A346328 Numbers that are the sum of eight fifth powers in exactly three ways.

52417, 54518, 69634, 70954, 84458, 84489, 84700, 85481, 87582, 92233, 101264, 102890, 112574, 117225, 119326, 134473, 143264, 143442, 143506, 149781, 151448, 158719, 159465, 165634, 166998, 167029, 167196, 167240, 168021, 170122, 174773, 183804, 184457

Offset

1,1

Comments

Differs from A345611 at term 105 because 391250 = 2^5 + 3^5 + 5^5 + 5^5 + 5^5 + 8^5 + 10^5 + 12^5 = 1^5 + 1^5 + 4^5 + 7^5 + 8^5 + 8^5 + 9^5 + 12^5 = 2^5 + 3^5 + 4^5 + 4^5 + 6^5 + 9^5 + 11^5 + 11^5 = 1^5 + 3^5 + 3^5 + 5^5 + 8^5 + 8^5 + 11^5 + 11^5.

Links

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

Example

52417 is a term 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.

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, 8):
    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])

Crossrefs

Cf. A345611, A345835, A346280, A346327, A346329, A346338.

Sequence in context: A186849 A107654 A345611 * A345620 A346338 A015359

Adjacent sequences: A346325 A346326 A346327 * A346329 A346330 A346331

Keyword

nonn

Author

David Consiglio, Jr., Jul 13 2021

Status

approved