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A345614 Numbers that are the sum of eight fifth powers in six or more ways.

%I A345614 #6 Jul 31 2021 16:17:16
%S A345614 1431397,2593811,3329119,3345410,3609912,3800722,3932480,4093604,
%T A345614 4096697,4104553,4114187,4129433,4154031,4169869,4377714,4451412,
%U A345614 4475603,4484634,4501409,4730845,4756642,4882770,4912477,4915506,4970823,5003645,5112274,5259111,5449985
%N A345614 Numbers that are the sum of eight fifth powers in six or more ways.
%H A345614 Sean A. Irvine, <a href="/A345614/b345614.txt">Table of n, a(n) for n = 1..10000</a>
%e A345614 2593811 is a term because 2593811 = 1^5 + 1^5 + 4^5 + 9^5 + 13^5 + 13^5 + 13^5 + 17^5 = 1^5 + 1^5 + 8^5 + 8^5 + 8^5 + 14^5 + 14^5 + 17^5 = 1^5 + 6^5 + 6^5 + 8^5 + 9^5 + 9^5 + 14^5 + 18^5 = 2^5 + 5^5 + 6^5 + 6^5 + 6^5 + 15^5 + 15^5 + 16^5 = 3^5 + 3^5 + 6^5 + 7^5 + 9^5 + 12^5 + 13^5 + 18^5 = 4^5 + 4^5 + 4^5 + 6^5 + 11^5 + 11^5 + 13^5 + 18^5.
%o A345614 (Python)
%o A345614 from itertools import combinations_with_replacement as cwr
%o A345614 from collections import defaultdict
%o A345614 keep = defaultdict(lambda: 0)
%o A345614 power_terms = [x**5 for x in range(1, 1000)]
%o A345614 for pos in cwr(power_terms, 8):
%o A345614     tot = sum(pos)
%o A345614     keep[tot] += 1
%o A345614     rets = sorted([k for k, v in keep.items() if v >= 6])
%o A345614     for x in range(len(rets)):
%o A345614         print(rets[x])
%Y A345614 Cf. A345581, A345609, A345613, A345615, A345623, A346331.
%K A345614 nonn
%O A345614 1,1
%A A345614 _David Consiglio, Jr._, Jun 20 2021