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%I #7 Dec 14 2016 09:24:48
%S 1,1,1,1,1,2,2,1,1,2,3,1,1,2,2,2,1,3,5,2,2,5,4,1,1,2,4,1,0,7,3,1,1,1,
%T 4,6,4,5,4,4,3,3,5,2,3,5,4,2,1,4,8,5,1,5,12,1,1,2,6,3,4,3,3,9,1,6,4,2,
%U 3,8,8,2,3,7,7,7,3,8,14,3,2
%N Number of ways to write n as w^2 + x^2 + y^2 + z^2 with w + 2*x + 3*y + 5*z a square, where w,x,y,z are nonnegative integers.
%C Conjecture: (i) a(n) = 0 if and only if n = 16^k*28 for some k = 0,1,2,....
%C (ii) For any positive integers a,b,c,d, there are infinitely many positive integers which cannot be written as w^2 + x^2 + y^2 + z^2 with a*w + b*x + c*y + d*z a square, where w,x,y,z are nonnegative integers.
%H Zhi-Wei Sun, <a href="/A279522/b279522.txt">Table of n, a(n) for n = 0..5000</a>
%H Zhi-Wei Sun, <a href="http://arxiv.org/abs/1604.06723">Refining Lagrange's four-square theorem</a>, arXiv:1604.06723 [math.NT], 2016.
%e a(27) = 1 since 27 = 1^2 + 5^2 + 0^2 + 1^2 with 1 + 2*5 + 3*0 + 5*1 = 4^2.
%e a(31) = 1 since 31 = 1^2 + 2^2 + 5^2 + 1^2 with 1 + 2*2 + 3*5 + 5*1 = 5^2.
%e a(33) = 1 since 33 = 0^2 + 4^2 + 4^2 + 1^2 with 0 + 2*4 + 3*4 + 5*1 = 5^2.
%e a(52) = 1 since 52 = 4^2 + 6^2 + 0^2 + 0^2 with 4 + 2*6 + 3*0 + 5*0 = 4^2.
%e a(55) = 1 since 55 = 1^2 + 5^2 + 5^2 + 2^2 with 1 + 2*5 + 3*5 + 5*2 = 6^2.
%e a(56) = 1 since 56 = 0^2 + 4^2 + 6^2 + 2^2 with 0 + 2*4 + 3*6 + 5*2 = 6^2.
%e a(88) = 1 since 88 = 4^2 + 8^2 + 2^2 + 2^2 with 4 + 2*8 + 3*2 + 5*2 = 6^2.
%e a(137) = 1 since 137 = 10^2 + 6^2 + 1^2 + 0^2 with 10 + 2*6 + 3*1 + 5*0 = 5^2.
%e a(164) = 1 since 164 = 12^2 + 2^2 + 0^2 + 4^2 with 12 + 2*2 + 3*0 + 5*4 = 6^2.
%t SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
%t Do[r=0;Do[If[SQ[n-x^2-y^2-z^2]&&SQ[Sqrt[n-x^2-y^2-z^2]+2x+3y+5z],r=r+1],{x,0,Sqrt[n]},{y,0,Sqrt[n-x^2]},{z,0,Sqrt[n-x^2-y^2]}];Print[n," ",r];Continue,{n,0,80}]
%Y Cf. A000118, A000290, A270969, A271518, A273404.
%K nonn
%O 0,6
%A _Zhi-Wei Sun_, Dec 14 2016
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