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A345156 Numbers that are the sum of four third powers in exactly ten ways.

%I A345156 #6 Jul 31 2021 23:29:20
%S A345156 21896,36225,48825,51506,52416,53200,58338,58968,60480,66024,67851,
%T A345156 70434,70525,71155,72819,76923,78624,78912,85995,87507,88641,90181,
%U A345156 90783,91728,93555,97552,98280,98560,99008,99225,99792,100170,103040,104104,104265,104958
%N A345156 Numbers that are the sum of four third powers in exactly ten ways.
%C A345156 Differs from A345155 at term 3 because 46872 = 1^3 + 16^3 + 22^3 + 30^3  = 2^3 + 11^3 + 17^3 + 33^3  = 3^3 + 3^3 + 4^3 + 35^3  = 3^3 + 4^3 + 26^3 + 29^3  = 3^3 + 5^3 + 23^3 + 31^3  = 4^3 + 10^3 + 24^3 + 30^3  = 5^3 + 17^3 + 23^3 + 29^3  = 6^3 + 10^3 + 20^3 + 32^3  = 11^3 + 11^3 + 21^3 + 31^3  = 11^3 + 14^3 + 17^3 + 32^3  = 19^3 + 21^3 + 21^3 + 25^3.
%H A345156 David Consiglio, Jr., <a href="/A345156/b345156.txt">Table of n, a(n) for n = 1..10000</a>
%e A345156 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 A345156 (Python)
%o A345156 from itertools import combinations_with_replacement as cwr
%o A345156 from collections import defaultdict
%o A345156 keep = defaultdict(lambda: 0)
%o A345156 power_terms = [x**3 for x in range(1, 1000)]
%o A345156 for pos in cwr(power_terms, 4):
%o A345156     tot = sum(pos)
%o A345156     keep[tot] += 1
%o A345156 rets = sorted([k for k, v in keep.items() if v == 10])
%o A345156 for x in range(len(rets)):
%o A345156     print(rets[x])
%Y A345156 Cf. A025366, A344929, A345122, A345154, A345155, A345188.
%K A345156 nonn
%O A345156 1,1
%A A345156 _David Consiglio, Jr._, Jun 09 2021