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%I #6 Aug 05 2021 15:21:09
%S 1185,1243,1262,1288,1295,1299,1386,1393,1397,1400,1412,1419,1423,
%T 1448,1449,1451,1458,1460,1464,1467,1475,1477,1497,1501,1503,1504,
%U 1505,1512,1513,1514,1516,1521,1523,1539,1540,1541,1542,1553,1558,1559,1560,1565,1566
%N Numbers that are the sum of eight cubes in ten or more ways.
%H Sean A. Irvine, <a href="/A345540/b345540.txt">Table of n, a(n) for n = 1..10000</a>
%e 1243 is a term because 1243 = 1^3 + 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 5^3 + 9^3 = 1^3 + 1^3 + 1^3 + 3^3 + 3^3 + 4^3 + 7^3 + 7^3 = 1^3 + 1^3 + 2^3 + 2^3 + 3^3 + 6^3 + 6^3 + 7^3 = 1^3 + 1^3 + 2^3 + 2^3 + 3^3 + 5^3 + 5^3 + 8^3 = 1^3 + 1^3 + 4^3 + 4^3 + 5^3 + 5^3 + 5^3 + 6^3 = 1^3 + 2^3 + 3^3 + 5^3 + 5^3 + 5^3 + 5^3 + 6^3 = 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 9^3 = 2^3 + 3^3 + 3^3 + 3^3 + 4^3 + 6^3 + 6^3 + 6^3 = 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3 + 8^3 = 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 5^3 + 6^3 + 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, 8):
%o tot = sum(pos)
%o keep[tot] += 1
%o rets = sorted([k for k, v in keep.items() if v >= 10])
%o for x in range(len(rets)):
%o print(rets[x])
%Y Cf. A345497, A345506, A345539, A345549, A345585, A345792.
%K nonn
%O 1,1
%A _David Consiglio, Jr._, Jun 20 2021
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