A345639 Numbers that are the sum of ten fifth powers in seven or more ways.

555098, 674040, 683166, 707315, 763631, 777852, 778844, 780945, 783224, 893654, 896500, 897668, 920887, 926616, 927819, 928802, 936850, 937631, 944383, 945017, 952897, 953077, 953139, 953350, 953414, 955178, 963131, 975133, 979482, 984133, 985453, 985664

Offset: 1

Author

David Consiglio, Jr., Jun 20 2021

Keywords

• nonn

Examples

			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.
		

Links

• Sean A. Irvine, Table of n, a(n) for n = 1..10000

Crossrefs

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

Programs

Python

from itertools import combinations_with_replacement as cwr
from collections import defaultdict
keep = defaultdict(lambda: 0)
power_terms = [x**5 for x in range(1, 1000)]
for pos in cwr(power_terms, 10):
    tot = sum(pos)
    keep[tot] += 1
    rets = sorted([k for k, v in keep.items() if v >= 7])
    for x in range(len(rets)):
        print(rets[x])