A345566 - OEIS

%I #6 Jul 31 2021 18:05:28

%S 88595,122915,132546,134931,144835,146450,151556,161475,162355,162755,

%T 170275,171555,171795,172036,172835,173075,177380,177716,180770,

%U 183540,183620,184835,185315,185555,187700,187715,190100,190211,193635,195380,195780,196435

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

%H Sean A. Irvine, <a href="/A345566/b345566.txt">Table of n, a(n) for n = 1..10000</a>

%e 122915 is a term because 122915 = 1^4 + 3^4 + 6^4 + 9^4 + 10^4 + 18^4 = 1^4 + 4^4 + 7^4 + 8^4 + 15^4 + 16^4 = 1^4 + 7^4 + 9^4 + 10^4 + 14^4 + 16^4 = 2^4 + 3^4 + 4^4 + 5^4 + 14^4 + 17^4 = 2^4 + 4^4 + 5^4 + 7^4 + 11^4 + 18^4 = 2^4 + 9^4 + 9^4 + 12^4 + 14^4 + 15^4 = 3^4 + 5^4 + 6^4 + 6^4 + 11^4 + 18^4 = 3^4 + 8^4 + 10^4 + 11^4 + 13^4 + 16^4 = 5^4 + 6^4 + 7^4 + 11^4 + 14^4 + 16^4 = 8^4 + 8^4 + 9^4 + 10^4 + 11^4 + 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 >= 9])

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

%o         print(rets[x])

%Y Cf. A341891, A345518, A345565, A345567, A345575, A345723, A345821.

%K nonn

%O 1,1

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