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A345506 Numbers that are the sum of seven cubes in ten or more ways.

%I A345506 #6 Aug 05 2021 15:22:12
%S A345506 1704,1711,1774,1800,1837,1863,1889,1893,1926,1938,1963,1982,1989,
%T A345506 2008,2015,2019,2045,2052,2053,2059,2078,2097,2106,2113,2143,2161,
%U A345506 2169,2171,2176,2197,2204,2217,2223,2224,2227,2230,2234,2241,2250,2260,2266,2267,2276
%N A345506 Numbers that are the sum of seven cubes in ten or more ways.
%H A345506 Sean A. Irvine, <a href="/A345506/b345506.txt">Table of n, a(n) for n = 1..10000</a>
%e A345506 1711 is a term because 1711 = 1^3 + 1^3 + 1^3 + 4^3 + 4^3 + 8^3 + 8^3 = 1^3 + 1^3 + 2^3 + 3^3 + 5^3 + 8^3 + 8^3 = 1^3 + 1^3 + 2^3 + 3^3 + 4^3 + 7^3 + 9^3 = 1^3 + 1^3 + 3^3 + 3^3 + 4^3 + 4^3 + 10^3 = 1^3 + 2^3 + 2^3 + 2^3 + 6^3 + 6^3 + 9^3 = 1^3 + 2^3 + 3^3 + 3^3 + 3^3 + 5^3 + 10^3 = 1^3 + 3^3 + 3^3 + 4^3 + 5^3 + 7^3 + 8^3 = 2^3 + 2^3 + 3^3 + 5^3 + 6^3 + 6^3 + 8^3 = 3^3 + 3^3 + 3^3 + 4^3 + 4^3 + 6^3 + 9^3 = 4^3 + 4^3 + 5^3 + 5^3 + 6^3 + 6^3 + 6^3.
%o A345506 (Python)
%o A345506 from itertools import combinations_with_replacement as cwr
%o A345506 from collections import defaultdict
%o A345506 keep = defaultdict(lambda: 0)
%o A345506 power_terms = [x**3 for x in range(1, 1000)]
%o A345506 for pos in cwr(power_terms, 7):
%o A345506     tot = sum(pos)
%o A345506     keep[tot] += 1
%o A345506     rets = sorted([k for k, v in keep.items() if v >= 10])
%o A345506     for x in range(len(rets)):
%o A345506         print(rets[x])
%Y A345506 Cf. A345487, A345519, A345527, A345540, A345576, A345782.
%K A345506 nonn
%O A345506 1,1
%A A345506 _David Consiglio, Jr._, Jun 20 2021