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A345796 Numbers that are the sum of nine cubes in exactly four ways.

%I A345796 #6 Jul 31 2021 22:32:44
%S A345796 224,257,264,283,320,348,355,372,374,376,381,383,390,400,402,407,411,
%T A345796 414,416,442,450,453,454,461,474,476,481,486,488,500,503,509,510,514,
%U A345796 519,528,529,537,542,543,544,545,548,550,552,554,555,557,564,572,573,574
%N A345796 Numbers that are the sum of nine cubes in exactly four ways.
%C A345796 Differs from A345543 at term 17 because 409 = 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 4^3 + 4^3 + 5^3 + 5^3  = 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 4^3 + 4^3 + 6^3  = 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 5^3 + 5^3 + 5^3  = 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 5^3 + 6^3  = 2^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3 + 4^3 + 4^3 + 4^3.
%C A345796 Likely finite.
%H A345796 Sean A. Irvine, <a href="/A345796/b345796.txt">Table of n, a(n) for n = 1..124</a>
%e A345796 257 is a term because 257 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 4^3 + 4^3 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 5^3 = 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 = 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 4^3.
%o A345796 (Python)
%o A345796 from itertools import combinations_with_replacement as cwr
%o A345796 from collections import defaultdict
%o A345796 keep = defaultdict(lambda: 0)
%o A345796 power_terms = [x**3 for x in range(1, 1000)]
%o A345796 for pos in cwr(power_terms, 9):
%o A345796     tot = sum(pos)
%o A345796     keep[tot] += 1
%o A345796     rets = sorted([k for k, v in keep.items() if v == 4])
%o A345796     for x in range(len(rets)):
%o A345796         print(rets[x])
%Y A345796 Cf. A345543, A345786, A345795, A345797, A345806, A345846.
%K A345796 nonn
%O A345796 1,1
%A A345796 _David Consiglio, Jr._, Jun 26 2021