cp's OEIS Frontend

A345556 Numbers that are the sum of ten cubes in eight or more ways.

%I A345556 #6 Aug 05 2021 15:17:00
%S A345556 623,625,630,632,644,651,658,662,665,677,684,688,695,697,699,708,714,
%T A345556 715,721,723,725,728,730,733,734,736,740,745,747,749,751,752,754,756,
%U A345556 757,758,759,760,764,766,769,771,773,775,776,777,778,780,782,785,786,787
%N A345556 Numbers that are the sum of ten cubes in eight or more ways.
%H A345556 Sean A. Irvine, <a href="/A345556/b345556.txt">Table of n, a(n) for n = 1..10000</a>
%e A345556 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 A345556 (Python)
%o A345556 from itertools import combinations_with_replacement as cwr
%o A345556 from collections import defaultdict
%o A345556 keep = defaultdict(lambda: 0)
%o A345556 power_terms = [x**3 for x in range(1, 1000)]
%o A345556 for pos in cwr(power_terms, 10):
%o A345556     tot = sum(pos)
%o A345556     keep[tot] += 1
%o A345556     rets = sorted([k for k, v in keep.items() if v >= 8])
%o A345556     for x in range(len(rets)):
%o A345556         print(rets[x])
%Y A345556 Cf. A345547, A345555, A345557, A345601, A345810, A346807.
%K A345556 nonn
%O A345556 1,1
%A A345556 _David Consiglio, Jr._, Jun 20 2021