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A346327 Numbers that are the sum of eight fifth powers in exactly two ways.

%I A346327 #6 Jul 31 2021 19:03:41
%S A346327 4100,4131,4162,4193,4342,4373,4404,4584,4615,4826,5123,5154,5185,
%T A346327 5365,5396,5607,6146,6177,6388,7169,7224,7255,7286,7466,7497,7708,
%U A346327 8247,8278,8489,9270,10348,10379,10590,11371,11875,11906,11937,12117,12148,12359,12898
%N A346327 Numbers that are the sum of eight fifth powers in exactly two ways.
%C A346327 Differs from A345610 at term 128 because 52417 = 3^5 + 3^5 + 3^5 + 3^5 + 5^5 + 6^5 + 6^5 + 8^5 = 1^5 + 4^5 + 4^5 + 4^5 + 4^5 + 6^5 + 6^5 + 8^5 = 3^5 + 3^5 + 3^5 + 3^5 + 4^5 + 7^5 + 7^5 + 7^5.
%H A346327 Sean A. Irvine, <a href="/A346327/b346327.txt">Table of n, a(n) for n = 1..10000</a>
%e A346327 4100 is a term because 4100 = 1^5 + 1^5 + 1^5 + 3^5 + 3^5 + 3^5 + 3^5 + 5^5 = 1^5 + 1^5 + 1^5 + 1^5 + 4^5 + 4^5 + 4^5 + 4^5.
%o A346327 (Python)
%o A346327 from itertools import combinations_with_replacement as cwr
%o A346327 from collections import defaultdict
%o A346327 keep = defaultdict(lambda: 0)
%o A346327 power_terms = [x**5 for x in range(1, 1000)]
%o A346327 for pos in cwr(power_terms, 8):
%o A346327     tot = sum(pos)
%o A346327     keep[tot] += 1
%o A346327     rets = sorted([k for k, v in keep.items() if v == 2])
%o A346327     for x in range(len(rets)):
%o A346327         print(rets[x])
%Y A346327 Cf. A345610, A345834, A346279, A346326, A346328, A346337.
%K A346327 nonn
%O A346327 1,1
%A A346327 _David Consiglio, Jr._, Jul 13 2021