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%I #6 Jul 31 2021 17:58:15
%S 19491,21267,21332,23652,31251,35427,36052,37812,38067,39891,40356,
%T 41732,41747,43267,43876,43891,43956,44131,44196,44532,44547,44612,
%U 45156,45171,45411,45651,45652,45827,45891,45892,45907,46276,46451,46516,47427,47667,47971
%N Numbers that are the sum of seven fourth powers in nine or more ways.
%H Sean A. Irvine, <a href="/A345575/b345575.txt">Table of n, a(n) for n = 1..10000</a>
%e 21267 is a term because 21267 = 1^4 + 1^4 + 1^4 + 2^4 + 4^4 + 4^4 + 12^4 = 1^4 + 2^4 + 2^4 + 2^4 + 2^4 + 9^4 + 11^4 = 1^4 + 2^4 + 7^4 + 8^4 + 8^4 + 8^4 + 9^4 = 2^4 + 2^4 + 2^4 + 3^4 + 7^4 + 8^4 + 11^4 = 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 4^4 + 12^4 = 2^4 + 2^4 + 4^4 + 6^4 + 9^4 + 9^4 + 9^4 = 2^4 + 4^4 + 4^4 + 6^4 + 7^4 + 7^4 + 11^4 = 3^4 + 4^4 + 6^4 + 6^4 + 6^4 + 7^4 + 11^4 = 3^4 + 7^4 + 7^4 + 8^4 + 8^4 + 8^4 + 8^4.
%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, 7):
%o tot = sum(pos)
%o keep[tot] += 1
%o rets = sorted([k for k, v in keep.items() if v >= 9])
%o for x in range(len(rets)):
%o print(rets[x])
%Y Cf. A345527, A345566, A345574, A345576, A345584, A345631, A345831.
%K nonn
%O 1,1
%A _David Consiglio, Jr._, Jun 20 2021
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