A345545 - OEIS

%I #6 Aug 05 2021 15:19:13

%S 472,498,505,507,524,596,598,605,624,629,631,636,643,650,655,657,661,

%T 662,669,672,676,681,687,688,690,692,694,696,706,707,713,718,720,722,

%U 725,727,728,729,731,732,737,739,742,744,746,748,749,750,751,753,755,756

%N Numbers that are the sum of nine cubes in six or more ways.

%H Sean A. Irvine, <a href="/A345545/b345545.txt">Table of n, a(n) for n = 1..10000</a>

%e 498 is a term because 498 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 4^3 + 4^3 + 5^3 = 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 5^3 + 5^3 = 1^3 + 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 6^3 = 1^3 + 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3 = 1^3 + 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 5^3 = 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 4^3 + 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, 9):

%o     tot = sum(pos)

%o     keep[tot] += 1

%o     rets = sorted([k for k, v in keep.items() if v >= 6])

%o     for x in range(len(rets)):

%o         print(rets[x])

%Y Cf. A345503, A345536, A345544, A345546, A345554, A345590, A345798.

%K nonn

%O 1,1

%A _David Consiglio, Jr._, Jun 20 2021