A272351 - OEIS

%I #24 Aug 03 2020 04:27:57

%S 1,1,1,1,2,2,1,1,3,3,2,1,2,3,2,1,4,5,2,2,3,3,2,2,3,5,4,1,5,6,1,1,5,4,

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

%U 6,5,1,5,5,10,6,2,6,8,1,2

%N Number of ordered ways to write n as w^2 + x^2 + y^2 + z^2 with x^4 + 8*y*z*(y^2+z^2) a fourth power, where w,x,y,z are nonnegative integers with w >= x and y > z.

%C Conjecture: (i) a(n) > 0 for all n > 0, and a(n) = 1 only for n = 2^k, 4^k*m (k = 0,1,2,... and m = 3, 7, 31, 55, 71, 79, 151, 191).

%C (ii) For (b,c) = (8,8), (16,64), any positive integer can be written as w^2 + x^2 + y^2 + z^2 with x^4 + b*y^3*z + c*y*z^3 a fourth power, where w is a positive integer and x,y,z are nonnegative integers.

%C (iii) For each triple (a,b,c) = (1,20,60), (1,24,56), (9,20,60), (9,32,96), any positive integer can be written as w^2 + x^2 + y^2 + z^2 with a*x^4 + b*y^3*z + c*y*z^3 a square, where w is a positive integer and x,y,z are nonnegative integers.

%C The author has proved part (ii) of the conjecture in arXiv:1604.06723. - _Zhi-Wei Sun_, May 09 2016

%H Zhi-Wei Sun, <a href="/A272351/b272351.txt">Table of n, a(n) for n = 1..10000</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.

%H Zhi-Wei Sun, <a href="http://listserv.nodak.edu/cgi-bin/wa.exe?A2=NMBRTHRY;852b9c4a.1604">Refine Lagrange's four-square theorem</a>, a message to Number Theory List, April 26, 2016.

%e a(1) = 1 since 1 = 0^2 + 0^2 + 1^2 + 0^2 with 0 = 0, 1 > 0 and 0^4 + 8*1*0*(1^2+0^2) = 0^4.

%e a(2) = 1 since 2 = 1^2 + 0^2 + 1^2 + 0^2 with 1 > 0 and 0^4 + 8*1*0*(1^2+0^2) = 0^4.

%e a(3) = 1 since 3 = 1^2 + 1^2 + 1^2 + 0^2 with 1 = 1, 1 > 0 and 1^4 + 8*1*0*(1^2+0^2) = 1^4.

%e a(7) = 1 since 7 = 1^2 + 1^2 + 2^2 + 1^2 with 1 = 1, 3 > 1 and 1^4 + 8*2*1*(2^2+1^2) = 3^4.

%e a(31) = 1 since 31 = 5^2 + 1^2 + 2^2 + 1^2 with 5 > 1, 2 > 1 and 1^4 + 8*2*1*(2^2+1^2) = 3^4.

%e a(55) = 1 since 55 = 7^2 + 1^2 + 2^2 + 1^2 with 7 > 1, 2 > 1 and 1^4 + 8*2*1*(2^2+1^2) = 3^4.

%e a(71) = 1 since 71 = 3^2 + 1^2 + 6^2 + 5^2 with 3 > 1, 6 > 5 and 1^4 + 8*6*5*(6^2+5^2) = 11^4.

%e a(79) = 1 since 79 = 5^2 + 3^2 + 6^2 + 3^2 with 5 > 3, 6 > 3 and 3^4 + 8*6*3*(6^2+3^2) = 9^4.

%e a(151) = 1 since 151 = 5^2 + 3^2 + 9^2 + 6^2 with 5 > 3, 9 > 6 and 3^4 + 8*9*6*(9^2+6^2) = 15^4.

%e a(191) = 1 since 191 = 3^2 + 1^2 + 10^2 + 9^2 with 3 > 1, 10 > 9 and 1^4 + 8*10*9*(10^2+9^2) = 19^4.

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

%t QQ[n_]:=QQ[n]=IntegerQ[n^(1/4)]

%t Do[r=0;Do[If[SQ[n-x^2-y^2-z^2]&&QQ[x^4+8y*z*(y^2+z^2)],r=r+1],{z,0,(Sqrt[2n-1]-1)/2},{y,z+1,Sqrt[n-z^2]},{x,0,Sqrt[(n-y^2-z^2)/2]}];Print[n," ",r];Continue,{n,1,80}]

%Y Cf. A000118, A000290, A000583, A262357, A268507, A269400, A271510, A271513, A271518, A271608, A271665, A271714, A271721, A271724, A271775, A271778, A271824, A272084, A272332.

%K nonn

%O 1,5

%A _Zhi-Wei Sun_, Apr 27 2016