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%I #7 Aug 05 2021 15:22:35
%S 1385,1496,1515,1552,1557,1585,1587,1603,1613,1622,1648,1655,1665,
%T 1674,1681,1704,1711,1718,1719,1720,1737,1739,1741,1746,1753,1755,
%U 1765,1767,1772,1774,1781,1782,1793,1800,1802,1805,1809,1811,1818,1819,1826,1828,1830
%N Numbers that are the sum of seven cubes in eight or more ways.
%H Sean A. Irvine, <a href="/A345526/b345526.txt">Table of n, a(n) for n = 1..10000</a>
%e 1496 is a term because 1496 = 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 8^3 + 8^3 = 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 3^3 + 10^3 = 1^3 + 1^3 + 3^3 + 3^3 + 4^3 + 7^3 + 8^3 = 1^3 + 2^3 + 2^3 + 3^3 + 6^3 + 6^3 + 8^3 = 1^3 + 4^3 + 4^3 + 5^3 + 6^3 + 6^3 + 6^3 = 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 4^3 + 10^3 = 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 6^3 + 9^3 = 2^3 + 3^3 + 5^3 + 5^3 + 6^3 + 6^3 + 6^3 = 3^3 + 3^3 + 3^3 + 3^3 + 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 >= 8])
%o for x in range(len(rets)):
%o print(rets[x])
%Y Cf. A345485, A345517, A345525, A345527, A345538, A345574, A345780.
%K nonn
%O 1,1
%A _David Consiglio, Jr._, Jun 20 2021
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