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A345085 Numbers that are the sum of three third powers in exactly seven ways.

%I A345085 #17 Jul 29 2023 14:35:43
%S A345085 2016496,4525632,4783680,5268024,6366816,7451352,7457120,8275392,
%T A345085 9063144,9086104,9931167,10036872,10266138,10371024,10973880,12002472,
%U A345085 12452049,12983517,13639816,13641480,13818384,13832729,14090112,15081984,15212016,15685704,16131968
%N A345085 Numbers that are the sum of three third powers in exactly seven ways.
%C A345085 Differs from A345086 at term 2 because 2562624 = 7^3 + 35^3 + 135^3 = 7^3 + 63^3 + 131^3 = 11^3 + 99^3 + 115^3 = 16^3 + 45^3 + 134^3 = 29^3 + 102^3 + 112^3 = 35^3 + 59^3 + 131^3 = 50^3 + 84^3 + 121^3 = 68^3 + 71^3 + 122^3.
%H A345085 David Consiglio, Jr., <a href="/A345085/b345085.txt">Table of n, a(n) for n = 1..100</a>
%e A345085 2016496 is a term because 2016496 = 5^3 + 71^3 + 117^3 = 9^3 + 65^3 + 119^3 = 18^3 + 20^3 + 125^3 = 46^3 + 96^3 + 99^3 = 53^3 + 59^3 + 117^3 = 65^3 + 89^3 + 99^3 = 82^3 + 84^3 + 93^3.
%o A345085 (Python)
%o A345085 from itertools import combinations_with_replacement as cwr
%o A345085 from collections import defaultdict
%o A345085 keep = defaultdict(lambda: 0)
%o A345085 power_terms = [x**3 for x in range(1, 1000)]
%o A345085 for pos in cwr(power_terms, 3):
%o A345085     tot = sum(pos)
%o A345085     keep[tot] += 1
%o A345085 rets = sorted([k for k, v in keep.items() if v == 7])
%o A345085 for x in range(len(rets)):
%o A345085     print(rets[x])
%Y A345085 Cf. A025327, A344730, A345084, A345086, A345088, A345151.
%K A345085 nonn
%O A345085 1,1
%A A345085 _David Consiglio, Jr._, Jun 07 2021