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%I #13 Jul 22 2016 20:16:08
%S 1,2,1,1,2,2,1,1,3,4,2,1,2,2,2,1,3,5,2,3,4,3,1,1,5,5,4,2,3,6,3,3,3,6,
%T 3,4,6,3,3,1,6,7,3,2,3,5,1,2,3,5,6,7,7,5,4,2,5,4,2,4,6,7,4,3,6,8,5,5,
%U 7,7,1,3,6,4,5,6,6,4,3,5
%N Number of ordered ways to write n as w^3 + x^2 + y^2 + z^2 with x - w a square, where x,y,z,w are nonnegative integers with y <= z > w.
%C Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 1, 3, 4, 7, 8, 12, 16, 23, 24, 40, 47, 71, 167, 311, 599.
%C (ii) For each triple (a,b,c) = (1,1,1), (2,1,1), (2,1,2), (2,2,2), (3,1,2), any natural number can be written as x^2 + y^2 + z^2 + w^3 with x,y,z,w nonnegative integers such that a*y - b*z - c*w is a square.
%C See also A275297 and A275299 for similar conjectures.
%H Zhi-Wei Sun, <a href="/A275298/b275298.txt">Table of n, a(n) for n = 1..7000</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 = 0^3 + 0^2 + 0^2 + 1^2 with 0 - 0 = 0^2 and 0 < 1 > 0.
%e a(3) = 1 since 3 = 0^3 + 1^2 + 1^2 + 1^2 with 1 - 0 = 1^2 and 1 = 1 > 0.
%e a(4) = 1 since 4 = 0^3 + 0^2 + 0^2 + 2^2 with 0 - 0 = 0^2 and 0 < 2 > 0.
%e a(7) = 1 since 7 = 1^3 + 1^2 + 1^2 + 2^2 with 1 - 1 = 0^2 and 1 < 2 > 1.
%e a(8) = 1 since 8 = 0^3 + 0^2 + 2^2 + 2^2 with 0 - 0 = 0^2 and 2 = 2 > 0.
%e a(12) = 1 since 12 = 1^3 + 1^2 + 1^2 + 3^2 with 1 - 1 = 0^2 and 1 < 3 > 1.
%e a(16) = 1 since 16 = 0^3 + 0^2 + 0^2 + 4^2 with 0 - 0 = 0^2 and 0 < 4 > 0.
%e a(23) = 1 since 23 = 1^3 + 2^2 + 3^2 + 3^2 with 2 - 1 = 1^2 and 3 = 3 > 1.
%e a(24) = 1 since 24 = 0^3 + 4^2 + 2^2 + 2^2 with 4 - 0 = 2^2 and 2 = 2 > 0.
%e a(40) = 1 since 40 = 0^3 + 0^2 + 2^2 + 6^2 with 0 - 0 = 0^2 and 2 < 6 > 0.
%e a(47) = 1 since 47 = 1^3 + 1^2 + 3^2 + 6^2 with 1 - 1 = 0^2 and 3 < 6 > 1.
%e a(71) = 1 since 71 = 1^3 + 5^2 + 3^2 + 6^2 with 5 - 1 = 2^2 and 3 < 6 > 1.
%e a(167) = 1 since 167 = 1^3 + 2^2 + 9^2 + 9^2 with 2 - 1 = 1^2 and 9 = 9 > 1.
%e a(311) = 1 since 311 = 1^3 + 2^2 + 9^2 + 15^2 with 2 - 1 = 1^2 and 9 < 15 > 1.
%e a(599) = 1 since 599 = 5^3 + 5^2 + 7^2 + 20^2 with 5 - 5 = 0^2 and 7 < 20 > 5.
%t SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]]
%t CQ[n_]:=CQ[n]=IntegerQ[n^(1/3)]
%t Do[r=0;Do[If[CQ[n-x^2-y^2-z^2]&&SQ[x-(n-x^2-y^2-z^2)^(1/3)]&&(n-x^2-y^2-z^2)^(1/3)<z,r=r+1],{x,0,Sqrt[n]},{y,0,Sqrt[(n-x^2)/2]},{z,Max[y,Ceiling[(n-x^2-y^2)^(1/3)]],Sqrt[n-x^2-y^2]}];Print[n," ",r];Continue,{n,1,80}]
%Y Cf. A000290, A000578, A271518, A275297, A275299.
%K nonn
%O 1,2
%A _Zhi-Wei Sun_, Jul 22 2016
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