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A345155 Numbers that are the sum of four third powers in ten or more ways.

%I A345155 #6 Aug 05 2021 15:27:37
%S A345155 21896,36225,46872,48321,48825,51506,52416,53200,55575,58338,58968,
%T A345155 59059,60480,62244,66024,67536,67851,70434,70525,71155,72819,73808,
%U A345155 76384,76923,77896,78624,78912,81081,81991,85995,87507,88641,90181,90783,91448,91728,92008
%N A345155 Numbers that are the sum of four third powers in ten or more ways.
%H A345155 David Consiglio, Jr., <a href="/A345155/b345155.txt">Table of n, a(n) for n = 1..10000</a>
%e A345155 21896 is a term because 21896 = 1^3 + 11^3 + 19^3 + 22^3  = 2^3 + 2^3 + 12^3 + 26^3  = 2^3 + 3^3 + 19^3 + 23^3  = 2^3 + 5^3 + 15^3 + 25^3  = 3^3 + 10^3 + 16^3 + 24^3  = 3^3 + 17^3 + 19^3 + 19^3  = 4^3 + 6^3 + 20^3 + 22^3  = 5^3 + 8^3 + 14^3 + 25^3  = 7^3 + 11^3 + 17^3 + 23^3  = 8^3 + 9^3 + 19^3 + 22^3.
%o A345155 (Python)
%o A345155 from itertools import combinations_with_replacement as cwr
%o A345155 from collections import defaultdict
%o A345155 keep = defaultdict(lambda: 0)
%o A345155 power_terms = [x**3 for x in range(1, 1000)]
%o A345155 for pos in cwr(power_terms, 4):
%o A345155     tot = sum(pos)
%o A345155     keep[tot] += 1
%o A345155 rets = sorted([k for k, v in keep.items() if v >= 10])
%o A345155 for x in range(len(rets)):
%o A345155     print(rets[x])
%Y A345155 Cf. A025375, A344928, A345121, A345146, A345156, A345187.
%K A345155 nonn
%O A345155 1,1
%A A345155 _David Consiglio, Jr._, Jun 09 2021