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%I #12 Jan 05 2024 13:18:13
%S 3104,3215,3222,3267,3313,3339,3374,3430,3465,3491,3493,3528,3547,
%T 3584,3645,3654,3698,3708,3736,3745,3752,3754,3761,3771,3780,3789,
%U 3817,3824,3843,3862,3869,3878,3880,3888,3906,3915,3923,3943,3950,3969,3995,4004,4014
%N Numbers that are the sum of six cubes in ten or more ways.
%H Sean A. Irvine, <a href="/A345519/b345519.txt">Table of n, a(n) for n = 1..10000</a>
%e 3215 is a term because 3215 = 1^3 + 1^3 + 1^3 + 4^3 + 6^3 + 13^3 = 1^3 + 1^3 + 2^3 + 2^3 + 9^3 + 12^3 = 1^3 + 1^3 + 3^3 + 8^3 + 8^3 + 11^3 = 1^3 + 2^3 + 3^3 + 5^3 + 8^3 + 12^3 = 1^3 + 3^3 + 3^3 + 3^3 + 10^3 + 11^3 = 1^3 + 3^3 + 8^3 + 8^3 + 8^3 + 9^3 = 2^3 + 3^3 + 6^3 + 6^3 + 8^3 + 11^3 = 3^3 + 3^3 + 3^3 + 8^3 + 9^3 + 10^3 = 3^3 + 5^3 + 5^3 + 5^3 + 6^3 + 12^3 = 6^3 + 6^3 + 6^3 + 6^3 + 7^3 + 10^3.
%o (Python)
%o from itertools import combinations_with_replacement as cwr
%o from collections import defaultdict
%o keep = defaultdict(lambda: 0)
%o power_terms = [x**3 for x in range(1, 1000)]
%o for pos in cwr(power_terms, 6):
%o tot = sum(pos)
%o keep[tot] += 1
%o rets = sorted([k for k, v in keep.items() if v >= 10])
%o for x in range(len(rets)):
%o print(rets[x])
%Y Cf. A345187, A345477, A345506, A345518, A345567, A345772.
%K nonn
%O 1,1
%A _David Consiglio, Jr._, Jun 20 2021
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