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

1,1

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.

Links

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

Example

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.

Prog

(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])

Crossrefs

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

Sequence in context: A273753 A116173 A345608 * A088238 A079015 A372085

Adjacent sequences: A346279 A346280 A346281 * A346283 A346284 A346285

Keyword

nonn

Author

David Consiglio, Jr., Jul 13 2021

Status

approved