A343326 Number of ways to write n as the integral part of (a^3+b^3)/2 + (c^3+d^3)/6, where a,b,c,d are nonnegative integers with a >= max{b,1} and c >= max{d,1}.

2, 3, 3, 2, 4, 7, 4, 1, 4, 6, 3, 4, 3, 6, 5, 6, 5, 3, 7, 5, 2, 4, 6, 4, 5, 7, 5, 2, 6, 7, 1, 2, 8, 4, 6, 5, 9, 10, 7, 4, 6, 7, 6, 2, 5, 8, 4, 6, 5, 5, 6, 4, 2, 7, 7, 2, 3, 9, 5, 3, 4, 6, 5, 7, 9, 7, 8, 8, 12, 5, 5, 6, 9, 10, 7, 5, 7, 7, 5, 4, 3, 6, 4, 5, 6, 8, 9, 7, 5, 10, 5, 5, 3, 7, 10, 3, 3, 8, 5, 10, 9

Offset

0,1

Comments

Conjecture: a(n) > 0 for any nonnegative integer n.

This has been verified for all n = 0..10^5.

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.

Example

a(0) = 2 with 0 = floor((1^3+0^3)/2 + (1^3+0^3)/6) = floor((1^3+0^3)/2 + (1^3+1^3)/6).
a(7) = 1 with 7 = floor((3^3+1^3)/2 + (2^3+2^3)/6).
a(30) = 1 with 30 = floor((2^3+2^3)/2 + (5^3+2^3)/6).
a(111) = 1 with 111 = floor((6^3+1^3)/2 + (2^3+2^3)/6).
a(163) = 1 with 163 = floor((6^3+3^3)/2 + (5^3+5^3)/6).
a(219) = 1 with 219 = floor((4^3+0^3)/2 + (10^3+5^3)/6).

Mathematica

CQ[n_]:=CQ[n]=IntegerQ[n^(1/3)]
tab={}; Do[r=0; Do[If[CQ[6n+s-3(x^3+y^3)-z^3], r=r+1], {s, Boole[n==0], 5}, {x, 1, ((6n+s-1)/3)^(1/3)}, {y, 0, Min[x, ((6n+s-1)/3-x^3)^(1/3)]}, {z, 0, ((6n+s-3(x^3+y^3))/2)^(1/3)}]; tab=Append[tab, r], {n, 0, 100}]; Print[tab]

Crossrefs

Cf. A000578, A004999, A343368.

Sequence in context: A182865 A131469 A309076 * A073078 A360296 A034799

Adjacent sequences: A343323 A343324 A343325 * A343327 A343328 A343329

Keyword

nonn

Author

Zhi-Wei Sun, Apr 11 2021

Status

approved