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%I #6 Jul 31 2021 23:16:42
%S 2430,2979,3214,3249,3312,3492,3520,3737,3753,3788,3816,3842,3942,
%T 3968,4121,4185,4213,4267,4355,4411,4418,4446,4453,4456,4465,4482,
%U 4509,4563,4626,4663,4670,4723,4753,4896,4905,4924,4938,4941,4950,4960,4976,4987,4994
%N Numbers that are the sum of five third powers in exactly six ways.
%C Differs from A345174 at term 20 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 David Consiglio, Jr., <a href="/A345175/b345175.txt">Table of n, a(n) for n = 1..10000</a>
%e 2430 is a term because 2430 = 1^3 + 2^3 + 2^3 + 5^3 + 12^3 = 1^3 + 3^3 + 4^3 + 7^3 + 11^3 = 2^3 + 2^3 + 6^3 + 6^3 + 11^3 = 2^3 + 3^3 + 3^3 + 9^3 + 10^3 = 3^3 + 5^3 + 8^3 + 8^3 + 8^3 = 3^3 + 4^3 + 7^3 + 8^3 + 9^3.
%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 == 6])
%o for x in range(len(rets)):
%o print(rets[x])
%Y Cf. A294740, A343988, A344941, A345149, A345174, A345181, A345768.
%K nonn
%O 1,1
%A _David Consiglio, Jr._, Jun 10 2021
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