A345788 - OEIS

%I #6 Jul 31 2021 22:37:21

%S 628,719,769,776,778,795,832,839,846,858,860,865,872,875,876,882,886,

%T 891,893,895,901,907,912,927,928,931,945,946,947,951,954,956,964,965,

%U 972,989,992,998,999,1001,1012,1014,1015,1021,1034,1035,1036,1038,1040,1045

%N Numbers that are the sum of eight cubes in exactly six ways.

%C Differs from A345536 at term 22 because 902 = 1^3 + 1^3 + 2^3 + 2^3 + 3^3 + 4^3 + 4^3 + 9^3  = 1^3 + 1^3 + 3^3 + 4^3 + 5^3 + 5^3 + 6^3 + 7^3  = 1^3 + 2^3 + 3^3 + 3^3 + 4^3 + 6^3 + 6^3 + 7^3  = 1^3 + 2^3 + 4^3 + 4^3 + 4^3 + 4^3 + 5^3 + 8^3  = 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 5^3 + 9^3  = 2^3 + 2^3 + 2^3 + 4^3 + 4^3 + 4^3 + 7^3 + 7^3  = 3^3 + 5^3 + 5^3 + 5^3 + 5^3 + 5^3 + 5^3 + 5^3.

%C Likely finite.

%H Sean A. Irvine, <a href="/A345788/b345788.txt">Table of n, a(n) for n = 1..173</a>

%e 719 is a term because 719 = 1^3 + 1^3 + 1^3 + 4^3 + 4^3 + 4^3 + 4^3 + 5^3 = 1^3 + 1^3 + 2^3 + 3^3 + 4^3 + 4^3 + 5^3 + 5^3 = 1^3 + 2^3 + 2^3 + 3^3 + 3^3 + 5^3 + 5^3 + 5^3 = 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 5^3 + 6^3 = 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 7^3 = 3^3 + 3^3 + 3^3 + 3^3 + 4^3 + 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, 8):

%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. A345536, A345778, A345787, A345789, A345798, A345838.

%K nonn

%O 1,1

%A _David Consiglio, Jr._, Jun 26 2021