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Numbers that are the sum of nine fourth powers in six or more ways.
8

%I #6 Jul 31 2021 17:39:01

%S 4469,4484,5444,5459,5524,5589,5699,5764,6629,6659,6674,6694,6724,

%T 6739,6789,6804,6854,6869,6884,6899,6914,6934,6949,6964,6979,7014,

%U 7029,7044,7094,7109,7154,7219,7269,7284,7334,7348,7349,7413,7459,7478,7494,7523,7524

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

%H Sean A. Irvine, <a href="/A345590/b345590.txt">Table of n, a(n) for n = 1..10000</a>

%e 4484 is a term because 4484 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 2^4 + 3^4 + 4^4 + 8^4 = 1^4 + 1^4 + 1^4 + 2^4 + 4^4 + 4^4 + 4^4 + 6^4 + 7^4 = 1^4 + 1^4 + 1^4 + 3^4 + 4^4 + 4^4 + 6^4 + 6^4 + 6^4 = 2^4 + 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 3^4 + 8^4 = 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 4^4 + 4^4 + 6^4 + 7^4 = 2^4 + 3^4 + 3^4 + 3^4 + 3^4 + 4^4 + 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 >= 6])

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

%o print(rets[x])

%Y Cf. A345545, A345581, A345589, A345591, A345599, A345623, A345848.

%K nonn

%O 1,1

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