A345637 - OEIS

%I #6 Jul 31 2021 15:58:36

%S 200009,220350,235658,329271,329810,330052,359211,359453,359498,

%T 360298,367314,368529,374519,374847,375089,375870,376620,376651,

%U 377159,377643,380283,382622,384395,384934,387035,388933,391736,392064,392095,392275,392306,392339

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

%H Sean A. Irvine, <a href="/A345637/b345637.txt">Table of n, a(n) for n = 1..10000</a>

%e 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 (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, 10):

%o     tot = sum(pos)

%o     keep[tot] += 1

%o     rets = sorted([k for k, v in keep.items() if v >= 5])

%o     for x in range(len(rets)):

%o         print(rets[x])

%Y Cf. A345598, A345622, A345636, A345638, A346350.

%K nonn

%O 1,1

%A _David Consiglio, Jr._, Jun 20 2021