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A345611 Numbers that are the sum of eight fifth powers in three or more ways.

%I A345611 #6 Jul 31 2021 16:17:02
%S A345611 52417,54518,69634,70954,84458,84489,84700,85481,87582,92233,101264,
%T A345611 102890,112574,117225,119326,134473,143264,143442,143506,149781,
%U A345611 151448,158719,159465,165634,166998,167029,167196,167240,168021,170122,174773,183804,184457
%N A345611 Numbers that are the sum of eight fifth powers in three or more ways.
%H A345611 Sean A. Irvine, <a href="/A345611/b345611.txt">Table of n, a(n) for n = 1..10000</a>
%e A345611 54518 is a term because 54518 = 1^5 + 4^5 + 4^5 + 4^5 + 4^5 + 7^5 + 7^5 + 7^5 = 1^5 + 4^5 + 4^5 + 4^5 + 5^5 + 6^5 + 6^5 + 8^5 = 3^5 + 3^5 + 3^5 + 3^5 + 5^5 + 7^5 + 7^5 + 7^5.
%o A345611 (Python)
%o A345611 from itertools import combinations_with_replacement as cwr
%o A345611 from collections import defaultdict
%o A345611 keep = defaultdict(lambda: 0)
%o A345611 power_terms = [x**5 for x in range(1, 1000)]
%o A345611 for pos in cwr(power_terms, 8):
%o A345611     tot = sum(pos)
%o A345611     keep[tot] += 1
%o A345611     rets = sorted([k for k, v in keep.items() if v >= 3])
%o A345611     for x in range(len(rets)):
%o A345611         print(rets[x])
%Y A345611 Cf. A345578, A345606, A345610, A345612, A345620, A346328.
%K A345611 nonn
%O A345611 1,1
%A A345611 _David Consiglio, Jr._, Jun 20 2021