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%I #6 Jul 31 2021 16:17:20
%S 4104553,4915506,6011150,6027989,6323394,6563733,6622231,6776363,
%T 6785394,7982834,8181481,8288806,8625619,8658144,8710484,8742208,
%U 8773477,8932244,8996669,9252219,9253706,9311478,9773236,9904983,9976120,10036233,10045233,10053008
%N Numbers that are the sum of eight fifth powers in seven or more ways.
%H Sean A. Irvine, <a href="/A345615/b345615.txt">Table of n, a(n) for n = 1..1000</a>
%e 4915506 is a term because 4915506 = 1^5 + 3^5 + 5^5 + 5^5 + 8^5 + 8^5 + 15^5 + 21^5 = 1^5 + 8^5 + 12^5 + 12^5 + 14^5 + 14^5 + 17^5 + 18^5 = 1^5 + 9^5 + 9^5 + 13^5 + 14^5 + 16^5 + 17^5 + 17^5 = 2^5 + 4^5 + 4^5 + 5^5 + 6^5 + 9^5 + 15^5 + 21^5 = 4^5 + 8^5 + 8^5 + 14^5 + 14^5 + 14^5 + 15^5 + 19^5 = 4^5 + 8^5 + 10^5 + 12^5 + 12^5 + 15^5 + 16^5 + 19^5 = 9^5 + 9^5 + 10^5 + 10^5 + 10^5 + 12^5 + 16^5 + 20^5.
%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**5 for x in range(1, 1000)]
%o for pos in cwr(power_terms, 8):
%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. A345582, A345614, A345616, A345624, A345629, A346332.
%K nonn
%O 1,1
%A _David Consiglio, Jr._, Jun 20 2021
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