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%I #12 Mar 31 2016 13:52:47
%S 1,1,2,2,3,3,2,2,2,2,1,2,2,2,1,1,3,1,3,3,3,3,1,2,2,2,3,4,4,3,4,2,5,3,
%T 4,5,2,4,1,1,4,2,4,3,4,1,2,1,3,2,1,4,1,2,4,2,7,4,5,5,2,3,2,3,3,4,2,5,
%U 4,3,6
%N Number of ordered ways to write n as u^3 + v^3 + 2*x^3 + 2*y^3 + 3*z^3, where u, v, x, y and z are nonnegative integers with u <= v and x <= y.
%C Conjecture: (i) a(n) > 0 for all n = 0,1,2,..., and a(n) = 1 only for n = 0, 1, 10, 14, 15, 17, 22, 38, 39, 45, 47, 50, 52, 76, 102, 103, 188, 295, 366, 534.
%C (ii) Any natural number n can be written as s^4 + t^4 + 2*u^4 + 2*v^4 + 3*x^4 + 3*y^4 + 7*z^4, where s, t, u, v, x, y and z are nonnegative integers. Also, each natural number n can be written as r^5 + s^5 + t^5 + u^5 + 2*v^5 + 4*w^5 + 6*x^5 + 9*y^5 +12*z^5, where r, s, t, u, v, w, x, y and z are nonnegative integers.
%C (iii) In general, for any integer k > 2, there are 2*k-1 positive integers c(1), c(2), ..., c(2k-1) such that {c(1)*x(1)^k + c(2)*x(2)^k + ... + c(2k-1)*x(2k-1)^k: x(1),x(2),...,x(k) = 0,1,2,...} = {0,1,2,3,...} and that c(1)+c(2)+...+c(2k-1) = g(k), where g(k) = 2^k+floor((3/2)^k)-2 as given by A002804.
%C This conjecture is stronger than the classical Waring problem on sums of k-th powers. Concerning parts (i) and (ii) of the conjecture, we note that 1+1+2+2+3 = 9 = g(3), 1+1+2+2+3+3+7 = 19 = g(4) and 1+1+1+1+2+4+6+9+12 = 37 = g(5).
%C We have verified that a(n) > 0 for all n = 0..10^6, and that part (ii) of the conjecture holds for n up to 10^5. Concerning part (iii) for k = 6, we conjecture that any natural number can be written as x(1)^6+x(2)^6+x(3)^6+x(4)^6+x(5)^6+3*x(6)^6+5*x(7)^6+6*x(8)^6+10*x(9)^6+18*x(10)^6+26*x(11)^6 with x(1),x(2),...,x(11) nonnegative integers. Note that 1+1+1+1+1+3+5+6+10+18+26 = 73 = g(6). - _Zhi-Wei Sun_, Mar 31 2016
%D M. B. Nathanson, Additive Number Theory: The Classical Bases, Grad. Texts in Math., Vol 164, Springer, 1996, Chapters 2 and 3.
%H Zhi-Wei Sun, <a href="/A271099/b271099.txt">Table of n, a(n) for n = 0..10000</a>
%e a(1) = 1 since 1 = 0^3 + 1^3 + 2*0^3 + 2*0^3 + 3*0^3.
%e a(10) = 1 since 10 = 0^3 + 2^3 + 2*0^3 + 2*1^3 + 3*0^3.
%e a(14) = 1 since 14 = 1^3 + 2^3 + 2*0^3 + 2*1^3 + 3*1^3.
%e a(15) = 1 since 15 = 0^3 + 2^3 + 2*1^3 + 2*1^3 + 3*1^3.
%e a(17) = 1 since 17 = 0^3 + 1^3 + 2*0^3 + 2*2^3 + 3*0^3.
%e a(22) = 1 since 22 = 0^3 + 1^3 + 2*1^3 + 2*2^3 + 3*1^3.
%e a(38) = 1 since 38 = 2^3 + 3^3 + 2*0^3 + 2*0^3 + 3*1^3.
%e a(39) = 1 since 39 = 2^3 + 3^3 + 2*1^3 + 2*1^3 + 3*0^3.
%e a(45) = 1 since 45 = 0^3 + 3^3 + 2*1^3 + 2*2^3 + 3*0^3.
%e a(47) = 1 since 47 = 1^3 + 3^3 + 2*0^3 + 2*2^3 + 3*1^3.
%e a(50) = 1 since 50 = 0^3 + 2^3 + 2*1^3 + 2*2^3 + 3*2^3.
%e a(52) = 1 since 52 = 1^3 + 3^3 + 2*0^3 + 2*0^3 + 3*2^3.
%e a(76) = 1 since 76 = 2^3 + 4^3 + 2*1^3 +2*1^3 + 3*0^3.
%e a(102) = 1 since 102 = 0^3 + 2^3 + 2*2^3 + 2*3^3 + 3*2^3.
%e a(103) = 1 since 103 = 1^3 + 2^3 + 2*2^3 + 2*3^3 + 3*2^3.
%e a(188) = 1 since 188 = 3^3 + 4^3 + 2*0^3 + 2*2^3 + 3*3^3.
%e a(295) = 1 since 295 = 1^3 + 6^3 + 2*0^3 + 2*3^3 + 3*2^3.
%e a(366) = 1 since 366 = 2^3 + 3^3 + 2*0^3 + 2*5^3 + 3*3^3.
%e a(534) = 1 since 534 = 1^3 + 8^3 + 2*1^3 + 2*2^3 + 3*1^3.
%t CQ[n_]:=CQ[n]=IntegerQ[n^(1/3)]
%t Do[r=0;Do[If[CQ[n-3z^3-2x^3-2y^3-u^3],r=r+1],{z,0,(n/3)^(1/3)},{x,0,((n-3z^3)/4)^(1/3)},{y,x,((n-3z^3-2x^3)/2)^(1/3)},{u,0,((n-3z^3-2x^3-2y^3)/2)^(1/3)}];Print[n," ",r];Continue,{n,0,70}]
%Y Cf. A000578, A000583, A000584, A001014, A002804.
%K nonn
%O 0,3
%A _Zhi-Wei Sun_, Mar 30 2016
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