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A345186 Numbers that are the sum of five third powers in exactly nine ways.

%I A345186 #6 Jul 31 2021 23:16:53
%S A345186 6112,6138,6462,6497,7001,7038,7057,7064,7099,7190,7316,7328,7372,
%T A345186 7433,7561,7587,7703,7759,7841,7902,8163,8352,8443,8560,8630,8632,
%U A345186 8928,8991,9017,9136,9143,9171,9288,9316,9379,9505,9566,9647,9658,9675,9684,9745,9773
%N A345186 Numbers that are the sum of five third powers in exactly nine ways.
%C A345186 Differs from A345185 at term 1 because 5860 = 1^3 + 1^3 + 5^3 + 8^3 + 16^3  = 1^3 + 2^3 + 3^3 + 11^3 + 15^3  = 1^3 + 3^3 + 8^3 + 11^3 + 14^3  = 1^3 + 5^3 + 5^3 + 10^3 + 15^3  = 1^3 + 9^3 + 10^3 + 10^3 + 12^3  = 2^3 + 3^3 + 8^3 + 9^3 + 15^3  = 2^3 + 3^3 + 5^3 + 12^3 + 14^3  = 2^3 + 8^3 + 8^3 + 12^3 + 12^3  = 3^3 + 8^3 + 8^3 + 9^3 + 14^3  = 3^3 + 6^3 + 7^3 + 12^3 + 13^3.
%H A345186 David Consiglio, Jr., <a href="/A345186/b345186.txt">Table of n, a(n) for n = 1..10000</a>
%e A345186 6112 is a term because 6112 = 1^3 + 2^3 + 9^3 + 11^3 + 14^3  = 1^3 + 3^3 + 7^3 + 12^3 + 14^3  = 1^3 + 6^3 + 6^3 + 7^3 + 16^3  = 2^3 + 2^3 + 9^3 + 9^3 + 15^3  = 2^3 + 3^3 + 5^3 + 11^3 + 15^3  = 2^3 + 8^3 + 9^3 + 9^3 + 14^3  = 3^3 + 3^3 + 3^3 + 4^3 + 17^3  = 3^3 + 5^3 + 8^3 + 11^3 + 14^3  = 8^3 + 8^3 + 8^3 + 11^3 + 12^3.
%o A345186 (Python)
%o A345186 from itertools import combinations_with_replacement as cwr
%o A345186 from collections import defaultdict
%o A345186 keep = defaultdict(lambda: 0)
%o A345186 power_terms = [x**3 for x in range(1, 1000)]
%o A345186 for pos in cwr(power_terms, 5):
%o A345186     tot = sum(pos)
%o A345186     keep[tot] += 1
%o A345186 rets = sorted([k for k, v in keep.items() if v == 9])
%o A345186 for x in range(len(rets)):
%o A345186     print(rets[x])
%Y A345186 Cf. A294743, A341892, A345154, A345184, A345185, A345188, A345771.
%K A345186 nonn
%O A345186 1,1
%A A345186 _David Consiglio, Jr._, Jun 10 2021