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A345637 Numbers that are the sum of ten fifth powers in five or more ways.

%I A345637 #6 Jul 31 2021 15:58:36
%S A345637 200009,220350,235658,329271,329810,330052,359211,359453,359498,
%T A345637 360298,367314,368529,374519,374847,375089,375870,376620,376651,
%U A345637 377159,377643,380283,382622,384395,384934,387035,388933,391736,392064,392095,392275,392306,392339
%N A345637 Numbers that are the sum of ten fifth powers in five or more ways.
%H A345637 Sean A. Irvine, <a href="/A345637/b345637.txt">Table of n, a(n) for n = 1..10000</a>
%e A345637 220350 is a term because 220350 = 1^5 + 1^5 + 1^5 + 1^5 + 1^5 + 1^5 + 1^5 + 3^5 + 9^5 + 11^5 = 1^5 + 3^5 + 4^5 + 5^5 + 7^5 + 7^5 + 7^5 + 8^5 + 8^5 + 10^5 = 1^5 + 3^5 + 5^5 + 5^5 + 6^5 + 6^5 + 8^5 + 8^5 + 8^5 + 10^5 = 2^5 + 4^5 + 4^5 + 4^5 + 6^5 + 7^5 + 7^5 + 7^5 + 9^5 + 10^5 = 2^5 + 4^5 + 4^5 + 5^5 + 6^5 + 6^5 + 6^5 + 8^5 + 9^5 + 10^5.
%o A345637 (Python)
%o A345637 from itertools import combinations_with_replacement as cwr
%o A345637 from collections import defaultdict
%o A345637 keep = defaultdict(lambda: 0)
%o A345637 power_terms = [x**5 for x in range(1, 1000)]
%o A345637 for pos in cwr(power_terms, 10):
%o A345637     tot = sum(pos)
%o A345637     keep[tot] += 1
%o A345637     rets = sorted([k for k, v in keep.items() if v >= 5])
%o A345637     for x in range(len(rets)):
%o A345637         print(rets[x])
%Y A345637 Cf. A345598, A345622, A345636, A345638, A346350.
%K A345637 nonn
%O A345637 1,1
%A A345637 _David Consiglio, Jr._, Jun 20 2021