%I #6 Jul 31 2021 18:54:12
%S 200009,220350,235658,329271,329810,330052,359211,359453,359498,
%T 360298,367314,368529,374519,374847,375089,375870,376620,376651,
%U 377159,377643,380283,382622,384395,384934,387035,388933,391736,392064,392275,392339,392517,392581
%N Numbers that are the sum of ten fifth powers in exactly five ways.
%C Differs from A345637 at term 29 because 392095 = 2^5 + 2^5 + 2^5 + 4^5 + 5^5 + 5^5 + 5^5 + 8^5 + 10^5 + 12^5 = 1^5 + 1^5 + 1^5 + 5^5 + 6^5 + 6^5 + 6^5 + 7^5 + 10^5 + 12^5 = 2^5 + 2^5 + 2^5 + 3^5 + 3^5 + 6^5 + 7^5 + 9^5 + 9^5 + 12^5 = 2^5 + 2^5 + 2^5 + 4^5 + 4^5 + 4^5 + 6^5 + 9^5 + 11^5 + 11^5 = 1^5 + 2^5 + 2^5 + 3^5 + 4^5 + 5^5 + 8^5 + 8^5 + 11^5 + 11^5 = 1^5 + 1^5 + 1^5 + 2^5 + 3^5 + 8^5 + 9^5 + 10^5 + 10^5 + 10^5.
%H Sean A. Irvine, <a href="/A346350/b346350.txt">Table of n, a(n) for n = 1..10000</a>
%e 200009 is a term because 200009 = 2^5 + 4^5 + 4^5 + 6^5 + 6^5 + 6^5 + 6^5 + 6^5 + 9^5 + 10^5 = 1^5 + 3^5 + 5^5 + 6^5 + 6^5 + 6^5 + 6^5 + 8^5 + 8^5 + 10^5 = 1^5 + 3^5 + 4^5 + 6^5 + 6^5 + 7^5 + 7^5 + 7^5 + 8^5 + 10^5 = 2^5 + 2^5 + 4^5 + 4^5 + 6^5 + 8^5 + 8^5 + 8^5 + 8^5 + 9^5 = 1^5 + 2^5 + 3^5 + 5^5 + 8^5 + 8^5 + 8^5 + 8^5 + 8^5 + 8^5.
%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, 10):
%o tot = sum(pos)
%o keep[tot] += 1
%o rets = sorted([k for k, v in keep.items() if v == 5])
%o for x in range(len(rets)):
%o print(rets[x])
%Y Cf. A345637, A345857, A346340, A346349, A346351.
%K nonn
%O 1,1
%A _David Consiglio, Jr._, Jul 13 2021