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A345825 Numbers that are the sum of seven fourth powers in exactly three ways. 8
2677, 2692, 2757, 2852, 2867, 2917, 2997, 3107, 3172, 3301, 3476, 3541, 3972, 4132, 4227, 4242, 4257, 4307, 4322, 4372, 4437, 4452, 4482, 4497, 4562, 4627, 4737, 4756, 4851, 4866, 4867, 4931, 4996, 5077, 5106, 5107, 5122, 5187, 5252, 5282, 5317, 5347, 5362 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,1

COMMENTS

Differs from A345569 at term 7 because 2932 = 1^4 + 1^4 + 1^4 + 2^4 + 4^4 + 4^4 + 7^4  = 1^4 + 1^4 + 1^4 + 3^4 + 4^4 + 6^4 + 6^4  = 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 4^4 + 7^4  = 2^4 + 3^4 + 3^4 + 3^4 + 3^4 + 6^4 + 6^4.

LINKS

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

EXAMPLE

2692 is a term because 2692 = 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 4^4 + 7^4 = 1^4 + 1^4 + 1^4 + 2^4 + 3^4 + 6^4 + 6^4 = 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 7^4.

PROG

(Python)

from itertools import combinations_with_replacement as cwr

from collections import defaultdict

keep = defaultdict(lambda: 0)

power_terms = [x**4 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])

CROSSREFS

Cf. A345569, A345775, A345815, A345824, A345826, A345835, A346280.

Sequence in context: A345815 A166513 A345569 * A258505 A258512 A254992

Adjacent sequences:  A345822 A345823 A345824 * A345826 A345827 A345828

KEYWORD

nonn

AUTHOR

David Consiglio, Jr., Jun 26 2021

STATUS

approved

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Last modified September 18 15:57 EDT 2021. Contains 347527 sequences. (Running on oeis4.)