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A345805 Numbers that are the sum of ten cubes in exactly three ways.

%I A345805 #6 Jul 31 2021 22:27:17
%S A345805 197,239,246,253,260,267,277,279,281,293,295,298,300,302,303,305,309,
%T A345805 312,316,317,319,324,326,329,330,335,336,338,340,343,344,345,351,352,
%U A345805 354,358,361,362,364,365,368,370,379,386,387,388,392,394,395,396,402,406
%N A345805 Numbers that are the sum of ten cubes in exactly three ways.
%C A345805 Differs from A345551 at term 2 because 225 = 1^3 + 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 + 1^3 + 3^3 + 4^3 + 4^3 + 4^3  = 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 4^3 + 5^3  = 1^3 + 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3.
%C A345805 Likely finite.
%H A345805 Sean A. Irvine, <a href="/A345805/b345805.txt">Table of n, a(n) for n = 1..93</a>
%e A345805 225 is a term because 225 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 5^3 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 3^3 + 3^3 + 3^3 = 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 4^3 = 1^3 + 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3.
%o A345805 (Python)
%o A345805 from itertools import combinations_with_replacement as cwr
%o A345805 from collections import defaultdict
%o A345805 keep = defaultdict(lambda: 0)
%o A345805 power_terms = [x**3 for x in range(1, 1000)]
%o A345805 for pos in cwr(power_terms, 10):
%o A345805     tot = sum(pos)
%o A345805     keep[tot] += 1
%o A345805     rets = sorted([k for k, v in keep.items() if v == 3])
%o A345805     for x in range(len(rets)):
%o A345805         print(rets[x])
%Y A345805 Cf. A345551, A345795, A345804, A345806, A345855.
%K A345805 nonn
%O A345805 1,1
%A A345805 _David Consiglio, Jr._, Jun 26 2021