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A343368 Number of ways to write n as floor((a^3+b^3)/3) + floor((c^3+d^3)/5), where a,b,c,d are nonnegative integers with a > b and c >= d. 7

%I

%S 3,2,3,6,2,3,1,3,2,7,6,3,6,2,7,2,6,1,2,2,1,10,6,3,6,6,5,6,6,4,4,5,1,4,

%T 9,6,4,4,1,5,2,4,7,5,6,5,13,6,4,6,6,7,6,5,6,8,4,4,4,5,3,2,2,4,7,4,4,8,

%U 8,5,6,6,9,8,7,8,3,15,2,10,3,8,4,3,7,6,8,4,7,9,5,4,7,8,6,6,2,8,10,4,6

%N Number of ways to write n as floor((a^3+b^3)/3) + floor((c^3+d^3)/5), where a,b,c,d are nonnegative integers with a > b and c >= d.

%C Conjecture 1: a(n) > 0 for all n >= 0.

%C Conjecture 2: Each n = 0,1,... can be written as floor((a^3+b^3)/4) + floor((c^3+d^3)/5) with a,b,c,d nonnegative integers.

%C Both conjectures have been verified for all n = 0..10^5.

%C We also conjecture that the pair (4,5) of denominators in Conjecture 2 can be replaced by some other pairs such as (4,6), (5,6), (3,7), (4,7), (5,7), (6,7).

%H Zhi-Wei Sun, <a href="/A343368/b343368.txt">Table of n, a(n) for n = 0..10000</a>

%H Zhi-Wei Sun, <a href="http://arxiv.org/abs/1504.01608">Natural numbers represented by floor(x^2/a) + floor(y^2/b) + floor(z^2/c)</a>, arXiv:1504.01608 [math.NT], 2015.

%e a(1) = 2 with 1 = floor((1^3+0^3)/3) + floor((2^3+0^3)/5) = floor((1^3+0^3)/3) + floor((2^3+1^3)/5).

%e a(17) = 1 with 17 = floor((2^3+1^3)/3) + floor((4^3+2^3)/5).

%e a(20) = 1 with 20 = floor((2^3+0^3)/3) + floor((4^3+3^3)/5).

%e a(38) = 1 with 38 = floor((4^3+2^3)/3) + floor((4^3+2^3)/5).

%e a(103) = 1 with 103 = floor((6^3+4^3)/3) + floor((3^3+3^3)/5).

%e a(304) = 1 with 304 = floor((2^3+0^3)/3) + floor((10^3+8^3)/5).

%t CQ[n_]:=CQ[n]=IntegerQ[n^(1/3)];

%t tab={};Do[r=0;Do[If[CQ[5(n-Floor[(x^3+y^3)/3])+s-z^3],r=r+1],{s,0,4},{x,1,(3n+2)^(1/3)},{y,0,Min[x-1,(3n+2-x^3)^(1/3)]},{z,0,((5(n-Floor[(x^3+y^3)/3])+s)/2)^(1/3)}];tab=Append[tab,r],{n,0,100}];Print[tab]

%Y Cf. A000578, A343326.

%K nonn

%O 0,1

%A _Zhi-Wei Sun_, Apr 12 2021

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Last modified November 27 03:56 EST 2021. Contains 349345 sequences. (Running on oeis4.)