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A345629 Numbers that are the sum of seven fifth powers in seven or more ways.

%I A345629 #6 Jul 31 2021 16:25:26
%S A345629 28608832,35663099,36090526,36620574,46998599,51095638,52541851,
%T A345629 54233651,54827543,54886349,61263643,61634374,63514593,64810976,
%U A345629 65198607,66708676,67887843,70979107,72970305,74002457,74115801,74132607,74487093,75044651,75378359
%N A345629 Numbers that are the sum of seven fifth powers in seven or more ways.
%H A345629 Sean A. Irvine, <a href="/A345629/b345629.txt">Table of n, a(n) for n = 1..10000</a>
%e A345629 35663099 is a term because 35663099 = 1^5 + 9^5 + 16^5 + 17^5 + 24^5 + 24^5 + 28^5 = 2^5 + 3^5 + 17^5 + 23^5 + 24^5 + 24^5 + 26^5 = 2^5 + 10^5 + 15^5 + 17^5 + 23^5 + 23^5 + 29^5 = 4^5 + 8^5 + 13^5 + 19^5 + 21^5 + 27^5 + 27^5 = 4^5 + 11^5 + 13^5 + 19^5 + 20^5 + 22^5 + 30^5 = 5^5 + 6^5 + 19^5 + 19^5 + 20^5 + 20^5 + 30^5 = 7^5 + 9^5 + 12^5 + 18^5 + 18^5 + 27^5 + 28^5.
%o A345629 (Python)
%o A345629 from itertools import combinations_with_replacement as cwr
%o A345629 from collections import defaultdict
%o A345629 keep = defaultdict(lambda: 0)
%o A345629 power_terms = [x**5 for x in range(1, 1000)]
%o A345629 for pos in cwr(power_terms, 7):
%o A345629     tot = sum(pos)
%o A345629     keep[tot] += 1
%o A345629     rets = sorted([k for k, v in keep.items() if v >= 7])
%o A345629     for x in range(len(rets)):
%o A345629         print(rets[x])
%Y A345629 Cf. A345573, A345609, A345615, A345630, A345721, A346284.
%K A345629 nonn
%O A345629 1,1
%A A345629 _David Consiglio, Jr._, Jun 22 2021