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A345151 Numbers that are the sum of four third powers in exactly seven ways.

%I A345151 #7 Jul 31 2021 23:29:06
%S A345151 13104,18928,19376,20755,21203,22743,24544,24570,24787,25172,25928,
%T A345151 27755,27846,28917,29582,31031,31248,31528,32858,34056,34713,35289,
%U A345151 35317,35441,35497,35712,36190,36288,36610,36890,36946,38080,39221,39440,39464,39851,39942
%N A345151 Numbers that are the sum of four third powers in exactly seven ways.
%C A345151 Differs from A345150 at term 6 because 21896 = 1^3 + 11^3 + 19^3 + 22^3  = 2^3 + 2^3 + 12^3 + 26^3  = 2^3 + 3^3 + 19^3 + 23^3  = 2^3 + 5^3 + 15^3 + 25^3  = 3^3 + 10^3 + 16^3 + 24^3  = 3^3 + 17^3 + 19^3 + 19^3  = 4^3 + 6^3 + 20^3 + 22^3  = 5^3 + 8^3 + 14^3 + 25^3  = 7^3 + 11^3 + 17^3 + 23^3  = 8^3 + 9^3 + 19^3 + 22^3.
%H A345151 David Consiglio, Jr., <a href="/A345151/b345151.txt">Table of n, a(n) for n = 1..10000</a>
%e A345151 13104 is a term because 13104 = 1^3 + 10^3 + 16^3 + 18^3  = 1^3 + 11^3 + 14^3 + 19^3  = 2^3 + 9^3 + 15^3 + 19^3  = 4^3 + 6^3 + 14^3 + 20^3  = 4^3 + 9^3 + 10^3 + 21^3  = 5^3 + 7^3 + 11^3 + 21^3  = 8^3 + 9^3 + 14^3 + 19^3.
%o A345151 (Python)
%o A345151 from itertools import combinations_with_replacement as cwr
%o A345151 from collections import defaultdict
%o A345151 keep = defaultdict(lambda: 0)
%o A345151 power_terms = [x**3 for x in range(1, 1000)]
%o A345151 for pos in cwr(power_terms, 4):
%o A345151     tot = sum(pos)
%o A345151     keep[tot] += 1
%o A345151 rets = sorted([k for k, v in keep.items() if v == 7])
%o A345151 for x in range(len(rets)):
%o A345151     print(rets[x])
%Y A345151 Cf. A025363, A344923, A345085, A345149, A345150, A345153, A345181.
%K A345151 nonn
%O A345151 1,1
%A A345151 _David Consiglio, Jr._, Jun 09 2021