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