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%I #6 Jul 31 2021 17:51:40
%S 6723,6788,6853,6898,6963,7028,7938,8003,8068,8178,8243,8308,8483,
%T 8963,9043,9173,9218,9283,9348,9413,9493,9523,9668,9763,9828,10003,
%U 10132,10258,10277,10307,10372,10628,10708,10738,10788,10803,10868,10933,10948,10978
%N Numbers that are the sum of eight fourth powers in six or more ways.
%H Sean A. Irvine, <a href="/A345581/b345581.txt">Table of n, a(n) for n = 1..10000</a>
%e 6788 is a term because 6788 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 4^4 + 7^4 + 8^4 = 1^4 + 1^4 + 1^4 + 2^4 + 3^4 + 6^4 + 6^4 + 8^4 = 1^4 + 2^4 + 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 9^4 = 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 7^4 + 8^4 = 2^4 + 3^4 + 3^4 + 4^4 + 4^4 + 6^4 + 7^4 + 7^4 = 3^4 + 3^4 + 3^4 + 4^4 + 6^4 + 6^4 + 6^4 + 7^4.
%o (Python)
%o from itertools import combinations_with_replacement as cwr
%o from collections import defaultdict
%o keep = defaultdict(lambda: 0)
%o power_terms = [x**4 for x in range(1, 1000)]
%o for pos in cwr(power_terms, 8):
%o tot = sum(pos)
%o keep[tot] += 1
%o rets = sorted([k for k, v in keep.items() if v >= 6])
%o for x in range(len(rets)):
%o print(rets[x])
%Y Cf. A345536, A345572, A345580, A345582, A345590, A345614, A345838.
%K nonn
%O 1,1
%A _David Consiglio, Jr._, Jun 20 2021
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