A346280 Numbers that are the sum of seven fifth powers in exactly three ways.

84457, 166997, 324860, 326199, 358482, 359327, 391007, 391999, 408158, 455146, 455749, 486468, 502429, 572054, 595519, 614505, 622280, 648319, 671210, 672022, 696468, 696499, 696710, 697491, 699592, 704243, 713274, 729235, 755516, 796467, 857518, 877645

Offset: 1

Author

David Consiglio, Jr., Jul 13 2021

Keywords

• nonn

Comments

Differs from A345606 at term 39 because 893604 = 5^5 + 6^5 + 6^5 + 6^5 + 6^5 + 10^5 + 15^5 = 2^5 + 5^5 + 8^5 + 8^5 + 8^5 + 8^5 + 15^5 = 2^5 + 2^5 + 7^5 + 7^5 + 11^5 + 11^5 + 14^5 = 2^5 + 2^5 + 6^5 + 7^5 + 12^5 + 12^5 + 13^5.

Examples

			84457 is a term because 84457 = 2^5 + 4^5 + 4^5 + 6^5 + 6^5 + 6^5 + 9^5 = 1^5 + 3^5 + 5^5 + 6^5 + 6^5 + 8^5 + 8^5 = 1^5 + 3^5 + 4^5 + 7^5 + 7^5 + 7^5 + 8^5.
		

Links

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

Crossrefs

Cf. A345606, A345825, A346279, A346281, A346328, A346358.

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, 7):
    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])