%I #6 Jul 31 2021 18:53:59
%S 4102,4133,4164,4195,4226,4257,4344,4375,4406,4437,4468,4586,4617,
%T 4648,4679,4828,4859,4890,5070,5101,5125,5156,5187,5218,5249,5312,
%U 5367,5398,5429,5460,5609,5640,5671,5851,5882,6093,6148,6179,6210,6241,6390,6421,6452
%N Numbers that are the sum of ten fifth powers in exactly two ways.
%C Differs from A345634 at term 67 because 8194 = 3^5 + 3^5 + 3^5 + 3^5 + 3^5 + 3^5 + 3^5 + 3^5 + 5^5 + 5^5 = 1^5 + 3^5 + 3^5 + 3^5 + 3^5 + 4^5 + 4^5 + 4^5 + 4^5 + 5^5 = 1^5 + 1^5 + 4^5 + 4^5 + 4^5 + 4^5 + 4^5 + 4^5 + 4^5 + 4^5.
%H Sean A. Irvine, <a href="/A346347/b346347.txt">Table of n, a(n) for n = 1..10000</a>
%e 4102 is a term because 4102 = 1^5 + 1^5 + 1^5 + 1^5 + 1^5 + 3^5 + 3^5 + 3^5 + 3^5 + 5^5 = 1^5 + 1^5 + 1^5 + 1^5 + 1^5 + 1^5 + 4^5 + 4^5 + 4^5 + 4^5.
%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 == 2])
%o for x in range(len(rets)):
%o print(rets[x])
%Y Cf. A345634, A345854, A346337, A346346, A346348.
%K nonn
%O 1,1
%A _David Consiglio, Jr._, Jul 13 2021