%I #11 Feb 07 2017 09:23:48
%S 1,3,2,2,5,4,2,2,4,6,4,3,4,6,3,1,9,7,5,6,7,7,1,4,8,11,7,1,11,10,2,3,8,
%T 9,6,9,8,11,5,5,15,7,4,5,13,9,2,2,8,15,10,8,10,17,3,7,12,4,10,4,11,16,
%U 3,2,18,16,6,9,15,11,4,6,8,16,12,3,13,13,1,5
%N Number of ways to write n as x^2 + y^2 + z^2 + w^2 with x and (12*x)^2 + (5*y-10*z)^2 both squares, where x,y,z are nonnegative integers and w is a positive integer.
%C Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 16^k*m (k = 0,1,2,... and m = 1, 23, 28, 79, 119, 191, 223, 263, 463, 703, 860, 1052).
%C (ii) Any positive integer n can be written as x^2 + y^2 + z^2 + w^2 with x and (35*x)^2 + (12*y-24*z)^2 both squares, where x,y,z are nonnegative integers and w is a positive integer.
%C The author has proved that any nonnegative integer can be written as the sum of a fourth power and three squares.
%C See also A281976, A281977, A282013 and A282014 for similar conjectures.
%H Zhi-Wei Sun, <a href="/A282161/b282161.txt">Table of n, a(n) for n = 1..10000</a>
%H Zhi-Wei Sun, <a href="http://dx.doi.org/10.1016/j.jnt.2016.11.008">Refining Lagrange's four-square theorem</a>, J. Number Theory 175(2017), 167-190.
%H Zhi-Wei Sun, <a href="http://arxiv.org/abs/1701.05868">Restricted sums of four squares</a>, arXiv:1701.05868 [math.NT], 2017.
%e a(1) = 1 since 1 = 0^2 + 0^2 + 0^2 + 1^2 with 0 = 0^2 and (12*0)^2 + (5*0-10*0)^2 = 0^2.
%e a(23) = 1 since 23 = 1^2 + 3^2 + 2^2 + 3^2 with 1 = 1^2 and (12*1)^2 + (5*3-10*2)^2 = 13^2.
%e a(28) = 1 since 28 = 1^2 + 1^2 + 1^2 + 5^2 with 1 = 1^2 and (12*1)^2 + (5*1-10*1)^2 = 13^2.
%e a(79) = 1 since 79 = 1^2 + 5^2 + 2^2 + 7^2 with 1 = 1^2 and (12*1)^2 + (5*5-10*2)^2 = 13^2.
%e a(119) = 1 since 119 = 1^2 + 9^2 + 1^2 + 6^2 with 1 = 1^2 and (12*1)^2 + (5*9-10*1)^2 = 37^2.
%e a(191) = 1 since 191 = 9^2 + 5^2 + 7^2 + 6^2 with 9 = 3^2 and (12*9)^2 + (5*5-10*7)^2 = 117^2.
%e a(223) = 1 since 223 = 1^2 + 13^2 + 7^2 + 2^2 with 1 = 1^2 and (12*1)^2 + (5*13-10*7)^2 = 13^2.
%e a(263) = 1 since 263 = 9^2 + 13^2 + 2^2 + 3^2 with 9 = 3^2 and (12*9)^2 + (5*13-10*2)^2 = 117^2.
%e a(463) = 1 since 463 = 1^2 + 19^2 + 10^2 + 1^2 with 1 = 1^2 and (12*1)^2 + (5*19-10*10)^2 = 13^2.
%e a(703) = 1 since 703 = 1^2 + 13^2 + 7^2 + 22^2 with 1 = 1^2 and (12*1)^2 + (5*13-10*7)^2 = 13^2.
%e a(860) = 1 since 860 = 4^2 + 18^2 + 18^2 + 14^2 with 4 = 2^2 and (12*4)^2 + (5*18-10*18)^2 = 102^2.
%e a(1052) = 1 since 1052 = 4^2 + 30^2 + 6^2 + 10^2 with 4 = 2^2 and (12*4)^2 + (5*30-10*6)^2 = 102^2.
%t SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
%t Do[r=0;Do[If[SQ[n-x^4-y^2-z^2]&&SQ[144x^4+(5y-10z)^2],r=r+1],{x,0,(n-1)^(1/4)},{y,0,Sqrt[n-1-x^4]},{z,0,Sqrt[n-1-x^4-y^2]}];Print[n," ",r];Continue,{n,1,80}]
%Y Cf. A000118, A000290, A270969, A271714, A273108, A281939, A281941, A281975, A281976, A281977, A282013, A282014.
%K nonn
%O 1,2
%A _Zhi-Wei Sun_, Feb 07 2017