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%I #15 Jun 01 2019 23:52:55
%S 1,2,3,3,2,1,3,2,1,4,4,2,2,2,1,2,2,4,8,4,3,2,3,2,3,5,5,7,3,2,5,1,3,7,
%T 6,5,5,3,5,3,2,3,9,5,2,6,3,1,3,5,5,10,6,2,8,4,3,5,6,3,3,3,4,4,2,5,9,8,
%U 5,4,6,1,5,6,5,9,2,3,7,1,1
%N Number of ways to write n as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers such that both x and 49*x + 48*(y-z) are squares.
%C Conjecture: a(n) > 0 for all n = 0,1,2,..., and a(n) = 1 only for n = 0, 16^k*m (k = 0,1,2,... and m = 5, 8, 14, 31, 47, 71, 79, 143, 248, 463, 1039).
%C The author has proved that any nonnegative integer can be written as the sum of a fourth power and three squares.
%C We have verified a(n) > 0 for all n = 0..10^7.
%C See also A281976, A281977 and A282014 for similar conjectures.
%C Qing-Hu Hou at Tianjin University verified a(n) > 0 for n up to 10^9. - _Zhi-Wei Sun_, Jun 02 2019
%H Zhi-Wei Sun, <a href="/A282013/b282013.txt">Table of n, a(n) for n = 0..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(5) = 1 since 5 = 1^2 + 0^2 + 0^2 + 2^2 with 1 = 1^2 and 49*1 + 48*(0-0) = 7^2.
%e a(8) = 1 since 8 = 0^2 + 2^2 + 2^2 + 0^2 with 0 = 0^2 and 49*0 + 48*(2-2) = 0^2.
%e a(14) = 1 since 14 = 1^2 + 2^2 + 3^2 + 0^2 with 1 = 1^2 and 49*1 + 48*(2-3) = 1^2.
%e a(31) = 1 since 31 = 1^2 + 1^2 + 2^2 + 5^2 with 1 = 1^2 and 49*1 + 48*(1-2) = 1^2.
%e a(47) = 1 since 47 = 1^2 + 6^2 + 1^2 + 3^2 with 1 = 1^2 and 49*1 + 48*(6-1) = 17^2.
%e a(71) = 1 since 71 = 1^2 + 5^2 + 6^2 + 3^2 with 1 = 1^2 and 49*1 + 48*(5-6) = 1^2.
%e a(79) = 1 since 79 = 1^2 + 7^2 + 2^2 + 5^2 with 1 = 1^2 and 49*1 + 48*(7-2) = 17^2.
%e a(143) = 1 since 143 = 1^2 + 5^2 + 6^2 + 9^2 with 1 = 1^2 and 49*1 + 48*(5-6) = 1^2.
%e a(248) = 1 since 248 = 4^2 + 6^2 + 0^2 + 14^2 with 4 = 2^2 and 49*4 + 48*(6-0) = 22^2.
%e a(463) = 1 since 463 = 9^2 + 6^2 + 15^2 + 11^2 with 9 = 3^2 and 49*9 + 48*(6-15) = 3^2.
%e a(1039) = 1 since 1039 = 1^2 + 22^2 + 23^2 + 5^2 with 1 = 1^2 and 49*1 + 48*(22-23) = 1^2.
%t SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
%t Do[r=0;Do[If[SQ[n-x^4-y^2-z^2]&&SQ[49x^2+48(y-z)],r=r+1],{x,0,n^(1/4)},{y,0,Sqrt[n-x^4]},{z,0,Sqrt[n-x^4-y^2]}];Print[n," ",r];Continue,{n,0,80}]
%Y Cf. A000118, A000290, A270969, A271518, A281939, A281941, A281975, A281976, A281977, A282014.
%K nonn
%O 0,2
%A _Zhi-Wei Sun_, Feb 04 2017
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