A346338 Numbers that are the sum of nine fifth powers in exactly three ways.

52418, 52449, 52660, 53441, 54519, 54550, 54761, 55690, 57643, 60193, 62294, 69224, 69635, 69666, 69877, 70658, 70955, 70986, 71197, 71325, 71978, 72759, 73001, 74079, 76031, 77410, 78730, 84162, 84459, 84490, 84521, 84701, 84732, 84943, 85185, 85482, 85513

Offset: 1

Author

David Consiglio, Jr., Jul 13 2021

Keywords

• nonn

Comments

Differs from A345620 at term 8 because 55542 = 3^5 + 3^5 + 3^5 + 3^5 + 5^5 + 5^5 + 6^5 + 6^5 + 8^5 = 1^5 + 4^5 + 4^5 + 4^5 + 4^5 + 5^5 + 6^5 + 6^5 + 8^5 = 3^5 + 3^5 + 3^5 + 3^5 + 4^5 + 5^5 + 7^5 + 7^5 + 7^5 = 1^5 + 4^5 + 4^5 + 4^5 + 4^5 + 4^5 + 7^5 + 7^5 + 7^5.

Examples

			52418 is a term because 52418 = 1^5 + 3^5 + 3^5 + 3^5 + 3^5 + 5^5 + 6^5 + 6^5 + 8^5 = 1^5 + 1^5 + 4^5 + 4^5 + 4^5 + 4^5 + 6^5 + 6^5 + 8^5 = 1^5 + 3^5 + 3^5 + 3^5 + 3^5 + 4^5 + 7^5 + 7^5 + 7^5.
		

Links

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

Crossrefs

Cf. A345620, A345845, A346328, A346337, A346339, A346348.

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, 9):
    tot = sum(pos)
    keep[tot] += 1
    rets = sorted([k for k, v in keep.items() if v == 3])
    for x in range(len(rets)):
        print(rets[x])