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A345767
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Numbers that are the sum of six cubes in exactly five ways.
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7
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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
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OFFSET
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1,1
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COMMENTS
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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.
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LINKS
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EXAMPLE
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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.
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PROG
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(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])
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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