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A345149 Numbers that are the sum of four third powers in exactly six ways.

%I A345149 #7 Jul 31 2021 23:29:03
%S A345149 6883,12411,13923,14112,14581,14896,14904,15561,15876,16317,16640,
%T A345149 17208,17479,17992,18739,18865,19035,19080,19665,19712,19763,19880,
%U A345149 20007,20384,20979,21231,21420,21707,22409,22617,23149,23940,24355,25515,25984,26208,26334
%N A345149 Numbers that are the sum of four third powers in exactly six ways.
%C A345149 Differs from A345148 at term 3 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.
%H A345149 David Consiglio, Jr., <a href="/A345149/b345149.txt">Table of n, a(n) for n = 1..10000</a>
%e A345149 6883 is a term because 6883 = 2^3 + 2^3 + 2^3 + 18^3  = 2^3 + 4^3 + 14^3 + 14^3  = 3^3 + 7^3 + 7^3 + 17^3  = 3^3 + 10^3 + 13^3 + 13^3  = 4^3 + 10^3 + 10^3 + 15^3  = 7^3 + 8^3 + 8^3 + 16^3.
%o A345149 (Python)
%o A345149 from itertools import combinations_with_replacement as cwr
%o A345149 from collections import defaultdict
%o A345149 keep = defaultdict(lambda: 0)
%o A345149 power_terms = [x**3 for x in range(1, 1000)]
%o A345149 for pos in cwr(power_terms, 4):
%o A345149     tot = sum(pos)
%o A345149     keep[tot] += 1
%o A345149 rets = sorted([k for k, v in keep.items() if v == 6])
%o A345149 for x in range(len(rets)):
%o A345149     print(rets[x])
%Y A345149 Cf. A025362, A343986, A344921, A345084, A345148, A345151, A345175.
%K A345149 nonn
%O A345149 1,1
%A A345149 _David Consiglio, Jr._, Jun 09 2021