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A345855 Numbers that are the sum of ten fourth powers in exactly three ways. 7
520, 535, 550, 600, 615, 680, 775, 790, 855, 1030, 1144, 1159, 1224, 1365, 1380, 1399, 1445, 1540, 1555, 1605, 1635, 1685, 1700, 1768, 1795, 1815, 1830, 1860, 1875, 1895, 1989, 2070, 2164, 2229, 2244, 2439, 2485, 2580, 2595, 2645, 2675, 2680, 2695, 2710, 2755 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,1

COMMENTS

Differs from A345596 at term 21 because 1620 = 1^4 + 1^4 + 1^4 + 3^4 + 4^4 + 4^4 + 4^4 + 4^4 + 4^4 + 4^4 = 2^4 + 3^4 + 3^4 + 3^4 + 3^4 + 4^4 + 4^4 + 4^4 + 4^4 + 4^4.

LINKS

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

EXAMPLE

535 is a term because 535 = 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 1^4 + 2^4 + 4^4 + 4^4 = 1^4 + 1^4 + 1^4 + 1^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 4^4 = 1^4 + 2^4 + 2^4 + 2^4 + 3^4 + 3^4 + 3^4 + 3^4 + 3^4 + 3^4.

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, 10):

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. A345596, A345805, A345845, A345854, A345856, A346348.

Sequence in context: A250768 A043626 A345596 * A235274 A147854 A147856

Adjacent sequences: A345852 A345853 A345854 * A345856 A345857 A345858

KEYWORD

nonn

AUTHOR

David Consiglio, Jr., Jun 26 2021

STATUS

approved

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Last modified January 27 12:24 EST 2023. Contains 359840 sequences. (Running on oeis4.)