A346327 Numbers that are the sum of eight fifth powers in exactly two ways.

4100, 4131, 4162, 4193, 4342, 4373, 4404, 4584, 4615, 4826, 5123, 5154, 5185, 5365, 5396, 5607, 6146, 6177, 6388, 7169, 7224, 7255, 7286, 7466, 7497, 7708, 8247, 8278, 8489, 9270, 10348, 10379, 10590, 11371, 11875, 11906, 11937, 12117, 12148, 12359, 12898

Offset: 1

Author

David Consiglio, Jr., Jul 13 2021

Keywords

• nonn

Comments

Differs from A345610 at term 128 because 52417 = 3^5 + 3^5 + 3^5 + 3^5 + 5^5 + 6^5 + 6^5 + 8^5 = 1^5 + 4^5 + 4^5 + 4^5 + 4^5 + 6^5 + 6^5 + 8^5 = 3^5 + 3^5 + 3^5 + 3^5 + 4^5 + 7^5 + 7^5 + 7^5.

Examples

			4100 is a term because 4100 = 1^5 + 1^5 + 1^5 + 3^5 + 3^5 + 3^5 + 3^5 + 5^5 = 1^5 + 1^5 + 1^5 + 1^5 + 4^5 + 4^5 + 4^5 + 4^5.
		

Links

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

Crossrefs

Cf. A345610, A345834, A346279, A346326, A346328, A346337.

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, 8):
    tot = sum(pos)
    keep[tot] += 1
    rets = sorted([k for k, v in keep.items() if v == 2])
    for x in range(len(rets)):
        print(rets[x])