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%I #6 Jul 31 2021 17:39:05
%S 6739,6804,6854,6869,6979,7029,7044,7094,7109,7269,7284,7844,7909,
%T 7939,8004,8019,8084,8149,8194,8244,8259,8309,8324,8389,8434,8499,
%U 8564,8628,8739,8868,8979,9044,9059,9124,9189,9219,9234,9254,9284,9299,9364,9414,9429
%N Numbers that are the sum of nine fourth powers in seven or more ways.
%H Sean A. Irvine, <a href="/A345591/b345591.txt">Table of n, a(n) for n = 1..10000</a>
%e 6804 is a term because 6804 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 2^4 + 4^4 + 7^4 + 8^4 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 3^4 + 6^4 + 6^4 + 8^4 = 1^4 + 1^4 + 1^4 + 4^4 + 4^4 + 6^4 + 6^4 + 6^4 + 7^4 = 1^4 + 2^4 + 2^4 + 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 9^4 = 2^4 + 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 7^4 + 8^4 = 2^4 + 2^4 + 3^4 + 3^4 + 4^4 + 4^4 + 6^4 + 7^4 + 7^4 = 2^4 + 3^4 + 3^4 + 3^4 + 4^4 + 6^4 + 6^4 + 6^4 + 7^4 = 3^4 + 3^4 + 3^4 + 3^4 + 6^4 + 6^4 + 6^4 + 6^4 + 6^4.
%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, 9):
%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. A345546, A345582, A345590, A345592, A345600, A345624, A345849.
%K nonn
%O 1,1
%A _David Consiglio, Jr._, Jun 20 2021
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