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A345791
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Numbers that are the sum of eight cubes in exactly nine ways.
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7
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984, 1080, 1136, 1171, 1192, 1197, 1204, 1223, 1269, 1273, 1280, 1306, 1318, 1325, 1332, 1333, 1337, 1344, 1356, 1360, 1369, 1370, 1374, 1377, 1379, 1404, 1406, 1415, 1416, 1422, 1425, 1430, 1432, 1438, 1442, 1444, 1445, 1456, 1476, 1481, 1486, 1488, 1494
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OFFSET
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1,1
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COMMENTS
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Differs from A345539 at term 5 because 1185 = 1^3 + 1^3 + 1^3 + 3^3 + 3^3 + 4^3 + 4^3 + 10^3 = 1^3 + 1^3 + 1^3 + 4^3 + 6^3 + 6^3 + 7^3 + 7^3 = 1^3 + 2^3 + 2^3 + 2^3 + 2^3 + 3^3 + 5^3 + 10^3 = 1^3 + 2^3 + 2^3 + 2^3 + 6^3 + 6^3 + 6^3 + 8^3 = 1^3 + 2^3 + 2^3 + 4^3 + 5^3 + 5^3 + 5^3 + 9^3 = 1^3 + 5^3 + 5^3 + 5^3 + 5^3 + 5^3 + 6^3 + 7^3 = 2^3 + 2^3 + 2^3 + 3^3 + 4^3 + 5^3 + 6^3 + 9^3 = 2^3 + 2^3 + 3^3 + 3^3 + 3^3 + 4^3 + 8^3 + 8^3 = 2^3 + 3^3 + 5^3 + 5^3 + 5^3 + 6^3 + 6^3 + 7^3 = 3^3 + 4^3 + 4^3 + 4^3 + 4^3 + 6^3 + 7^3 + 7^3.
Likely finite.
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LINKS
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EXAMPLE
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1080 is a term because 1080 = 1^3 + 1^3 + 1^3 + 2^3 + 4^3 + 5^3 + 5^3 + 7^3 = 1^3 + 1^3 + 2^3 + 2^3 + 2^3 + 3^3 + 3^3 + 9^3 = 1^3 + 1^3 + 2^3 + 3^3 + 3^3 + 4^3 + 4^3 + 8^3 = 1^3 + 2^3 + 2^3 + 3^3 + 3^3 + 3^3 + 5^3 + 8^3 = 1^3 + 3^3 + 3^3 + 4^3 + 4^3 + 5^3 + 5^3 + 6^3 = 1^3 + 3^3 + 3^3 + 3^3 + 4^3 + 4^3 + 4^3 + 7^3 = 2^3 + 2^3 + 3^3 + 3^3 + 4^3 + 4^3 + 6^3 + 6^3 = 2^3 + 3^3 + 3^3 + 3^3 + 5^3 + 5^3 + 5^3 + 6^3 = 2^3 + 3^3 + 3^3 + 3^3 + 3^3 + 4^3 + 5^3 + 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, 8):
tot = sum(pos)
keep[tot] += 1
rets = sorted([k for k, v in keep.items() if v == 9])
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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