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A345775 Numbers that are the sum of seven cubes in exactly three ways.

%I A345775 #6 Jul 31 2021 22:39:15
%S A345775 222,229,248,255,262,281,283,285,318,346,370,374,377,379,381,396,400,
%T A345775 407,412,419,426,433,437,438,444,451,463,472,475,477,489,494,501,505,
%U A345775 507,510,522,529,533,536,559,564,566,568,570,577,578,584,585,592,594,596
%N A345775 Numbers that are the sum of seven cubes in exactly three ways.
%C A345775 Differs from A345521 at term 28 because 470 = 1^3 + 1^3 + 1^3 + 1^3 + 5^3 + 5^3 + 6^3  = 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 6^3 + 6^3  = 1^3 + 3^3 + 4^3 + 4^3 + 4^3 + 5^3 + 5^3  = 2^3 + 3^3 + 3^3 + 4^3 + 4^3 + 4^3 + 6^3.
%C A345775 Likely finite.
%H A345775 Sean A. Irvine, <a href="/A345775/b345775.txt">Table of n, a(n) for n = 1..390</a>
%e A345775 229 is a term because 229 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 5^3 = 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 3^3 + 3^3 = 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 4^3.
%o A345775 (Python)
%o A345775 from itertools import combinations_with_replacement as cwr
%o A345775 from collections import defaultdict
%o A345775 keep = defaultdict(lambda: 0)
%o A345775 power_terms = [x**3 for x in range(1, 1000)]
%o A345775 for pos in cwr(power_terms, 7):
%o A345775     tot = sum(pos)
%o A345775     keep[tot] += 1
%o A345775     rets = sorted([k for k, v in keep.items() if v == 3])
%o A345775     for x in range(len(rets)):
%o A345775         print(rets[x])
%Y A345775 Cf. A048931, A345521, A345774, A345776, A345785, A345825.
%K A345775 nonn
%O A345775 1,1
%A A345775 _David Consiglio, Jr._, Jun 26 2021