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A345181 Numbers that are the sum of five third powers in exactly seven ways.

%I A345181 #7 Jul 31 2021 23:16:46
%S A345181 4472,4544,4600,4957,5076,5113,5120,5132,5165,5174,5347,5354,5384,
%T A345181 5391,5410,5445,5474,5481,5507,5543,5617,5715,5760,5834,5895,5923,
%U A345181 5984,5986,6049,6128,6131,6245,6280,6373,6407,6434,6436,6544,6553,6733,6768,6831,6840
%N A345181 Numbers that are the sum of five third powers in exactly seven ways.
%C A345181 Differs from A345180 at term 1 because 4392 = 1^3 + 1^3 + 10^3 + 10^3 + 11^3  = 1^3 + 2^3 + 2^3 + 9^3 + 14^3  = 1^3 + 8^3 + 9^3 + 10^3 + 10^3  = 2^3 + 2^3 + 3^3 + 5^3 + 15^3  = 2^3 + 3^3 + 5^3 + 8^3 + 14^3  = 2^3 + 8^3 + 8^3 + 8^3 + 12^3  = 3^3 + 6^3 + 7^3 + 8^3 + 13^3  = 5^3 + 5^3 + 5^3 + 9^3 + 13^3.
%H A345181 David Consiglio, Jr., <a href="/A345181/b345181.txt">Table of n, a(n) for n = 1..10000</a>
%e A345181 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 A345181 (Python)
%o A345181 from itertools import combinations_with_replacement as cwr
%o A345181 from collections import defaultdict
%o A345181 keep = defaultdict(lambda: 0)
%o A345181 power_terms = [x**3 for x in range(1, 1000)]
%o A345181 for pos in cwr(power_terms, 5):
%o A345181     tot = sum(pos)
%o A345181     keep[tot] += 1
%o A345181 rets = sorted([k for k, v in keep.items() if v == 7])
%o A345181 for x in range(len(rets)):
%o A345181     print(rets[x])
%Y A345181 Cf. A294741, A344943, A345151, A345175, A345180, A345184, A345769.
%K A345181 nonn
%O A345181 1,1
%A A345181 _David Consiglio, Jr._, Jun 10 2021