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A345798 Numbers that are the sum of nine cubes in exactly six ways.

%I A345798 #6 Jul 31 2021 22:32:51
%S A345798 472,498,505,507,524,596,598,605,636,643,655,661,662,669,672,676,681,
%T A345798 688,690,692,696,706,718,722,725,728,729,731,732,737,739,742,748,749,
%U A345798 750,751,756,765,772,782,783,785,787,788,791,793,794,800,801,802,808,810
%N A345798 Numbers that are the sum of nine cubes in exactly six ways.
%C A345798 Differs from A345545 at term 9 because 624 = 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 8^3  = 1^3 + 1^3 + 1^3 + 3^3 + 4^3 + 4^3 + 5^3 + 5^3 + 6^3  = 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 4^3 + 4^3 + 6^3 + 6^3  = 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 5^3 + 5^3 + 5^3 + 6^3  = 1^3 + 2^3 + 2^3 + 2^3 + 4^3 + 4^3 + 4^3 + 4^3 + 7^3  = 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 5^3 + 6^3 + 6^3  = 3^3 + 3^3 + 4^3 + 4^3 + 4^3 + 4^3 + 4^3 + 5^3 + 5^3.
%C A345798 Likely finite.
%H A345798 Sean A. Irvine, <a href="/A345798/b345798.txt">Table of n, a(n) for n = 1..127</a>
%e A345798 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 A345798 (Python)
%o A345798 from itertools import combinations_with_replacement as cwr
%o A345798 from collections import defaultdict
%o A345798 keep = defaultdict(lambda: 0)
%o A345798 power_terms = [x**3 for x in range(1, 1000)]
%o A345798 for pos in cwr(power_terms, 9):
%o A345798     tot = sum(pos)
%o A345798     keep[tot] += 1
%o A345798     rets = sorted([k for k, v in keep.items() if v == 6])
%o A345798     for x in range(len(rets)):
%o A345798         print(rets[x])
%Y A345798 Cf. A345545, A345788, A345797, A345799, A345808, A345848.
%K A345798 nonn
%O A345798 1,1
%A A345798 _David Consiglio, Jr._, Jun 26 2021