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A345184 Numbers that are the sum of five third powers in exactly eight ways.

%I A345184 #6 Jul 31 2021 23:16:49
%S A345184 4392,4915,5139,5256,5321,5624,5643,5678,5741,5769,5797,5832,5914,
%T A345184 6075,6202,6499,6560,6616,6642,6677,6833,6884,7008,7111,7128,7155,
%U A345184 7218,7344,7395,7641,7696,7729,7785,7813,7820,7849,7883,8037,8100,8243,8282,8308,8315
%N A345184 Numbers that are the sum of five third powers in exactly eight ways.
%C A345184 Differs from A345183 at term 13 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 A345184 David Consiglio, Jr., <a href="/A345184/b345184.txt">Table of n, a(n) for n = 1..10000</a>
%e A345184 4915 is a term because 4915 = 1^3 + 2^3 + 7^3 + 12^3 + 12^3  = 1^3 + 3^3 + 7^3 + 9^3 + 14^3  = 1^3 + 8^3 + 8^3 + 11^3 + 11^3  = 2^3 + 4^3 + 6^3 + 6^3 + 15^3  = 3^3 + 3^3 + 5^3 + 7^3 + 15^3  = 3^3 + 3^3 + 10^3 + 11^3 + 11^3  = 4^3 + 6^3 + 6^3 + 8^3 + 14^3  = 8^3 + 8^3 + 8^3 + 9^3 + 11^3.
%o A345184 (Python)
%o A345184 from itertools import combinations_with_replacement as cwr
%o A345184 from collections import defaultdict
%o A345184 keep = defaultdict(lambda: 0)
%o A345184 power_terms = [x**3 for x in range(1, 1000)]
%o A345184 for pos in cwr(power_terms, 5):
%o A345184     tot = sum(pos)
%o A345184     keep[tot] += 1
%o A345184 rets = sorted([k for k, v in keep.items() if v == 8])
%o A345184 for x in range(len(rets)):
%o A345184     print(rets[x])
%Y A345184 Cf. A294742, A344945, A345153, A345181, A345183, A345186, A345770.
%K A345184 nonn
%O A345184 1,1
%A A345184 _David Consiglio, Jr._, Jun 10 2021