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A346328 Numbers that are the sum of eight fifth powers in exactly three ways. 7
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 (list; graph; refs; listen; history; text; internal format)
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

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Last modified October 1 04:31 EDT 2022. Contains 357134 sequences. (Running on oeis4.)