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%I #6 Aug 05 2021 15:24:07
%S 1981,2105,2168,2277,2368,2376,2431,2438,2457,2466,2494,2538,2555,
%T 2557,2583,2593,2646,2665,2672,2709,2746,2753,2763,2765,2772,2880,
%U 2881,2889,2916,2942,2961,2970,2977,2979,2980,2987,3007,3033,3040,3042,3043,3049,3068
%N Numbers that are the sum of six cubes in eight or more ways.
%H Sean A. Irvine, <a href="/A345517/b345517.txt">Table of n, a(n) for n = 1..10000</a>
%e 2105 is a term because 2105 = 1^3 + 1^3 + 4^3 + 4^3 + 4^3 + 11^3 = 1^3 + 2^3 + 3^3 + 4^3 + 5^3 + 11^3 = 1^3 + 2^3 + 6^3 + 7^3 + 7^3 + 8^3 = 1^3 + 4^3 + 4^3 + 4^3 + 8^3 + 9^3 = 1^3 + 4^3 + 5^3 + 5^3 + 5^3 + 10^3 = 2^3 + 3^3 + 4^3 + 5^3 + 8^3 + 9^3 = 3^3 + 3^3 + 3^3 + 7^3 + 7^3 + 9^3 = 5^3 + 5^3 + 5^3 + 5^3 + 7^3 + 8^3.
%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 >= 8])
%o for x in range(len(rets)):
%o print(rets[x])
%Y Cf. A344812, A345183, A345516, A345518, A345526, A345565, A345770.
%K nonn
%O 1,1
%A _David Consiglio, Jr._, Jun 20 2021
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