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%I #7 Aug 05 2021 15:22:39
%S 1496,1648,1704,1711,1720,1737,1772,1774,1781,1800,1802,1835,1837,
%T 1844,1863,1882,1889,1891,1893,1898,1900,1907,1912,1919,1926,1938,
%U 1945,1952,1954,1961,1963,1982,1989,1996,2000,2008,2012,2015,2019,2026,2045,2052,2053
%N Numbers that are the sum of seven cubes in nine or more ways.
%H Sean A. Irvine, <a href="/A345527/b345527.txt">Table of n, a(n) for n = 1..10000</a>
%e 1648 is a term because 1648 = 1^3 + 1^3 + 1^3 + 2^3 + 4^3 + 7^3 + 9^3 = 1^3 + 1^3 + 1^3 + 3^3 + 4^3 + 4^3 + 10^3 = 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 5^3 + 10^3 = 1^3 + 1^3 + 3^3 + 4^3 + 5^3 + 7^3 + 8^3 = 1^3 + 2^3 + 2^3 + 5^3 + 6^3 + 6^3 + 8^3 = 1^3 + 3^3 + 3^3 + 4^3 + 4^3 + 6^3 + 9^3 = 2^3 + 3^3 + 3^3 + 3^3 + 5^3 + 6^3 + 9^3 = 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 8^3 + 8^3 = 3^3 + 3^3 + 3^3 + 5^3 + 5^3 + 7^3 + 7^3.
%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, 7):
%o tot = sum(pos)
%o keep[tot] += 1
%o rets = sorted([k for k, v in keep.items() if v >= 9])
%o for x in range(len(rets)):
%o print(rets[x])
%Y Cf. A345486, A345506, A345518, A345526, A345539, A345575, A345781.
%K nonn
%O 1,1
%A _David Consiglio, Jr._, Jun 20 2021
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