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%I A345183 #6 Aug 05 2021 15:25:47
%S A345183 4392,4915,5139,5256,5321,5624,5643,5678,5741,5769,5797,5832,5860,
%T A345183 5914,6075,6112,6138,6202,6462,6497,6499,6560,6588,6616,6642,6651,
%U A345183 6677,6833,6859,6884,6947,7001,7008,7038,7057,7064,7099,7111,7128,7155,7190,7218,7316
%N A345183 Numbers that are the sum of five third powers in eight or more ways.
%H A345183 David Consiglio, Jr., <a href="/A345183/b345183.txt">Table of n, a(n) for n = 1..10000</a>
%e A345183 4915 is a term because 4915 = 1^3 + 2^3 + 7^3 + 12^3 + 12^3  = 1^3 + 3^3 + 7^3 + 9^3 + 14^3  = 1^3 + 8^3 + 8^3 + 11^3 + 11^3  = 2^3 + 4^3 + 6^3 + 6^3 + 15^3  = 3^3 + 3^3 + 5^3 + 7^3 + 15^3  = 3^3 + 3^3 + 10^3 + 11^3 + 11^3  = 4^3 + 6^3 + 6^3 + 8^3 + 14^3  = 8^3 + 8^3 + 8^3 + 9^3 + 11^3.
%o A345183 (Python)
%o A345183 from itertools import combinations_with_replacement as cwr
%o A345183 from collections import defaultdict
%o A345183 keep = defaultdict(lambda: 0)
%o A345183 power_terms = [x**3 for x in range(1, 1000)]
%o A345183 for pos in cwr(power_terms, 5):
%o A345183     tot = sum(pos)
%o A345183     keep[tot] += 1
%o A345183 rets = sorted([k for k, v in keep.items() if v >= 8])
%o A345183 for x in range(len(rets)):
%o A345183     print(rets[x])
%Y A345183 Cf. A344801, A344944, A345152, A345180, A345184, A345185, A345517.
%K A345183 nonn
%O A345183 1,1
%A A345183 _David Consiglio, Jr._, Jun 10 2021

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