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A345580 Numbers that are the sum of eight fourth powers in five or more ways.

%I A345580 #6 Jul 31 2021 17:51:37
%S A345580 4228,4403,4468,5443,5508,5683,6613,6643,6658,6708,6723,6773,6788,
%T A345580 6838,6853,6868,6883,6898,6948,6963,7013,7028,7093,7138,7203,7267,
%U A345580 7268,7332,7397,7478,7507,7572,7588,7828,7858,7923,7938,7988,8003,8068,8113,8133,8178
%N A345580 Numbers that are the sum of eight fourth powers in five or more ways.
%H A345580 Sean A. Irvine, <a href="/A345580/b345580.txt">Table of n, a(n) for n = 1..10000</a>
%e A345580 4403 is a term because 4403 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 2^4 + 4^4 + 8^4 = 1^4 + 1^4 + 1^4 + 4^4 + 4^4 + 6^4 + 6^4 + 6^4 = 2^4 + 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 8^4 = 2^4 + 2^4 + 3^4 + 3^4 + 4^4 + 4^4 + 6^4 + 7^4 = 2^4 + 3^4 + 3^4 + 3^4 + 4^4 + 6^4 + 6^4 + 6^4.
%o A345580 (Python)
%o A345580 from itertools import combinations_with_replacement as cwr
%o A345580 from collections import defaultdict
%o A345580 keep = defaultdict(lambda: 0)
%o A345580 power_terms = [x**4 for x in range(1, 1000)]
%o A345580 for pos in cwr(power_terms, 8):
%o A345580     tot = sum(pos)
%o A345580     keep[tot] += 1
%o A345580     rets = sorted([k for k, v in keep.items() if v >= 5])
%o A345580     for x in range(len(rets)):
%o A345580         print(rets[x])
%Y A345580 Cf. A345535, A345571, A345579, A345581, A345589, A345613, A345837.
%K A345580 nonn
%O A345580 1,1
%A A345580 _David Consiglio, Jr._, Jun 20 2021