}];table=Append[table,r];Continue,{n,1,80}]

%Y Cf. A000118, A000290, A271518, A272332, A272336, A272351, A272888.

%K nonn

%O 1,2

%A _Zhi-Wei Sun_, Dec 05 2016