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%I #18 May 21 2022 14:51:04
%S 0,1,1,1,2,1,2,1,4,1,4,3,2,3,4,5,2,7,2,2,2,5,4,5,5,3,3,3,3,7,6,3,5,5,
%T 6,2,11,3,6,2,6,6,8,10,2,9,2,5,5,10,5,2,6,4,4,7,5,7,2,2,4,6,7,3,12,3,
%U 7,4,9,6,5,10,4,15,4,8,5,11,4,8,14,6,4,6,10,7,8,9,5,6,4,4,13,5,7,3,10,2,7,11
%N Number of ways to write n as x^3 + [y^3/2] + [z^3/3] + 2^w, where [.] is the floor function, x,y,z are positive integers, and w is a nonnegative integer.
%C Conjecture: a(n) > 0 for all n > 1.
%C We have verified a(n) > 0 for all 1 < n <= 3*10^5.
%C Conjecture verified up to 10^10. - _Giovanni Resta_, Apr 14 2021
%H Zhi-Wei Sun, <a href="/A343411/b343411.txt">Table of n, a(n) for n = 1..10000</a>
%H Zhi-Wei Sun, <a href="http://arxiv.org/abs/1504.01608">Natural numbers represented by [x^2/a] + [y^2/b] + [z^2/c]</a>, arXiv:1504.01608 [math.NT], 2015.
%e a(2) = 1 with 2 = 1^3 + [1^3/2] + [1^3/3] + 2^0.
%e a(3) = 1 with 3 = 1^3 + [1^3/2] + [1^3/3] + 2^1.
%e a(4) = 1 with 4 = 1^3 + [1^3/2] + [2^3/3] + 2^0.
%e a(6) = 1 with 6 = 1^3 + [2^3/2] + [1^3/3] + 2^0.
%e a(8) = 1 with 8 = 1^3 + [2^3/2] + [2^3/3] + 2^0.
%e a(10) = 1 with 10 = 2^3 + [1^3/2] + [1^3/3] + 2^1.
%e a(103) = 1 with 103 = 3^3 + [1^3/2] + [6^3/3] + 2^2.
%t CQ[n_]:=CQ[n]=n>0&&IntegerQ[n^(1/3)];
%t tab={};Do[r=0;Do[If[CQ[n-Floor[x^3/2]-Floor[y^3/3]-2^z],r=r+1],{x,1,(2n-1)^(1/3)},{y,1,(3(n-Floor[x^3/2])-1)^(1/3)},{z,0,Log[2,n-Floor[x^3/2]-Floor[y^3/3]]}];tab=Append[tab,r],{n,1,100}];Print[tab]
%Y Cf. A000079, A000578, A343326, A343368, A343384, A343387, A343391, A343397.
%K nonn
%O 1,5
%A _Zhi-Wei Sun_, Apr 14 2021
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