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%I #6 Aug 05 2021 15:17:04
%S 632,651,658,688,695,714,721,736,740,745,752,754,756,771,773,778,780,
%T 782,790,792,795,797,799,801,806,808,812,813,815,816,818,819,821,823,
%U 825,832,834,841,843,845,847,848,849,850,851,852,853,855,856,857,858,860
%N Numbers that are the sum of ten cubes in nine or more ways.
%H Sean A. Irvine, <a href="/A345557/b345557.txt">Table of n, a(n) for n = 1..10000</a>
%e 651 is a term because 651 = 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 3^3 + 5^3 + 6^3 = 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 7^3 = 1^3 + 1^3 + 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3 + 5^3 = 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 4^3 + 4^3 + 4^3 + 4^3 + 4^3 = 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 5^3 + 5^3 = 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 4^3 + 4^3 + 4^3 + 5^3 = 1^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 6^3 = 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 4^3 + 5^3 + 5^3 = 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3.
%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, 10):
%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. A345548, A345556, A345558, A345602, A345811.
%K nonn
%O 1,1
%A _David Consiglio, Jr._, Jun 20 2021
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