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.

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, 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, 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

Offset: 0

Author

Zhi-Wei Sun, Apr 12 2021

Keywords

• nonn

Comments

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

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.

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

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).

Examples

			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).
a(17) = 1 with 17 = floor((2^3+1^3)/3) + floor((4^3+2^3)/5).
a(20) = 1 with 20 = floor((2^3+0^3)/3) + floor((4^3+3^3)/5).
a(38) = 1 with 38 = floor((4^3+2^3)/3) + floor((4^3+2^3)/5).
a(103) = 1 with 103 = floor((6^3+4^3)/3) + floor((3^3+3^3)/5).
a(304) = 1 with 304 = floor((2^3+0^3)/3) + floor((10^3+8^3)/5).
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 0..10000
• Zhi-Wei Sun, Natural numbers represented by floor(x^2/a) + floor(y^2/b) + floor(z^2/c), arXiv:1504.01608 [math.NT], 2015.

Crossrefs

Cf. A000578, A343326.

Programs

Mathematica

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