cp's OEIS Frontend

A345795 Numbers that are the sum of nine cubes in exactly three ways.

%I A345795 #6 Jul 31 2021 22:32:41
%S A345795 231,238,245,250,259,271,276,278,280,285,287,290,292,294,297,299,301,
%T A345795 302,309,311,313,315,316,318,322,327,334,335,337,339,341,346,350,353,
%U A345795 357,362,365,379,386,387,388,391,393,394,395,397,398,405,412,418,420,421
%N A345795 Numbers that are the sum of nine cubes in exactly three ways.
%C A345795 Differs from A345542 at term 1 because 224 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 6^3  = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 4^3 + 4^3 + 4^3  = 1^3 + 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 4^3 + 5^3  = 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3.
%C A345795 Likely finite.
%H A345795 Sean A. Irvine, <a href="/A345795/b345795.txt">Table of n, a(n) for n = 1..136</a>
%e A345795 231 is a term because 231 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 5^3 = 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 3^3 + 3^3 = 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 4^3.
%o A345795 (Python)
%o A345795 from itertools import combinations_with_replacement as cwr
%o A345795 from collections import defaultdict
%o A345795 keep = defaultdict(lambda: 0)
%o A345795 power_terms = [x**3 for x in range(1, 1000)]
%o A345795 for pos in cwr(power_terms, 9):
%o A345795     tot = sum(pos)
%o A345795     keep[tot] += 1
%o A345795     rets = sorted([k for k, v in keep.items() if v == 3])
%o A345795     for x in range(len(rets)):
%o A345795         print(rets[x])
%Y A345795 Cf. A345542, A345785, A345794, A345796, A345805, A345845.
%K A345795 nonn
%O A345795 1,1
%A A345795 _David Consiglio, Jr._, Jun 26 2021