cp's OEIS Frontend

A345811 Numbers that are the sum of ten cubes in exactly nine ways.

%I A345811 #6 Jul 31 2021 22:27:38
%S A345811 632,651,658,688,695,714,736,740,745,752,773,778,780,790,795,799,801,
%T A345811 812,813,815,816,818,821,823,825,841,843,849,851,852,853,855,856,857,
%U A345811 858,864,866,873,880,882,883,885,890,891,892,899,905,908,913,922,924,926
%N A345811 Numbers that are the sum of ten cubes in exactly nine ways.
%C A345811 Differs from A345557 at term 7 because 721 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 5^3 + 5^3 + 5^3 + 5^3 + 6^3  = 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 5^3 + 5^3 + 6^3 + 6^3  = 1^3 + 1^3 + 1^3 + 2^3 + 2^3 + 3^3 + 3^3 + 6^3 + 6^3 + 6^3  = 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 4^3 + 4^3 + 6^3 + 7^3  = 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3 + 8^3  = 1^3 + 1^3 + 3^3 + 4^3 + 4^3 + 4^3 + 5^3 + 5^3 + 5^3 + 5^3  = 1^3 + 2^3 + 3^3 + 3^3 + 4^3 + 4^3 + 4^3 + 5^3 + 5^3 + 6^3  = 2^3 + 2^3 + 2^3 + 2^3 + 4^3 + 5^3 + 5^3 + 5^3 + 5^3 + 5^3  = 2^3 + 2^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3 + 4^3 + 6^3 + 6^3  = 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3 + 5^3 + 7^3.
%C A345811 Likely finite.
%H A345811 Sean A. Irvine, <a href="/A345811/b345811.txt">Table of n, a(n) for n = 1..79</a>
%e A345811 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 A345811 (Python)
%o A345811 from itertools import combinations_with_replacement as cwr
%o A345811 from collections import defaultdict
%o A345811 keep = defaultdict(lambda: 0)
%o A345811 power_terms = [x**3 for x in range(1, 1000)]
%o A345811 for pos in cwr(power_terms, 10):
%o A345811     tot = sum(pos)
%o A345811     keep[tot] += 1
%o A345811     rets = sorted([k for k, v in keep.items() if v == 9])
%o A345811     for x in range(len(rets)):
%o A345811         print(rets[x])
%Y A345811 Cf. A345557, A345801, A345810, A345812, A345861.
%K A345811 nonn
%O A345811 1,1
%A A345811 _David Consiglio, Jr._, Jun 26 2021