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A345536 Numbers that are the sum of eight cubes in six or more ways.

%I A345536 #6 Aug 05 2021 15:20:56
%S A345536 628,719,769,776,778,795,832,839,846,858,860,865,872,875,876,882,886,
%T A345536 891,893,895,901,902,907,908,912,921,927,928,931,938,945,946,947,951,
%U A345536 954,956,958,963,964,965,970,972,977,982,984,989,991,992,996,998,999,1001
%N A345536 Numbers that are the sum of eight cubes in six or more ways.
%H A345536 Sean A. Irvine, <a href="/A345536/b345536.txt">Table of n, a(n) for n = 1..10000</a>
%e A345536 719 is a term because 719 = 1^3 + 1^3 + 1^3 + 4^3 + 4^3 + 4^3 + 4^3 + 5^3 = 1^3 + 1^3 + 2^3 + 3^3 + 4^3 + 4^3 + 5^3 + 5^3 = 1^3 + 2^3 + 2^3 + 3^3 + 3^3 + 5^3 + 5^3 + 5^3 = 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 + 3^3 + 3^3 + 7^3 = 3^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3 + 4^3 + 4^3.
%o A345536 (Python)
%o A345536 from itertools import combinations_with_replacement as cwr
%o A345536 from collections import defaultdict
%o A345536 keep = defaultdict(lambda: 0)
%o A345536 power_terms = [x**3 for x in range(1, 1000)]
%o A345536 for pos in cwr(power_terms, 8):
%o A345536     tot = sum(pos)
%o A345536     keep[tot] += 1
%o A345536     rets = sorted([k for k, v in keep.items() if v >= 6])
%o A345536     for x in range(len(rets)):
%o A345536         print(rets[x])
%Y A345536 Cf. A345493, A345524, A345535, A345537, A345545, A345581, A345788.
%K A345536 nonn
%O A345536 1,1
%A A345536 _David Consiglio, Jr._, Jun 20 2021