A345567 - OEIS

%I #6 Jul 31 2021 18:05:33

%S 122915,151556,161475,162755,173075,183620,185315,197795,199106,

%T 199940,201875,201955,202275,204275,204340,204595,206115,207395,

%U 209795,211075,212420,213731,217620,217826,217891,218515,221250,223715,223955,224180,224451,225875

%N Numbers that are the sum of six fourth powers in ten or more ways.

%H Sean A. Irvine, <a href="/A345567/b345567.txt">Table of n, a(n) for n = 1..10000</a>

%e 151556 is a term because 151556 = 1^4 + 2^4 + 2^4 + 9^4 + 11^4 + 19^4 = 1^4 + 2^4 + 3^4 + 7^4 + 16^4 + 17^4 = 1^4 + 8^4 + 11^4 + 12^4 + 13^4 + 17^4 = 2^4 + 3^4 + 7^4 + 8^4 + 11^4 + 19^4 = 3^4 + 3^4 + 3^4 + 4^4 + 12^4 + 19^4 = 3^4 + 4^4 + 11^4 + 11^4 + 14^4 + 17^4 = 3^4 + 4^4 + 13^4 + 13^4 + 13^4 + 16^4 = 4^4 + 6^4 + 9^4 + 9^4 + 9^4 + 19^4 = 4^4 + 7^4 + 11^4 + 11^4 + 11^4 + 18^4 = 4^4 + 8^4 + 9^4 + 13^4 + 13^4 + 17^4.

%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, 6):

%o     tot = sum(pos)

%o     keep[tot] += 1

%o     rets = sorted([k for k, v in keep.items() if v >= 10])

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

%o         print(rets[x])

%Y Cf. A341897, A344196, A345519, A345566, A345576, A345822.

%K nonn

%O 1,1

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