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A345767 Numbers that are the sum of six cubes in exactly five ways. 7
1045, 1169, 1241, 1260, 1384, 1432, 1440, 1495, 1530, 1539, 1549, 1556, 1558, 1584, 1594, 1602, 1612, 1617, 1640, 1654, 1657, 1675, 1703, 1712, 1715, 1719, 1729, 1736, 1745, 1747, 1754, 1771, 1780, 1792, 1801, 1803, 1806, 1810, 1818, 1825, 1827, 1834, 1843 (list; graph; refs; listen; history; text; internal format)
OFFSET

1,1

COMMENTS

Differs from A345514 at term 5 because 1377 = 1^3 + 1^3 + 2^3 + 7^3 + 8^3 + 8^3  = 1^3 + 1^3 + 5^3 + 5^3 + 5^3 + 10^3  = 1^3 + 2^3 + 3^3 + 5^3 + 6^3 + 10^3  = 1^3 + 6^3 + 6^3 + 6^3 + 6^3 + 8^3  = 3^3 + 3^3 + 5^3 + 7^3 + 7^3 + 8^3  = 3^3 + 4^3 + 5^3 + 6^3 + 6^3 + 9^3.

LINKS

Sean A. Irvine, Table of n, a(n) for n = 1..1227

EXAMPLE

1169 is a term because 1169 = 1^3 + 2^3 + 2^3 + 3^3 + 4^3 + 9^3 = 1^3 + 2^3 + 5^3 + 5^3 + 5^3 + 7^3 = 1^3 + 3^3 + 4^3 + 4^3 + 4^3 + 8^3 = 2^3 + 3^3 + 3^3 + 4^3 + 5^3 + 8^3 = 3^3 + 3^3 + 3^3 + 3^3 + 7^3 + 7^3.

PROG

(Python)

from itertools import combinations_with_replacement as cwr

from collections import defaultdict

keep = defaultdict(lambda: 0)

power_terms = [x**3 for x in range(1, 1000)]

for pos in cwr(power_terms, 6):

    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. A343988, A345514, A345766, A345768, A345777, A345817.

Sequence in context: A224604 A045027 A345514 * A324320 A334012 A344376

Adjacent sequences:  A345764 A345765 A345766 * A345768 A345769 A345770

KEYWORD

nonn

AUTHOR

David Consiglio, Jr., Jun 26 2021

STATUS

approved

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Last modified October 25 01:40 EDT 2021. Contains 348233 sequences. (Running on oeis4.)