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A349943 Number of ways to write n as a^4 + (b^4 + c^2 + d^2)/9, where a,b,c,d are nonnegative integers with c <= d.

%I A349943 #34 Jun 13 2022 15:14:49
%S A349943 1,3,5,4,3,4,3,1,1,6,9,6,2,4,7,3,3,7,9,7,7,5,4,2,3,10,11,8,2,10,10,1,
%T A349943 5,9,15,14,6,5,5,1,4,9,12,8,2,11,7,1,4,11,21,8,6,9,8,3,3,7,9,9,4,11,9,
%U A349943 2,3,13,14,7,7,10,10,4,3,10,18,16,3,10,7,1,4,10,15,12,11,12,11,3,3,16,29,17,5,6,14,10,3,10,18,15,14
%N A349943 Number of ways to write n as a^4 + (b^4 + c^2 + d^2)/9, where a,b,c,d are nonnegative integers with c <= d.
%C A349943 Conjecture 1: a(n) > 0 for all n >= 0, and a(n) = 1 only for n = 0, 16^k*m (k = 0,1,2,... and m = 7, 8, 31, 39, 47, 79, 519).
%C A349943 This implies that each nonnegative rational number can be written as x^4 + 9*y^4 + z^2 + w^2 with x,y,z,w rational numbers.
%C A349943 Conjecture 2: Each n = 0,1,2,... can be written as a^4 + (4*b^4 + c^2 + d^2)/81 with a,b,c,d nonnegative integers.
%C A349943 This implies that each nonnegative rational number can be written as x^4 + 4*y^4 + z^2 + w^2 with x,y,z,w rational numbers.
%C A349943 We have verified Conjectures 1 and 2 for n <= 10^5.
%C A349943 It seems that each n = 0,1,2,... can be written as a^4 + (b^4 + c^2 + d^2)/m^2 with a,b,c,d nonnegative integers, provided that m is among the odd numbers 7, 11, 15, 17, 19, 21, ....
%C A349943 See also A349942 for a similar conjecture.
%H A349943 Zhi-Wei Sun, <a href="/A349943/b349943.txt">Table of n, a(n) for n = 0..10000</a>
%H A349943 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 A349943 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 A349943 a(7) = 1 with 7 = 1^4 + (1^4 + 2^2 + 7^2)/9.
%e A349943 a(8) = 1 with 8 = 0^4 + (0^4 + 6^2 + 6^2)/9.
%e A349943 a(31) = 1 with 31 = 1^4 + (1^4 + 10^2 + 13^2)/9.
%e A349943 a(39) = 1 with 39 = 1^4 + (3^4 + 6^2 + 15^2)/9.
%e A349943 a(47) = 1 with 47 = 1^4 + (3^4 + 3^2 + 18^2)/9.
%e A349943 a(79) = 1 with 79 = 1^4 + (1^4 + 5^2 + 26^2)/9.
%e A349943 a(519) = 1 with 519 = 1^4 + (3^4 + 15^2 + 66^2)/9.
%t A349943 SQ[n_]:=SQ[n]=IntegerQ[Sqrt[n]];
%t A349943 tab={};Do[r=0;Do[If[SQ[9(n-x^4)-y^4-z^2],r=r+1],{x,0,n^(1/4)},{y,0,(9(n-x^4))^(1/4)},{z,0,Sqrt[(9(n-x^4)-y^4)/2]}];tab=Append[tab,r],{n,0,100}];Print[tab]
%Y A349943 Cf. A000290, A000583, A349942.
%K A349943 nonn
%O A349943 0,2
%A A349943 _Zhi-Wei Sun_, Dec 05 2021