%I #13 Dec 09 2021 08:16:15
%S 1,3,4,4,4,4,4,4,4,5,5,5,5,2,2,2,4,8,7,7,6,5,6,6,6,8,7,8,6,1,4,2,6,8,
%T 6,7,5,7,6,6,6,7,7,8,7,3,5,3,4,6,6,6,7,5,3,5,4,9,8,9,8,2,4,1,2,9,8,10,
%U 8,4,6,4,9,6,6,6,4,2,2,1,2,10,10,13,8,9,7,9,9,7,10,6,10,4,3,4,3,11,10,9
%N Number of ways to write n as x^4 + y^2 + (z^2 + 2*4^w)/3, where x, y, z are nonnegative integers, and w is 0 or 1.
%C Conjecture 1: a(n) > 0 for all n > 0, and a(n) = 1 only for n = 1, 30, 64, 80, 302, 350, 472, 480, 847, 3497, 13582, 25630, 38064.
%C This has been verified for n up to 10^6.
%C Conjecture 2: If (a,b,c,m) is one of the ordered tuples (1,1,11,12), (1,1,11,60), (1,1,14,15), (1,1,23,24), (1,1,23,32), (1,1,23,48), (1,2,23,96), (2,1,11,60), (2,1,23,24), (2,1,23,48), (4,1,23,48), then each n = 1 2,3,... can be written as a*x^4 + b*y^2 + (z^2 + c*4^w)/m, where x,y,z are nonnegative integers, and w is 0 or 1.
%C We have verified Conjecture 2 for n up to 2*10^5.
%H Zhi-Wei Sun, <a href="/A349992/b349992.txt">Table of n, a(n) for n = 1..10000</a>
%H Zhi-Wei Sun, <a href="http://hitpress.hit.edu.cn/2021/1015/c12593a261001/page.htm">New Conjectures in Number Theory and Combinatorics</a> (in Chinese), Harbin Institute of Technology Press, 2021.
%e a(30) = 1 with 30 = 1^4 + 5^2 + (2^2 + 2*4)/3.
%e a(480) = 1 with 480 = 1^4 + 14^2 + (29^2 + 2*4)/3.
%e a(847) = 1 with 847 = 0^4 + 29^2 + (4^2 + 2*4^0)/3.
%e a(3497) = 1 with 3497 = 4^4 + 48^2 + (53^2 + 2*4^0)/3.
%e a(13582) = 1 with 13582 = 9^4 + 28^2 + (53^2 + 2*4^0)/3.
%e a(25630) = 1 with 25630 = 5^4 + 158^2 + (11^2 + 2*4^0)/3.
%e a(38064) = 1 with 38064 = 3^4 + 157^2 + (200^2 + 2*4^0)/3.
%t SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
%t tab={};Do[r=0;Do[If[SQ[3(n-x^4-y^2)-2*4^z],r=r+1],{x,0,(n-1)^(1/4)},{y,0,Sqrt[n-1-x^4]},{z,0,1}];tab=Append[tab,r],{n,1,100}];Print[tab]
%Y Cf. A000290, A000583, A349942, A349943, A349957.
%K nonn
%O 1,2
%A _Zhi-Wei Sun_, Dec 08 2021