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%I #7 Jul 31 2021 17:51:44
%S 8003,8243,9043,9218,9283,9523,10372,10803,10868,10948,11043,11412,
%T 11557,11587,12083,12692,12932,13188,13268,13333,13508,13972,14147,
%U 14212,14387,14788,14883,14933,14948,14963,15013,15028,15093,15173,15268,15317,15332,15397
%N Numbers that are the sum of eight fourth powers in seven or more ways.
%H Sean A. Irvine, <a href="/A345582/b345582.txt">Table of n, a(n) for n = 1..10000</a>
%e 8243 is a term because 8243 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 2^4 + 8^4 + 8^4 = 1^4 + 1^4 + 1^4 + 4^4 + 6^4 + 6^4 + 6^4 + 8^4 = 1^4 + 2^4 + 2^4 + 2^4 + 3^4 + 4^4 + 6^4 + 9^4 = 2^4 + 2^4 + 3^4 + 3^4 + 4^4 + 6^4 + 7^4 + 8^4 = 2^4 + 3^4 + 3^4 + 3^4 + 6^4 + 6^4 + 6^4 + 8^4 = 2^4 + 4^4 + 4^4 + 4^4 + 4^4 + 7^4 + 7^4 + 7^4 = 3^4 + 4^4 + 4^4 + 4^4 + 6^4 + 6^4 + 7^4 + 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 >= 7])
%o for x in range(len(rets)):
%o print(rets[x])
%Y Cf. A345537, A345573, A345581, A345583, A345591, A345615, A345839.
%K nonn
%O 1,1
%A _David Consiglio, Jr._, Jun 20 2021
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