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A345807 Numbers that are the sum of ten cubes in exactly five ways.

%I A345807 #6 Jul 31 2021 22:27:24
%S A345807 288,349,382,384,401,403,408,410,414,415,417,421,429,443,454,455,462,
%T A345807 475,482,487,496,501,520,543,544,545,546,548,555,556,558,565,566,570,
%U A345807 573,574,575,576,578,580,582,586,587,591,592,593,594,600,609,610,611,617
%N A345807 Numbers that are the sum of ten cubes in exactly five ways.
%C A345807 Differs from A345553 at term 14 because 436 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 4^3 + 7^3  = 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 3^3 + 4^3 + 4^3 + 5^3 + 5^3  = 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3 + 6^3  = 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 5^3 + 5^3 + 5^3  = 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 5^3 + 6^3  = 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3 + 4^3 + 4^3 + 4^3.
%C A345807 Likely finite.
%H A345807 Sean A. Irvine, <a href="/A345807/b345807.txt">Table of n, a(n) for n = 1..86</a>
%e A345807 349 is a term because 349 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 4^3 + 5^3 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3 = 1^3 + 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 4^3 + 4^3 = 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 5^3 = 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3.
%o A345807 (Python)
%o A345807 from itertools import combinations_with_replacement as cwr
%o A345807 from collections import defaultdict
%o A345807 keep = defaultdict(lambda: 0)
%o A345807 power_terms = [x**3 for x in range(1, 1000)]
%o A345807 for pos in cwr(power_terms, 10):
%o A345807     tot = sum(pos)
%o A345807     keep[tot] += 1
%o A345807     rets = sorted([k for k, v in keep.items() if v == 5])
%o A345807     for x in range(len(rets)):
%o A345807         print(rets[x])
%Y A345807 Cf. A345553, A345797, A345806, A345808, A345857.
%K A345807 nonn
%O A345807 1,1
%A A345807 _David Consiglio, Jr._, Jun 26 2021