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%I #6 Jul 31 2021 19:03:45
%S 52417,54518,69634,70954,84458,84489,84700,85481,87582,92233,101264,
%T 102890,112574,117225,119326,134473,143264,143442,143506,149781,
%U 151448,158719,159465,165634,166998,167029,167196,167240,168021,170122,174773,183804,184457
%N Numbers that are the sum of eight fifth powers in exactly three ways.
%C Differs from A345611 at term 105 because 391250 = 2^5 + 3^5 + 5^5 + 5^5 + 5^5 + 8^5 + 10^5 + 12^5 = 1^5 + 1^5 + 4^5 + 7^5 + 8^5 + 8^5 + 9^5 + 12^5 = 2^5 + 3^5 + 4^5 + 4^5 + 6^5 + 9^5 + 11^5 + 11^5 = 1^5 + 3^5 + 3^5 + 5^5 + 8^5 + 8^5 + 11^5 + 11^5.
%H Sean A. Irvine, <a href="/A346328/b346328.txt">Table of n, a(n) for n = 1..10000</a>
%e 52417 is a term because 52417 = 3^5 + 3^5 + 3^5 + 3^5 + 5^5 + 6^5 + 6^5 + 8^5 = 1^5 + 4^5 + 4^5 + 4^5 + 4^5 + 6^5 + 6^5 + 8^5 = 3^5 + 3^5 + 3^5 + 3^5 + 4^5 + 7^5 + 7^5 + 7^5.
%o (Python)
%o from itertools import combinations_with_replacement as cwr
%o from collections import defaultdict
%o keep = defaultdict(lambda: 0)
%o power_terms = [x**5 for x in range(1, 1000)]
%o for pos in cwr(power_terms, 8):
%o tot = sum(pos)
%o keep[tot] += 1
%o rets = sorted([k for k, v in keep.items() if v == 3])
%o for x in range(len(rets)):
%o print(rets[x])
%Y Cf. A345611, A345835, A346280, A346327, A346329, A346338.
%K nonn
%O 1,1
%A _David Consiglio, Jr._, Jul 13 2021
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