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A345623 Numbers that are the sum of nine fifth powers in six or more ways.

%I A345623 #6 Jul 31 2021 16:09:11
%S A345623 926404,936607,952896,985421,993574,993605,993816,1075779,1123321,
%T A345623 1133344,1134367,1151406,1160105,1166111,1177144,1206514,1209669,
%U A345623 1209847,1215545,1225630,1251130,1264929,1265320,1278611,1414834,1422367,1422609,1430384,1431367
%N A345623 Numbers that are the sum of nine fifth powers in six or more ways.
%H A345623 Sean A. Irvine, <a href="/A345623/b345623.txt">Table of n, a(n) for n = 1..10000</a>
%e A345623 936607 is a term because 936607 = 1^5 + 1^5 + 2^5 + 7^5 + 10^5 + 11^5 + 11^5 + 12^5 + 12^5 = 1^5 + 3^5 + 4^5 + 7^5 + 7^5 + 8^5 + 12^5 + 12^5 + 13^5 = 1^5 + 3^5 + 5^5 + 6^5 + 8^5 + 8^5 + 11^5 + 11^5 + 14^5 = 2^5 + 4^5 + 4^5 + 6^5 + 6^5 + 9^5 + 11^5 + 11^5 + 14^5 = 2^5 + 5^5 + 5^5 + 5^5 + 6^5 + 8^5 + 10^5 + 12^5 + 14^5 = 4^5 + 4^5 + 4^5 + 7^5 + 8^5 + 8^5 + 8^5 + 9^5 + 15^5.
%o A345623 (Python)
%o A345623 from itertools import combinations_with_replacement as cwr
%o A345623 from collections import defaultdict
%o A345623 keep = defaultdict(lambda: 0)
%o A345623 power_terms = [x**5 for x in range(1, 1000)]
%o A345623 for pos in cwr(power_terms, 9):
%o A345623     tot = sum(pos)
%o A345623     keep[tot] += 1
%o A345623     rets = sorted([k for k, v in keep.items() if v >= 6])
%o A345623     for x in range(len(rets)):
%o A345623         print(rets[x])
%Y A345623 Cf. A345590, A345614, A345622, A345624, A345638, A346341.
%K A345623 nonn
%O A345623 1,1
%A A345623 _David Consiglio, Jr._, Jun 20 2021