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A345187
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Numbers that are the sum of five third powers in ten or more ways.
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7
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5860, 6588, 6651, 6859, 6947, 8056, 8289, 8371, 8506, 8569, 8758, 9045, 9080, 9099, 9108, 9227, 9414, 9612, 9801, 9829, 9864, 10009, 10018, 10044, 10277, 10466, 10485, 10522, 10529, 10800, 10963, 10970, 10979, 11008, 11017, 11061, 11089, 11152, 11241, 11385
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listen;
history;
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internal format)
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OFFSET
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1,1
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LINKS
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EXAMPLE
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6588 is a term because 6588 = 1^3 + 3^3 + 5^3 + 7^3 + 17^3 = 1^3 + 4^3 + 6^3 + 13^3 + 14^3 = 1^3 + 5^3 + 8^3 + 8^3 + 16^3 = 1^3 + 10^3 + 10^3 + 11^3 + 12^3 = 2^3 + 2^3 + 9^3 + 12^3 + 14^3 = 2^3 + 3^3 + 8^3 + 11^3 + 15^3 = 3^3 + 8^3 + 8^3 + 11^3 + 14^3 = 3^3 + 3^3 + 5^3 + 10^3 + 16^3 = 5^3 + 5^3 + 8^3 + 10^3 + 15^3 = 8^3 + 9^3 + 10^3 + 10^3 + 12^3.
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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, 5):
tot = sum(pos)
keep[tot] += 1
rets = sorted([k for k, v in keep.items() if v >= 10])
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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