%I #6 Jul 31 2021 19:11:56
%S 893604,1117071,1182534,1414559,1629244,1933328,2280543,2586035,
%T 2867074,3050644,3055295,3055977,3256432,3329360,3369543,3436058,
%U 3551890,3576363,3896969,3914877,3930849,4055954,4087746,4088690,4093572,4096665,4098161,4104068,4104310
%N Numbers that are the sum of seven fifth powers in exactly four ways.
%C Differs from A345607 at term 92 because 6768576 = 4^5 + 6^5 + 6^5 + 6^5 + 9^5 + 12^5 + 23^5 = 1^5 + 3^5 + 4^5 + 8^5 + 11^5 + 17^5 + 22^5 = 6^5 + 12^5 + 13^5 + 14^5 + 15^5 + 15^5 + 21^5 = 8^5 + 10^5 + 12^5 + 12^5 + 16^5 + 18^5 + 20^5 = 8^5 + 8^5 + 14^5 + 14^5 + 14^5 + 18^5 + 20^5.
%H Sean A. Irvine, <a href="/A346281/b346281.txt">Table of n, a(n) for n = 1..10000</a>
%e 893604 is a term because 893604 = 5^5 + 6^5 + 6^5 + 6^5 + 6^5 + 10^5 + 15^5 = 2^5 + 5^5 + 8^5 + 8^5 + 8^5 + 8^5 + 15^5 = 2^5 + 2^5 + 7^5 + 7^5 + 11^5 + 11^5 + 14^5 = 2^5 + 2^5 + 6^5 + 7^5 + 12^5 + 12^5 + 13^5.
%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, 7):
%o tot = sum(pos)
%o keep[tot] += 1
%o rets = sorted([k for k, v in keep.items() if v == 4])
%o for x in range(len(rets)):
%o print(rets[x])
%Y Cf. A345607, A345826, A346280, A346282, A346329, A346359.
%K nonn
%O 1,1
%A _David Consiglio, Jr._, Jul 13 2021