cp's OEIS Frontend

A345770 Numbers that are the sum of six cubes in exactly eight ways.

%I A345770 #6 Jul 31 2021 22:50:00
%S A345770 1981,2105,2168,2277,2368,2376,2431,2466,2538,2557,2583,2646,2665,
%T A345770 2672,2746,2753,2763,2765,2880,2881,2916,2961,2970,2977,2979,2987,
%U A345770 3007,3040,3042,3049,3068,3088,3141,3159,3169,3185,3248,3278,3311,3312,3367,3384,3393
%N A345770 Numbers that are the sum of six cubes in exactly eight ways.
%C A345770 Differs from A345517 at term 8 because 2438 = 1^3 + 2^3 + 2^3 + 2^3 + 6^3 + 13^3  = 1^3 + 2^3 + 4^3 + 5^3 + 8^3 + 12^3  = 1^3 + 5^3 + 5^3 + 9^3 + 9^3 + 9^3  = 2^3 + 2^3 + 2^3 + 7^3 + 7^3 + 12^3  = 2^3 + 2^3 + 3^3 + 4^3 + 10^3 + 11^3  = 2^3 + 3^3 + 6^3 + 9^3 + 9^3 + 9^3  = 2^3 + 4^3 + 5^3 + 8^3 + 9^3 + 10^3  = 4^3 + 4^3 + 5^3 + 5^3 + 9^3 + 11^3  = 6^3 + 7^3 + 7^3 + 8^3 + 8^3 + 8^3.
%H A345770 Sean A. Irvine, <a href="/A345770/b345770.txt">Table of n, a(n) for n = 1..1347</a>
%e A345770 2105 is a term because 2105 = 1^3 + 1^3 + 4^3 + 4^3 + 4^3 + 11^3 = 1^3 + 2^3 + 3^3 + 4^3 + 5^3 + 11^3 = 1^3 + 2^3 + 6^3 + 7^3 + 7^3 + 8^3 = 1^3 + 4^3 + 4^3 + 4^3 + 8^3 + 9^3 = 1^3 + 4^3 + 5^3 + 5^3 + 5^3 + 10^3 = 2^3 + 3^3 + 4^3 + 5^3 + 8^3 + 9^3 = 3^3 + 3^3 + 3^3 + 7^3 + 7^3 + 9^3 = 5^3 + 5^3 + 5^3 + 5^3 + 7^3 + 8^3.
%o A345770 (Python)
%o A345770 from itertools import combinations_with_replacement as cwr
%o A345770 from collections import defaultdict
%o A345770 keep = defaultdict(lambda: 0)
%o A345770 power_terms = [x**3 for x in range(1, 1000)]
%o A345770 for pos in cwr(power_terms, 6):
%o A345770     tot = sum(pos)
%o A345770     keep[tot] += 1
%o A345770     rets = sorted([k for k, v in keep.items() if v == 8])
%o A345770     for x in range(len(rets)):
%o A345770         print(rets[x])
%Y A345770 Cf. A345184, A345517, A345769, A345771, A345780, A345820.
%K A345770 nonn
%O A345770 1,1
%A A345770 _David Consiglio, Jr._, Jun 26 2021