OFFSET
1,1
COMMENTS
It appears that this sequence has 11062 terms, the last of which is 1685758. This means that all numbers greater than 1685758 can be written as the sum of five positive cubes in at least two ways. - T. D. Noe, Dec 13 2006
LINKS
T. D. Noe, Table of n, a(n) for n=1..11062
Eric Weisstein's World of Mathematics, Cubic Number.
MATHEMATICA
Select[ Range[200], Count[ PowersRepresentations[#, 5, 3], r_ /; FreeQ[r, 0]] == 1 &] (* Jean-François Alcover, Oct 23 2012 *)
PROG
(Python)
from collections import Counter
from itertools import combinations_with_replacement as combs_with_rep
def aupto(lim):
s = filter(lambda x: x<=lim, (i**3 for i in range(1, int(lim**(1/3))+2)))
s2 = filter(lambda x: x<=lim, (sum(c) for c in combs_with_rep(s, 5)))
s2counts = Counter(s2)
return sorted(k for k in s2counts if s2counts[k] == 1)
print(aupto(196)) # Michael S. Branicky, May 12 2021
CROSSREFS
KEYWORD
nonn,fini
AUTHOR
STATUS
approved