cp's OEIS Frontend

A345810 Numbers that are the sum of ten cubes in exactly eight ways.

%I A345810 #6 Jul 31 2021 22:27:34
%S A345810 623,625,630,644,662,665,677,684,697,699,708,715,723,725,728,730,733,
%T A345810 734,747,749,751,757,758,759,760,764,766,769,775,776,777,785,786,787,
%U A345810 789,793,794,796,804,810,811,814,817,820,826,827,828,829,830,831,836,838
%N A345810 Numbers that are the sum of ten cubes in exactly eight ways.
%C A345810 Differs from A345556 at term 4 because 632 = 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 8^3  = 1^3 + 1^3 + 1^3 + 1^3 + 4^3 + 4^3 + 5^3 + 5^3 + 5^3 + 5^3  = 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 4^3 + 4^3 + 5^3 + 5^3 + 6^3  = 1^3 + 1^3 + 2^3 + 2^3 + 3^3 + 3^3 + 4^3 + 4^3 + 6^3 + 6^3  = 1^3 + 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 5^3 + 7^3  = 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 5^3 + 5^3 + 5^3 + 6^3  = 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 4^3 + 4^3 + 4^3 + 4^3 + 7^3  = 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 5^3 + 6^3 + 6^3  = 2^3 + 3^3 + 3^3 + 4^3 + 4^3 + 4^3 + 4^3 + 4^3 + 5^3 + 5^3.
%C A345810 Likely finite.
%H A345810 Sean A. Irvine, <a href="/A345810/b345810.txt">Table of n, a(n) for n = 1..75</a>
%e A345810 625 is a term because 625 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 7^3 = 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3 + 5^3 = 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 5^3 + 5^3 = 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 4^3 + 4^3 + 4^3 + 5^3 = 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 6^3 = 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 4^3 + 5^3 + 5^3 = 1^3 + 3^3 + 3^3 + 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 + 3^3 + 3^3 + 5^3.
%o A345810 (Python)
%o A345810 from itertools import combinations_with_replacement as cwr
%o A345810 from collections import defaultdict
%o A345810 keep = defaultdict(lambda: 0)
%o A345810 power_terms = [x**3 for x in range(1, 1000)]
%o A345810 for pos in cwr(power_terms, 10):
%o A345810     tot = sum(pos)
%o A345810     keep[tot] += 1
%o A345810     rets = sorted([k for k, v in keep.items() if v == 8])
%o A345810     for x in range(len(rets)):
%o A345810         print(rets[x])
%Y A345810 Cf. A345556, A345800, A345809, A345811, A345860.
%K A345810 nonn
%O A345810 1,1
%A A345810 _David Consiglio, Jr._, Jun 26 2021