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%I #6 Jul 31 2021 16:17:25
%S 8625619,8742208,9773236,10036233,10071050,12247994,13180706,13377868,
%T 13662501,14584992,14591744,14611077,15251119,15539667,16112362,
%U 16374250,16391025,16472544,16588000,16667851,17059075,17216298,17405300,17917097,18107564,18392902
%N Numbers that are the sum of eight fifth powers in eight or more ways.
%H Sean A. Irvine, <a href="/A345616/b345616.txt">Table of n, a(n) for n = 1..100</a>
%e 8742208 is a term because 8742208 = 1^5 + 1^5 + 2^5 + 3^5 + 5^5 + 7^5 + 15^5 + 24^5 = 1^5 + 1^5 + 9^5 + 9^5 + 11^5 + 17^5 + 18^5 + 22^5 = 1^5 + 3^5 + 7^5 + 12^5 + 12^5 + 13^5 + 17^5 + 23^5 = 2^5 + 5^5 + 6^5 + 7^5 + 15^5 + 15^5 + 15^5 + 23^5 = 3^5 + 3^5 + 7^5 + 9^5 + 12^5 + 12^5 + 21^5 + 21^5 = 4^5 + 4^5 + 4^5 + 11^5 + 11^5 + 12^5 + 21^5 + 21^5 = 4^5 + 4^5 + 8^5 + 8^5 + 9^5 + 15^5 + 17^5 + 23^5 = 8^5 + 13^5 + 14^5 + 14^5 + 14^5 + 16^5 + 19^5 + 20^5 = 10^5 + 12^5 + 12^5 + 13^5 + 16^5 + 16^5 + 19^5 + 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 >= 8])
%o for x in range(len(rets)):
%o print(rets[x])
%Y Cf. A345583, A345615, A345617, A345625, A345630, A346333.
%K nonn
%O 1,1
%A _David Consiglio, Jr._, Jun 20 2021
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