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%I #6 Jul 31 2021 19:00:50
%S 52418,52449,52660,53441,54519,54550,54761,55690,57643,60193,62294,
%T 69224,69635,69666,69877,70658,70955,70986,71197,71325,71978,72759,
%U 73001,74079,76031,77410,78730,84162,84459,84490,84521,84701,84732,84943,85185,85482,85513
%N Numbers that are the sum of nine fifth powers in exactly three ways.
%C Differs from A345620 at term 8 because 55542 = 3^5 + 3^5 + 3^5 + 3^5 + 5^5 + 5^5 + 6^5 + 6^5 + 8^5 = 1^5 + 4^5 + 4^5 + 4^5 + 4^5 + 5^5 + 6^5 + 6^5 + 8^5 = 3^5 + 3^5 + 3^5 + 3^5 + 4^5 + 5^5 + 7^5 + 7^5 + 7^5 = 1^5 + 4^5 + 4^5 + 4^5 + 4^5 + 4^5 + 7^5 + 7^5 + 7^5.
%H Sean A. Irvine, <a href="/A346338/b346338.txt">Table of n, a(n) for n = 1..10000</a>
%e 52418 is a term because 52418 = 1^5 + 3^5 + 3^5 + 3^5 + 3^5 + 5^5 + 6^5 + 6^5 + 8^5 = 1^5 + 1^5 + 4^5 + 4^5 + 4^5 + 4^5 + 6^5 + 6^5 + 8^5 = 1^5 + 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, 9):
%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. A345620, A345845, A346328, A346337, A346339, A346348.
%K nonn
%O 1,1
%A _David Consiglio, Jr._, Jul 13 2021
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