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%I #6 Jul 31 2021 17:25:58
%S 4485,5445,5460,5525,5540,5590,5605,5670,5700,5715,5765,5780,5830,
%T 5845,6645,6675,6710,6740,6755,6775,6805,6820,6855,6870,6885,6900,
%U 6915,6930,6935,6950,6965,6980,6995,7015,7030,7045,7060,7095,7110,7125,7175,7190,7235
%N Numbers that are the sum of ten fourth powers in seven or more ways.
%H Sean A. Irvine, <a href="/A345600/b345600.txt">Table of n, a(n) for n = 1..10000</a>
%e 5445 is a term because 5445 = 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 2^4 + 6^4 + 8^4 = 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 4^4 + 6^4 + 6^4 + 6^4 + 6^4 = 1^4 + 1^4 + 2^4 + 2^4 + 2^4 + 3^4 + 4^4 + 4^4 + 7^4 + 7^4 = 1^4 + 1^4 + 2^4 + 2^4 + 3^4 + 3^4 + 4^4 + 6^4 + 6^4 + 7^4 = 1^4 + 1^4 + 2^4 + 3^4 + 3^4 + 3^4 + 6^4 + 6^4 + 6^4 + 6^4 = 1^4 + 3^4 + 3^4 + 3^4 + 3^4 + 4^4 + 4^4 + 4^4 + 4^4 + 8^4 = 4^4 + 4^4 + 4^4 + 4^4 + 5^4 + 5^4 + 5^4 + 5^4 + 5^4 + 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, 10):
%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. A345555, A345591, A345599, A345601, A345639, A345859.
%K nonn
%O 1,1
%A _David Consiglio, Jr._, Jun 20 2021