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A345555 Numbers that are the sum of ten cubes in seven or more ways.

%I A345555 #6 Aug 05 2021 15:16:57
%S A345555 440,473,499,506,525,532,534,567,571,584,588,597,599,604,606,623,625,
%T A345555 627,630,632,637,639,640,644,651,656,658,660,662,663,665,669,670,673,
%U A345555 677,680,682,684,688,689,691,693,695,696,697,699,701,702,704,707,708,714
%N A345555 Numbers that are the sum of ten cubes in seven or more ways.
%H A345555 Sean A. Irvine, <a href="/A345555/b345555.txt">Table of n, a(n) for n = 1..10000</a>
%e A345555 473 is a term because 473 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 4^3 + 4^3 + 5^3 = 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 5^3 + 5^3 = 1^3 + 1^3 + 1^3 + 1^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3 = 1^3 + 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 5^3 = 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 4^3 + 4^3 + 4^3 = 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 4^3 + 5^3 = 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3 + 3^3.
%o A345555 (Python)
%o A345555 from itertools import combinations_with_replacement as cwr
%o A345555 from collections import defaultdict
%o A345555 keep = defaultdict(lambda: 0)
%o A345555 power_terms = [x**3 for x in range(1, 1000)]
%o A345555 for pos in cwr(power_terms, 10):
%o A345555     tot = sum(pos)
%o A345555     keep[tot] += 1
%o A345555     rets = sorted([k for k, v in keep.items() if v >= 7])
%o A345555     for x in range(len(rets)):
%o A345555         print(rets[x])
%Y A345555 Cf. A345546, A345554, A345556, A345600, A345809, A346806.
%K A345555 nonn
%O A345555 1,1
%A A345555 _David Consiglio, Jr._, Jun 20 2021