|
%I #6 Aug 05 2021 15:20:53
%S 471,497,504,597,623,628,630,635,642,649,654,661,667,680,686,691,693,
%T 712,717,719,723,728,736,738,741,743,752,754,755,762,769,774,776,778,
%U 780,781,783,784,785,788,791,793,795,797,800,802,804,810,813,814,815,817
%N Numbers that are the sum of eight cubes in five or more ways.
%H Sean A. Irvine, <a href="/A345535/b345535.txt">Table of n, a(n) for n = 1..10000</a>
%e 497 is a term because 497 = 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 4^3 + 4^3 + 5^3 = 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 5^3 + 5^3 = 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 6^3 = 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3 = 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 5^3.
%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 >= 5])
%o for x in range(len(rets)):
%o print(rets[x])
%Y Cf. A345492, A345523, A345534, A345536, A345544, A345580, A345787.
%K nonn
%O 1,1
%A _David Consiglio, Jr._, Jun 20 2021
|
