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A345180 Numbers that are the sum of five third powers in seven or more ways.

%I A345180 #8 Aug 05 2021 15:25:42
%S A345180 4392,4472,4544,4600,4915,4957,5076,5113,5120,5132,5139,5165,5174,
%T A345180 5256,5321,5347,5354,5384,5391,5410,5445,5474,5481,5507,5543,5617,
%U A345180 5624,5643,5678,5715,5741,5760,5769,5797,5832,5834,5860,5895,5914,5923,5984,5986,6049
%N A345180 Numbers that are the sum of five third powers in seven or more ways.
%H A345180 David Consiglio, Jr., <a href="/A345180/b345180.txt">Table of n, a(n) for n = 1..10000</a>
%e A345180 4472 is a term because 4472 = 1^3 + 4^3 + 4^3 + 4^3 + 15^3  = 2^3 + 2^3 + 9^3 + 11^3 + 11^3  = 2^3 + 3^3 + 4^3 + 5^3 + 15^3  = 2^3 + 3^3 + 7^3 + 11^3 + 12^3  = 3^3 + 3^3 + 6^3 + 10^3 + 13^3  = 3^3 + 4^3 + 5^3 + 8^3 + 14^3  = 5^3 + 5^3 + 7^3 + 10^3 + 12^3.
%o A345180 (Python)
%o A345180 from itertools import combinations_with_replacement as cwr
%o A345180 from collections import defaultdict
%o A345180 keep = defaultdict(lambda: 0)
%o A345180 power_terms = [x**3 for x in range(1, 1000)]
%o A345180 for pos in cwr(power_terms, 5):
%o A345180     tot = sum(pos)
%o A345180     keep[tot] += 1
%o A345180 rets = sorted([k for k, v in keep.items() if v >= 7])
%o A345180 for x in range(len(rets)):
%o A345180     print(rets[x])
%Y A345180 Cf. A344800, A344942, A345150, A345174, A345181, A345183, A345516.
%K A345180 nonn
%O A345180 1,1
%A A345180 _David Consiglio, Jr._, Jun 10 2021