cp's OEIS Frontend

A345767 Numbers that are the sum of six cubes in exactly five ways.

%I A345767 #6 Jul 31 2021 22:49:49
%S A345767 1045,1169,1241,1260,1384,1432,1440,1495,1530,1539,1549,1556,1558,
%T A345767 1584,1594,1602,1612,1617,1640,1654,1657,1675,1703,1712,1715,1719,
%U A345767 1729,1736,1745,1747,1754,1771,1780,1792,1801,1803,1806,1810,1818,1825,1827,1834,1843
%N A345767 Numbers that are the sum of six cubes in exactly five ways.
%C A345767 Differs from A345514 at term 5 because 1377 = 1^3 + 1^3 + 2^3 + 7^3 + 8^3 + 8^3  = 1^3 + 1^3 + 5^3 + 5^3 + 5^3 + 10^3  = 1^3 + 2^3 + 3^3 + 5^3 + 6^3 + 10^3  = 1^3 + 6^3 + 6^3 + 6^3 + 6^3 + 8^3  = 3^3 + 3^3 + 5^3 + 7^3 + 7^3 + 8^3  = 3^3 + 4^3 + 5^3 + 6^3 + 6^3 + 9^3.
%H A345767 Sean A. Irvine, <a href="/A345767/b345767.txt">Table of n, a(n) for n = 1..1227</a>
%e A345767 1169 is a term because 1169 = 1^3 + 2^3 + 2^3 + 3^3 + 4^3 + 9^3 = 1^3 + 2^3 + 5^3 + 5^3 + 5^3 + 7^3 = 1^3 + 3^3 + 4^3 + 4^3 + 4^3 + 8^3 = 2^3 + 3^3 + 3^3 + 4^3 + 5^3 + 8^3 = 3^3 + 3^3 + 3^3 + 3^3 + 7^3 + 7^3.
%o A345767 (Python)
%o A345767 from itertools import combinations_with_replacement as cwr
%o A345767 from collections import defaultdict
%o A345767 keep = defaultdict(lambda: 0)
%o A345767 power_terms = [x**3 for x in range(1, 1000)]
%o A345767 for pos in cwr(power_terms, 6):
%o A345767     tot = sum(pos)
%o A345767     keep[tot] += 1
%o A345767     rets = sorted([k for k, v in keep.items() if v == 5])
%o A345767     for x in range(len(rets)):
%o A345767         print(rets[x])
%Y A345767 Cf. A343988, A345514, A345766, A345768, A345777, A345817.
%K A345767 nonn
%O A345767 1,1
%A A345767 _David Consiglio, Jr._, Jun 26 2021