%I #6 Jul 31 2021 17:39:13

%S 6804,6869,8019,8084,8259,8324,8499,8564,9044,9124,9219,9234,9284,

%T 9299,9364,9429,9474,9494,9539,9604,9669,9749,9779,10148,10259,10293,

%U 10339,10388,10453,10514,10579,10628,10644,10709,10754,10789,10819,10884,10949,10964

%N Numbers that are the sum of nine fourth powers in eight or more ways.

%H Sean A. Irvine, <a href="/A345592/b345592.txt">Table of n, a(n) for n = 1..10000</a>

%e 6869 is a term because 6869 = 1^4 + 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 2^4 + 4^4 + 9^4 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 3^4 + 4^4 + 7^4 + 8^4 = 1^4 + 1^4 + 1^4 + 2^4 + 3^4 + 3^4 + 6^4 + 6^4 + 8^4 = 1^4 + 1^4 + 1^4 + 4^4 + 4^4 + 4^4 + 6^4 + 7^4 + 7^4 = 1^4 + 2^4 + 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 9^4 = 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 3^4 + 7^4 + 8^4 = 2^4 + 3^4 + 3^4 + 3^4 + 4^4 + 4^4 + 6^4 + 7^4 + 7^4 = 3^4 + 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, 9):

%o tot = sum(pos)

%o keep[tot] += 1

%o rets = sorted([k for k, v in keep.items() if v >= 8])

%o for x in range(len(rets)):

%o print(rets[x])

%Y Cf. A345547, A345583, A345591, A345593, A345601, A345625, A345850.

%K nonn

%O 1,1

%A _David Consiglio, Jr._, Jun 20 2021