|
%I #13 Jun 01 2019 23:54:05
%S 1,2,3,3,2,2,4,2,1,5,4,3,3,2,2,3,2,4,8,4,3,3,4,2,2,6,4,7,3,1,6,1,3,7,
%T 6,5,5,3,5,4,1,4,8,5,3,4,4,2,3,5,4,9,5,3,9,4,2,7,6,2,5,2,4,4,2,5,8,8,
%U 4,4,7,2,3,6,5,9,3,2,8,2,2
%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 121*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 = 8, 29, 31, 40, 94, 104, 143, 319, 671).
%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 A282013 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="/A282014/b282014.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(8) = 1 since 8 = 0^2 + 2^2 + 2^2 + 0^2 with 0 = 0^2 and 121*0 + 48*(2-2) = 0^2.
%e a(29) = 1 since 29 = 0^2 + 5^2 + 2^2 + 0^2 with 0 = 0^2 and 121*0 + 48*(5-2) = 12^2.
%e a(31) = 1 since 31 = 1^2 + 2^2 + 1^2 + 5^2 with 1 = 1^2 and 121*1 + 48*(2-1) = 13^2.
%e a(40) = 1 since 40 = 4^2 + 2^2 + 2^2 + 4^2 with 4 = 2^2 and 121*4 + 48*(2-2) = 22^2.
%e a(94) = 1 since 94 = 0^2 + 6^2 + 3^2 + 7^2 with 0 = 0^2 and 121*0 + 48*(6-3) = 12^2.
%e a(104) = 1 since 104 = 4^2 + 6^2 + 6^2 + 4^2 with 4 = 2^2 and 121*4 + 48*(6-6) = 22^2.
%e a(143) = 1 since 143 = 1^2 + 6^2 + 5^2 + 9^2 with 1 = 1^2 and 121*1 + 48*(6-5) = 13^2.
%e a(319) = 1 since 319 = 1^2 + 17^2 + 2^2 + 5^2 with 1 = 1^2 and 121*1 + 48*(17-2) = 29^2.
%e a(671) = 1 since 671 = 9^2 + 5^2 + 23^2 + 6^2 with 9 = 3^2 and 121*9 + 48*(5-23) = 15^2.
%t SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
%t Do[r=0;Do[If[SQ[n-x^4-y^2-z^2]&&SQ[121x^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, A282013.
%K nonn
%O 0,2
%A _Zhi-Wei Sun_, Feb 04 2017
|
