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A345545 Numbers that are the sum of nine cubes in six or more ways.

%I A345545 #6 Aug 05 2021 15:19:13
%S A345545 472,498,505,507,524,596,598,605,624,629,631,636,643,650,655,657,661,
%T A345545 662,669,672,676,681,687,688,690,692,694,696,706,707,713,718,720,722,
%U A345545 725,727,728,729,731,732,737,739,742,744,746,748,749,750,751,753,755,756
%N A345545 Numbers that are the sum of nine cubes in six or more ways.
%H A345545 Sean A. Irvine, <a href="/A345545/b345545.txt">Table of n, a(n) for n = 1..10000</a>
%e A345545 498 is a term because 498 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 4^3 + 4^3 + 5^3 = 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 5^3 + 5^3 = 1^3 + 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 6^3 = 1^3 + 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3 = 1^3 + 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 5^3 = 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 4^3 + 4^3 + 4^3.
%o A345545 (Python)
%o A345545 from itertools import combinations_with_replacement as cwr
%o A345545 from collections import defaultdict
%o A345545 keep = defaultdict(lambda: 0)
%o A345545 power_terms = [x**3 for x in range(1, 1000)]
%o A345545 for pos in cwr(power_terms, 9):
%o A345545     tot = sum(pos)
%o A345545     keep[tot] += 1
%o A345545     rets = sorted([k for k, v in keep.items() if v >= 6])
%o A345545     for x in range(len(rets)):
%o A345545         print(rets[x])
%Y A345545 Cf. A345503, A345536, A345544, A345546, A345554, A345590, A345798.
%K A345545 nonn
%O A345545 1,1
%A A345545 _David Consiglio, Jr._, Jun 20 2021