A345639 - OEIS

%I #6 Jul 31 2021 15:58:45

%S 555098,674040,683166,707315,763631,777852,778844,780945,783224,

%T 893654,896500,897668,920887,926616,927819,928802,936850,937631,

%U 944383,945017,952897,953077,953139,953350,953414,955178,963131,975133,979482,984133,985453,985664

%N Numbers that are the sum of ten fifth powers in seven or more ways.

%H Sean A. Irvine, <a href="/A345639/b345639.txt">Table of n, a(n) for n = 1..10000</a>

%e 674040 is a term because 674040 = 1^5 + 1^5 + 2^5 + 3^5 + 3^5 + 7^5 + 9^5 + 10^5 + 12^5 + 12^5 = 1^5 + 1^5 + 2^5 + 4^5 + 4^5 + 4^5 + 10^5 + 11^5 + 11^5 + 12^5 = 1^5 + 3^5 + 3^5 + 3^5 + 5^5 + 6^5 + 8^5 + 8^5 + 9^5 + 14^5 = 1^5 + 3^5 + 4^5 + 4^5 + 4^5 + 4^5 + 7^5 + 8^5 + 12^5 + 13^5 = 2^5 + 2^5 + 2^5 + 2^5 + 4^5 + 6^5 + 8^5 + 10^5 + 11^5 + 13^5 = 2^5 + 3^5 + 3^5 + 4^5 + 4^5 + 6^5 + 6^5 + 9^5 + 9^5 + 14^5 = 3^5 + 3^5 + 3^5 + 3^5 + 3^5 + 5^5 + 7^5 + 8^5 + 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, 10):

%o     tot = sum(pos)

%o     keep[tot] += 1

%o     rets = sorted([k for k, v in keep.items() if v >= 7])

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

%o         print(rets[x])

%Y Cf. A345600, A345624, A345638, A345640, A346352.

%K nonn

%O 1,1

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