%I
%S 534130,654754,663155,729219,737459,742770,758354,810034,813459,
%T 816579,831250,906034,930499,954930,1009954,1055619,1083955,1099459,
%U 1101859,1103554,1106019,1157634,1167794,1180003,1215394,1217539,1246354,1253074,1255539,1278690
%N Numbers that are the sum of five fourth powers in exactly eight ways.
%C Differs from A344944 at term 2 because 619090 = 1^4 + 2^4 + 18^4 + 22^4 + 23^4 = 1^4 + 3^4 + 4^4 + 8^4 + 28^4 = 1^4 + 11^4 + 14^4 + 22^4 + 24^4 = 2^4 + 2^4 + 8^4 + 17^4 + 27^4 = 2^4 + 13^4 + 13^4 + 18^4 + 26^4 = 3^4 + 6^4 + 12^4 + 16^4 + 27^4 = 4^4 + 12^4 + 14^4 + 23^4 + 23^4 = 9^4 + 12^4 + 16^4 + 21^4 + 24^4 = 14^4 + 16^4 + 18^4 + 19^4 + 23^4.
%H David Consiglio, Jr., <a href="/A344945/b344945.txt">Table of n, a(n) for n = 1..10000</a>
%e 534130 is a term because 534130 = 1^4 + 3^4 + 16^4 + 22^4 + 22^4 = 2^4 + 2^4 + 4^4 + 7^4 + 27^4 = 2^4 + 3^4 + 6^4 + 6^4 + 27^4 = 2^4 + 6^4 + 9^4 + 21^4 + 24^4 = 4^4 + 16^4 + 17^4 + 18^4 + 23^4 = 6^4 + 8^4 + 11^4 + 22^4 + 23^4 = 7^4 + 8^4 + 16^4 + 19^4 + 24^4 = 13^4 + 14^4 + 14^4 + 21^4 + 22^4.
%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**4 for x in range(1, 1000)]
%o for pos in cwr(power_terms, 5):
%o tot = sum(pos)
%o keep[tot] += 1
%o rets = sorted([k for k, v in keep.items() if v == 8])
%o for x in range(len(rets)):
%o print(rets[x])
%Y Cf. A341892, A344925, A344943, A344944, A345184, A345820.
%K nonn
%O 1,1
%A _David Consiglio, Jr._, Jun 03 2021
