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A345768 Numbers that are the sum of six cubes in exactly six ways.

%I A345768 #6 Jul 31 2021 22:49:52
%S A345768 1377,1488,1586,1595,1647,1673,1677,1738,1764,1799,1829,1836,1837,
%T A345768 1862,1881,1890,1911,1953,1955,2007,2011,2014,2018,2025,2044,2070,
%U A345768 2079,2097,2107,2108,2142,2153,2170,2177,2203,2214,2216,2222,2223,2226,2229,2252,2258
%N A345768 Numbers that are the sum of six cubes in exactly six ways.
%C A345768 Differs from A345515 at term 8 because 1710 = 1^3 + 1^3 + 5^3 + 5^3 + 9^3 + 9^3  = 1^3 + 2^3 + 3^3 + 6^3 + 9^3 + 9^3  = 1^3 + 2^3 + 4^3 + 5^3 + 8^3 + 10^3  = 1^3 + 4^3 + 4^3 + 5^3 + 5^3 + 11^3  = 2^3 + 2^3 + 2^3 + 7^3 + 7^3 + 10^3  = 2^3 + 3^3 + 4^3 + 4^3 + 6^3 + 11^3  = 4^3 + 4^3 + 5^3 + 6^3 + 8^3 + 9^3.
%H A345768 Sean A. Irvine, <a href="/A345768/b345768.txt">Table of n, a(n) for n = 1..1338</a>
%e A345768 1488 is a term because 1488 = 1^3 + 1^3 + 1^3 + 3^3 + 8^3 + 8^3 = 1^3 + 1^3 + 3^3 + 3^3 + 3^3 + 10^3 = 1^3 + 2^3 + 3^3 + 6^3 + 6^3 + 8^3 = 2^3 + 2^3 + 2^3 + 2^3 + 4^3 + 10^3 = 3^3 + 3^3 + 3^3 + 3^3 + 6^3 + 9^3 = 3^3 + 5^3 + 5^3 + 6^3 + 6^3 + 6^3.
%o A345768 (Python)
%o A345768 from itertools import combinations_with_replacement as cwr
%o A345768 from collections import defaultdict
%o A345768 keep = defaultdict(lambda: 0)
%o A345768 power_terms = [x**3 for x in range(1, 1000)]
%o A345768 for pos in cwr(power_terms, 6):
%o A345768     tot = sum(pos)
%o A345768     keep[tot] += 1
%o A345768     rets = sorted([k for k, v in keep.items() if v == 6])
%o A345768     for x in range(len(rets)):
%o A345768         print(rets[x])
%Y A345768 Cf. A345175, A345515, A345767, A345769, A345778, A345818.
%K A345768 nonn
%O A345768 1,1
%A A345768 _David Consiglio, Jr._, Jun 26 2021