OFFSET
1,1
COMMENTS
It appears that this sequence has 841 terms, the last of which is 19417. This means that all numbers greater than 19417 can be written as the sum of six 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..841
Eric Weisstein's World of Mathematics, Cubic Number.
MATHEMATICA
Select[ Range[200], Length[ Select[ PowersRepresentations[#, 6, 3], And @@ (Positive /@ #) &]] == 1 &] (* Jean-François Alcover, Oct 25 2012 *)
PROG
(Python)
from collections import Counter
from itertools import combinations_with_replacement as multi_combs
def aupto(lim):
c = filter(lambda x: x<=lim, (i**3 for i in range(1, int(lim**(1/3))+2)))
s = filter(lambda x: x<=lim, (sum(mc) for mc in multi_combs(c, 6)))
counts = Counter(s)
return sorted(k for k in counts if counts[k]==1)
print(aupto(20000)) # Michael S. Branicky, Jun 13 2021
CROSSREFS
KEYWORD
nonn,fini
AUTHOR
STATUS
approved