OFFSET
1,3
COMMENTS
Stated in Lee, p. 1: It is now known that when N is sufficiently large, the number of positive integers at most N that fail to be written in such a way (A022566) is slightly smaller than N^(37/42). Since any integer congruent to 4 (mod 9) is never a sum of three cubes, the number of summands here cannot in general be reduced. But of those four cubes, two of which (minicubes) need be at most N^theta, as long as theta >= 192/869. An asymptotic formula for the number of such representations is established when 1/4 < theta < 1/3. - Jonathan Vos Post, Jun 29 2010
LINKS
T. D. Noe, Table of n, a(n) for n = 1..1000
Siu-lun Alan Lee, On Waring's Problem: Two Cubes and Two Minicubes, arXiv:1006.5142 [math.NT], 2010.
G. Villemin's Almanach of Numbers, Sum of Four Cubes (0 through 100).
MATHEMATICA
Reap[For[k = 0, k <= 200, k++, If[PowersRepresentations[k, 4, 3] != {}, Print[k]; Sow[k]]]][[2, 1]] (* Jean-François Alcover, Oct 05 2018 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
STATUS
approved