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A345788 Numbers that are the sum of eight cubes in exactly six ways.

%I A345788 #6 Jul 31 2021 22:37:21
%S A345788 628,719,769,776,778,795,832,839,846,858,860,865,872,875,876,882,886,
%T A345788 891,893,895,901,907,912,927,928,931,945,946,947,951,954,956,964,965,
%U A345788 972,989,992,998,999,1001,1012,1014,1015,1021,1034,1035,1036,1038,1040,1045
%N A345788 Numbers that are the sum of eight cubes in exactly six ways.
%C A345788 Differs from A345536 at term 22 because 902 = 1^3 + 1^3 + 2^3 + 2^3 + 3^3 + 4^3 + 4^3 + 9^3  = 1^3 + 1^3 + 3^3 + 4^3 + 5^3 + 5^3 + 6^3 + 7^3  = 1^3 + 2^3 + 3^3 + 3^3 + 4^3 + 6^3 + 6^3 + 7^3  = 1^3 + 2^3 + 4^3 + 4^3 + 4^3 + 4^3 + 5^3 + 8^3  = 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 5^3 + 9^3  = 2^3 + 2^3 + 2^3 + 4^3 + 4^3 + 4^3 + 7^3 + 7^3  = 3^3 + 5^3 + 5^3 + 5^3 + 5^3 + 5^3 + 5^3 + 5^3.
%C A345788 Likely finite.
%H A345788 Sean A. Irvine, <a href="/A345788/b345788.txt">Table of n, a(n) for n = 1..173</a>
%e A345788 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 A345788 (Python)
%o A345788 from itertools import combinations_with_replacement as cwr
%o A345788 from collections import defaultdict
%o A345788 keep = defaultdict(lambda: 0)
%o A345788 power_terms = [x**3 for x in range(1, 1000)]
%o A345788 for pos in cwr(power_terms, 8):
%o A345788     tot = sum(pos)
%o A345788     keep[tot] += 1
%o A345788     rets = sorted([k for k, v in keep.items() if v == 6])
%o A345788     for x in range(len(rets)):
%o A345788         print(rets[x])
%Y A345788 Cf. A345536, A345778, A345787, A345789, A345798, A345838.
%K A345788 nonn
%O A345788 1,1
%A A345788 _David Consiglio, Jr._, Jun 26 2021