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%I #26 Jul 09 2023 08:28:19
%S 1,1,1,2,2,3,3,3,4,3,4,3,3,3,2,3,2,3,1,2,3,2,2,1,4,3,2,3,3,5,3,4,6,4,
%T 5,4,6,4,4,3,5,5,3,6,3,6,4,4,6,3,5,4,4,4,3,4,5,7,4,6,4,5,6,4,10,2,6,8,
%U 3,7,4,8,6,5,5,4,5,2,6,1,5,3,3,8,5,7,6,6,9,6,7,6,6,5,5,6,4,6,6,8,1
%N Number of ordered ways to write n as u^3 + 2*v^3 + 3*x^3 + 4*y^3 + 5*z^3, where u, v, x, y and z are nonnegative integers.
%C Conjecture: We have {u^3+a*v^3+b*x^3+c*y^3+d*z^3: u,v,x,y,z = 0,1,2,...} = {0,1,2,...} whenever (a,b,c,d) is among the following 32 quadruples: (1,2,2,3), (1,2,2,4), (1,2,3,4), (1,2,4,5), (1,2,4,6), (1,2,4,9), (1,2,4,10), (1,2,4,11), (1,2,4,18), (1,3,4,6), (1,3,4,9), (1,3,4,10), (2,2,4,5), (2,2,6,9), (2,3,4,5), (2,3,4,6), (2,3,4,7), (2,3,4,8), (2,3,4,9), (2,3,4,10), (2,3,4,12), (2,3,4,15), (2,3,4,18), (2,3,5,6), (2,3,6,12), (2,3,6,15), (2,4,5,6), (2,4,5,8), (2,4,5,9), (2,4,5,10), (2,4,6,7), (2,4,7,10).
%C In particular, this implies that a(n) > 0 for all n = 0,1,2,... We guess that a(n) = 1 only for n = 0, 1, 2, 18, 23, 79, 100.
%C If {m*u^3+a*v^3+b*x^3+c*y^3+d*z^3: u,v,x,y,z = 0,1,2,...} = {0,1,2,...} with 1 <= m <= a <= b <= c <= d, then m = 1, and we can show that (a,b,c,d) must be among the 32 quadruples listed in the conjecture (cf. Theorem 1.2 of the linked 2017 paper).
%C Conjecture verified for all the 32 quadruples up to 10^11. - _Mauro Fiorentini_, Jul 09 2023
%C It is known that there are exactly 54 quadruples (a,b,c,d) with 1 <= a <= b <= c <= d such that {a*w^2+b*x^2+c*y^2+d*z^2: w,x,y,z = 0,1,2,...} = {0,1,2,...}.
%C See also A271099 and A271169 for conjectures refining Waring's problem.
%C We also conjecture that if P(u,v,x,y,z) is one of the four polynomials u^6+v^3+2*x^3+4*y^3+5*z^3 and a*u^6+v^3+2*x^3+3*y^3+4*z^3 (a = 5,8,12) then any natural number can be written as P(u,v,x,y,z) with u,v,x,y,z nonnegative integers. - _Zhi-Wei Sun_, Apr 06 2016
%C Conjecture verified for all the 4 polynomials up to 10^11. - _Mauro Fiorentini_, Jul 09 2023
%D S. Ramanujan, On the expression of a number in the form a*x^2 + b*y^2 + c*z^2 + d*w^2, Proc. Cambridge Philos. Soc. 19(1917), 11-21.
%H Zhi-Wei Sun, <a href="/A271237/b271237.txt">Table of n, a(n) for n = 0..10000</a>
%H L. E. Dickson, <a href="http://www.jstor.org/stable/2370770">Quaternary quadratic forms representing all integers</a>, Amer. J. Math. 49(1927), 39-56.
%H Zhi-Wei Sun, <a href="http://dx.doi.org/10.1016/j.jnt.2015.10.014">A result similar to Lagrange's theorem</a>, J. Number Theory 162(2016), 190-211.
%H Zhi-Wei Sun, <a href="http://maths.nju.edu.cn/~zwsun/179b.pdf">New conjectures on representations of integers (I)</a>, Nanjing Univ. J. Math. Biquarterly 34(2017), no. 2, 97-120.
%e a(2) = 1 since 2 = 0^3 + 2*1^3 + 3*0^3 + 4*0^3 + 5*0^3.
%e a(18) = 1 since 18 = 2^3 + 2*1^3 + 3*1^3 + 4*0^3 + 5*1^3.
%e a(23) = 1 since 23 = 0^3 + 2*2^3 + 3*1^3 + 4*1^3 + 5*0^3.
%e a(79) = 1 since 79 = 1^3 + 2*3^3 + 3*2^3 + 4*0^3 + 5*0^3.
%e a(100) = 1 since 100 = 2^3 + 2*1^3 + 3*3^3 + 4*1^3 + 5*1^3.
%t CQ[n_]:=CQ[n]=IntegerQ[n^(1/3)]
%t Do[r=0;Do[If[CQ[n-5z^3-4y^3-3x^3-2v^3],r=r+1],{z,0,(n/5)^(1/3)},{y,0,((n-5z^3)/4)^(1/3)},{x,0,((n-5z^3-4y^3)/3)^(1/3)},{v,0,((n-5z^3-4y^3-3x^3)/2)^(1/3)}];Print[n," ",r];Continue,{n,0,100}]
%Y Cf. A000578, A002804, A001014, A271099, A271169.
%K nonn
%O 0,4
%A _Zhi-Wei Sun_, Apr 02 2016
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