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%I #6 Jul 31 2021 17:25:54
%S 3175,4150,4230,4390,4405,4455,4470,4485,4500,4550,4565,4630,4725,
%T 4740,4915,4980,5094,5109,5155,5190,5205,5220,5270,5285,5350,5365,
%U 5395,5430,5445,5460,5475,5525,5540,5590,5605,5635,5655,5670,5700,5715,5735,5765,5780
%N Numbers that are the sum of ten fourth powers in six or more ways.
%H Sean A. Irvine, <a href="/A345599/b345599.txt">Table of n, a(n) for n = 1..10000</a>
%e 4150 is a term because 4150 = 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 2^4 + 8^4 = 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 4^4 + 6^4 + 6^4 + 6^4 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 3^4 + 3^4 + 4^4 + 6^4 + 7^4 = 1^4 + 1^4 + 1^4 + 2^4 + 3^4 + 3^4 + 3^4 + 6^4 + 6^4 + 6^4 = 1^4 + 4^4 + 4^4 + 4^4 + 4^4 + 5^4 + 5^4 + 5^4 + 5^4 + 5^4 = 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 3^4 + 3^4 + 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, 10):
%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. A345554, A345590, A345598, A345600, A345638, A345858.
%K nonn
%O 1,1
%A _David Consiglio, Jr._, Jun 20 2021
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