cp's OEIS Frontend

A345787 Numbers that are the sum of eight cubes in exactly five ways.

%I A345787 #6 Jul 31 2021 22:37:17
%S A345787 471,497,504,597,623,630,635,642,649,654,661,667,680,686,691,693,712,
%T A345787 717,723,728,736,738,741,743,752,754,755,762,774,780,781,783,784,785,
%U A345787 788,791,793,797,800,802,804,810,813,814,815,817,819,820,821,830,834,837
%N A345787 Numbers that are the sum of eight cubes in exactly five ways.
%C A345787 Differs from A345535 at term 6 because 628 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 4^3 + 6^3 + 7^3  = 1^3 + 1^3 + 1^3 + 5^3 + 5^3 + 5^3 + 5^3 + 5^3  = 1^3 + 1^3 + 2^3 + 3^3 + 5^3 + 5^3 + 5^3 + 6^3  = 1^3 + 1^3 + 3^3 + 4^3 + 4^3 + 4^3 + 4^3 + 7^3  = 1^3 + 2^3 + 2^3 + 3^3 + 3^3 + 5^3 + 6^3 + 6^3  = 2^3 + 2^3 + 2^3 + 2^3 + 4^3 + 4^3 + 5^3 + 7^3.
%C A345787 Likely finite.
%H A345787 Sean A. Irvine, <a href="/A345787/b345787.txt">Table of n, a(n) for n = 1..180</a>
%e A345787 497 is a term because 497 = 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 4^3 + 4^3 + 5^3 = 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 5^3 + 5^3 = 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 6^3 = 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3 = 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 5^3.
%o A345787 (Python)
%o A345787 from itertools import combinations_with_replacement as cwr
%o A345787 from collections import defaultdict
%o A345787 keep = defaultdict(lambda: 0)
%o A345787 power_terms = [x**3 for x in range(1, 1000)]
%o A345787 for pos in cwr(power_terms, 8):
%o A345787     tot = sum(pos)
%o A345787     keep[tot] += 1
%o A345787     rets = sorted([k for k, v in keep.items() if v == 5])
%o A345787     for x in range(len(rets)):
%o A345787         print(rets[x])
%Y A345787 Cf. A345535, A345777, A345786, A345788, A345797, A345837.
%K A345787 nonn
%O A345787 1,1
%A A345787 _David Consiglio, Jr._, Jun 26 2021