%I #6 Jul 31 2021 15:58:51
%S 944383,953139,953414,985453,1118585,1151438,1185375,1192180,1198879,
%T 1206546,1209912,1216569,1217172,1218912,1223321,1225398,1226654,
%U 1234631,1241834,1242437,1251195,1251406,1252123,1259685,1265563,1265594,1267937,1275375,1281736
%N Numbers that are the sum of ten fifth powers in eight or more ways.
%H Sean A. Irvine, <a href="/A345640/b345640.txt">Table of n, a(n) for n = 1..1000</a>
%e 953139 is a term because 953139 = 1^5 + 1^5 + 1^5 + 3^5 + 8^5 + 10^5 + 10^5 + 10^5 + 12^5 + 13^5 = 1^5 + 2^5 + 2^5 + 6^5 + 6^5 + 8^5 + 9^5 + 9^5 + 12^5 + 14^5 = 2^5 + 2^5 + 3^5 + 3^5 + 6^5 + 7^5 + 9^5 + 12^5 + 12^5 + 13^5 = 2^5 + 2^5 + 3^5 + 3^5 + 7^5 + 7^5 + 9^5 + 11^5 + 11^5 + 14^5 = 2^5 + 2^5 + 4^5 + 4^5 + 4^5 + 6^5 + 11^5 + 11^5 + 12^5 + 13^5 = 2^5 + 3^5 + 3^5 + 5^5 + 8^5 + 8^5 + 8^5 + 8^5 + 9^5 + 15^5 = 3^5 + 3^5 + 3^5 + 4^5 + 4^5 + 6^5 + 10^5 + 10^5 + 13^5 + 13^5 = 3^5 + 3^5 + 5^5 + 6^5 + 6^5 + 6^5 + 6^5 + 9^5 + 10^5 + 15^5.
%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 >= 8])
%o for x in range(len(rets)):
%o print(rets[x])
%Y Cf. A345601, A345625, A345639, A345641, A346353.
%K nonn
%O 1,1
%A _David Consiglio, Jr._, Jun 20 2021