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A345777 Numbers that are the sum of seven cubes in exactly five ways.

%I A345777 #6 Jul 31 2021 22:39:21
%S A345777 627,768,838,845,857,864,874,881,894,900,920,937,950,962,976,981,983,
%T A345777 990,1002,1009,1011,1016,1027,1054,1060,1061,1063,1089,1096,1098,1102,
%U A345777 1105,1109,1115,1124,1128,1133,1135,1137,1140,1144,1151,1153,1154,1159,1163
%N A345777 Numbers that are the sum of seven cubes in exactly five ways.
%C A345777 Differs from A345523 at term 14 because 955 = 1^3 + 1^3 + 1^3 + 2^3 + 6^3 + 6^3 + 8^3  = 1^3 + 1^3 + 2^3 + 3^3 + 4^3 + 5^3 + 9^3  = 1^3 + 3^3 + 3^3 + 5^3 + 6^3 + 6^3 + 7^3  = 1^3 + 4^3 + 4^3 + 4^3 + 5^3 + 5^3 + 8^3  = 2^3 + 2^3 + 4^3 + 4^3 + 5^3 + 7^3 + 7^3  = 2^3 + 3^3 + 4^3 + 4^3 + 4^3 + 6^3 + 8^3.
%C A345777 Likely finite.
%H A345777 Sean A. Irvine, <a href="/A345777/b345777.txt">Table of n, a(n) for n = 1..351</a>
%e A345777 768 is a term because 768 = 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 8^3 = 1^3 + 1^3 + 1^3 + 3^3 + 3^3 + 4^3 + 7^3 = 1^3 + 1^3 + 2^3 + 2^3 + 3^3 + 6^3 + 6^3 = 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 5^3 + 7^3 = 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 5^3 + 6^3.
%o A345777 (Python)
%o A345777 from itertools import combinations_with_replacement as cwr
%o A345777 from collections import defaultdict
%o A345777 keep = defaultdict(lambda: 0)
%o A345777 power_terms = [x**3 for x in range(1, 1000)]
%o A345777 for pos in cwr(power_terms, 7):
%o A345777     tot = sum(pos)
%o A345777     keep[tot] += 1
%o A345777     rets = sorted([k for k, v in keep.items() if v == 5])
%o A345777     for x in range(len(rets)):
%o A345777         print(rets[x])
%Y A345777 Cf. A345523, A345767, A345776, A345778, A345787, A345827.
%K A345777 nonn
%O A345777 1,1
%A A345777 _David Consiglio, Jr._, Jun 26 2021