A346282 Numbers that are the sum of seven fifth powers in exactly five ways.

6768576, 6776120, 7883668, 8625376, 8740709, 10036201, 10604054, 12476826, 12618493, 13006575, 13060213, 13080706, 13174250, 13536416, 13550162, 13662500, 14110656, 15169276, 15247994, 16053313, 16060683, 16374218, 16573507, 16600001, 17735057, 17749152

Offset: 1

Author

David Consiglio, Jr., Jul 13 2021

Keywords

• nonn

Comments

Differs from A345608 at term 16 because 13562501 = 1^5 + 1^5 + 1^5 + 9^5 + 14^5 + 20^5 + 25^5 = 1^5 + 15^5 + 15^5 + 15^5 + 15^5 + 15^5 + 25^5 = 6^5 + 7^5 + 11^5 + 16^5 + 18^5 + 19^5 + 24^5 = 7^5 + 7^5 + 11^5 + 13^5 + 19^5 + 21^5 + 23^5 = 2^5 + 6^5 + 14^5 + 18^5 + 18^5 + 21^5 + 22^5 = 1^5 + 5^5 + 15^5 + 20^5 + 20^5 + 20^5 + 20^5.

Examples

			6768576 is a term because 6768576 = 4^5 + 6^5 + 6^5 + 6^5 + 9^5 + 12^5 + 23^5 = 1^5 + 3^5 + 4^5 + 8^5 + 11^5 + 17^5 + 22^5 = 6^5 + 12^5 + 13^5 + 14^5 + 15^5 + 15^5 + 21^5 = 8^5 + 10^5 + 12^5 + 12^5 + 16^5 + 18^5 + 20^5 = 8^5 + 8^5 + 14^5 + 14^5 + 14^5 + 18^5 + 20^5.
		

Links

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

Crossrefs

Cf. A345608, A345827, A346281, A346283, A346330, A346360.

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 == 5])
    for x in range(len(rets)):
        print(rets[x])