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A345612 Numbers that are the sum of eight fifth powers in four or more ways.

%I A345612 #6 Jul 31 2021 16:17:06
%S A345612 391250,392031,455750,519236,604822,622281,672023,672054,672265,
%T A345612 673554,697492,703978,707368,730259,763292,857761,893605,893636,
%U A345612 893816,893847,894027,894058,894452,894628,896729,897151,901380,903839,909124,909597,910411,911403
%N A345612 Numbers that are the sum of eight fifth powers in four or more ways.
%H A345612 Sean A. Irvine, <a href="/A345612/b345612.txt">Table of n, a(n) for n = 1..10000</a>
%e A345612 392031 is a term because 392031 = 1^5 + 3^5 + 4^5 + 5^5 + 8^5 + 8^5 + 11^5 + 11^5 = 2^5 + 3^5 + 3^5 + 6^5 + 7^5 + 9^5 + 9^5 + 12^5 = 2^5 + 4^5 + 4^5 + 4^5 + 6^5 + 9^5 + 11^5 + 11^5 = 2^5 + 4^5 + 5^5 + 5^5 + 5^5 + 8^5 + 10^5 + 12^5.
%o A345612 (Python)
%o A345612 from itertools import combinations_with_replacement as cwr
%o A345612 from collections import defaultdict
%o A345612 keep = defaultdict(lambda: 0)
%o A345612 power_terms = [x**5 for x in range(1, 1000)]
%o A345612 for pos in cwr(power_terms, 8):
%o A345612     tot = sum(pos)
%o A345612     keep[tot] += 1
%o A345612     rets = sorted([k for k, v in keep.items() if v >= 4])
%o A345612     for x in range(len(rets)):
%o A345612         print(rets[x])
%Y A345612 Cf. A345579, A345607, A345611, A345613, A345621, A346329.
%K A345612 nonn
%O A345612 1,1
%A A345612 _David Consiglio, Jr._, Jun 20 2021