A282494 - OEIS

%I #17 Feb 17 2017 07:48:19

%S 1,2,1,1,4,4,1,1,3,5,4,1,3,6,3,1,6,7,3,5,9,5,1,2,6,11,7,1,7,9,2,2,6,5,

%T 5,7,7,4,1,4,10,11,3,1,9,8,2,1,5,10,8,7,10,10,4,6,8,5,4,3,9,11,4,1,11,

%U 12,4,7,13,10,2,5,5,7,7,3,10,9,1,4

%N Number of ways to write n as x^4 + y^2 + z^2 + w^2 with y*(y+240*z) a positive square, where x,y,z,w are nonnegative integers.

%C Conjecture: a(n) > 0 for all n > 0.

%C By the linked JNT paper, any nonnegative integer can be expressed as the sum of a fourth power and three squares, and each n = 0,1,2,... can be written as x^2 + y^2 + z^2 + w^2 with x,y,z,w nonnegative integers and z*(y-2*z) = 0. Whether z = 0 or y = 2*z, the number y*(y+240*z) is definitely a square.

%C See also A282463 and A282495 for similar conjectures.

%H Zhi-Wei Sun, <a href="/A282494/b282494.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(3) = 1 since 3 = 1^4 + 1^2 + 0^2 + 1^2 with 1*(1+240*0) = 1^2.

%e a(4) = 1 since 4 = 0^4 + 2^2 + 0^2 + 0^2 with 2*(2+240*0) = 2^2.

%e a(39) = 1 since 39 = 1^4 + 2^2 + 3^2 + 5^2 with 2*(2+240*3) = 38^2.

%e a(188) = 1 since 188 = 3^4 + 5^2 + 1^2 + 9^2 with 5*(5+240*1) = 35^2.

%e a(399) = 1 since 399 = 3^4 + 10^2 + 7^2 + 13^2 with 10*(10+240*7) = 130^2.

%e a(428) = 1 since 428 = 0^4 + 10^2 + 2^2 + 18^2 with 10*(10+240*2) = 70^2.

%e a(439) = 1 since 439 = 1^4 + 10^2 + 7^2 + 17^2 with 10*(10+240*7) = 130^2.

%e a(508) = 1 since 508 = 1^4 + 5^2 + 11^2 + 19^2 with 5*(5+240*11) = 115^2.

%e a(748) = 1 since 748 = 3^4 + 1^2 + 21^2 + 15^2 with 1*(1+240*21) = 71^2.

%e a(1468) = 1 since 1468 = 2^4 + 10^2 + 26^2 + 26^2 with 10*(10+240*26) = 250^2.

%e a(2828) = 1 since 2828 = 3^4 + 5^2 + 11^2 + 51^2 with 5*(5+240*11) = 115^2.

%t SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];

%t Do[r=0;Do[If[SQ[n-x^4-y^2-z^2]&&SQ[y*(y+240*z)],r=r+1],{x,0,(n-1)^(1/4)},{y,1,Sqrt[n-x^4]},{z,0,Sqrt[n-x^4-y^2]}];Print[n," ",r];Continue,{n,1,80}]

%Y Cf. A000118, A000290, A000583, A270969, A271518, A281976, A281977, A282013, A282014, A282463, A282495.

%K nonn

%O 1,2

%A _Zhi-Wei Sun_, Feb 16 2017