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A345785 Numbers that are the sum of eight cubes in exactly three ways.

%I A345785 #6 Jul 31 2021 22:37:11
%S A345785 223,230,237,249,263,270,275,282,284,286,289,291,293,308,310,312,319,
%T A345785 326,345,349,354,364,371,373,375,378,380,385,386,387,389,397,399,404,
%U A345785 406,410,412,413,415,420,423,439,441,443,446,449,452,453,459,460,465,473
%N A345785 Numbers that are the sum of eight cubes in exactly three ways.
%C A345785 Differs from A345533 at term 5 because 256 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 5^3 + 5^3  = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 6^3  = 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 4^3 + 4^3 + 4^3  = 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 4^3 + 5^3.
%C A345785 Likely finite.
%H A345785 Sean A. Irvine, <a href="/A345785/b345785.txt">Table of n, a(n) for n = 1..198</a>
%e A345785 230 is a term because 230 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 5^3 = 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 3^3 + 3^3 = 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 4^3.
%o A345785 (Python)
%o A345785 from itertools import combinations_with_replacement as cwr
%o A345785 from collections import defaultdict
%o A345785 keep = defaultdict(lambda: 0)
%o A345785 power_terms = [x**3 for x in range(1, 1000)]
%o A345785 for pos in cwr(power_terms, 8):
%o A345785     tot = sum(pos)
%o A345785     keep[tot] += 1
%o A345785     rets = sorted([k for k, v in keep.items() if v == 3])
%o A345785     for x in range(len(rets)):
%o A345785         print(rets[x])
%Y A345785 Cf. A345533, A345775, A345784, A345786, A345795, A345835.
%K A345785 nonn
%O A345785 1,1
%A A345785 _David Consiglio, Jr._, Jun 26 2021