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A241489
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Least number k not divisible by 10 such that k^3 contains n zeros.
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0
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16, 52, 101, 252, 1002, 1001, 10003, 10002, 10001, 100003, 100002, 100001, 1000003, 1000002, 1000001, 10000003, 10000002, 10000001, 100000003, 100000002, 100000001, 1000000003, 1000000002, 1000000001, 10000000003, 10000000002, 10000000001, 100000000003, 100000000002
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OFFSET
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1,1
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COMMENTS
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It is believed that a(n) will have the pattern 1000...0003, 1000...0002, 1000...0001 after a(6).
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LINKS
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Table of n, a(n) for n=1..29.
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FORMULA
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For n > 6, a(n) = 10^(ceiling(n/3) + 1) + 3 - (n+2) mod 3.
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EXAMPLE
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16 is not divisible by 10 and 16^3 = 4096, has 1 zero. So a(1) = 16.
52 is not divisible by 10 and 52^3 = 140608, has 2 zeros. So a(2) = 52.
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PROG
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(Python)
def Cu(n):
..for k in range(10**100):
....if k % 10 != 0:
......if str(k**3).count("0") == n:
........return k
n = 1
while n < 100:
..print(Cu(n))
..n += 1
(PARI) a(n) = {k = 1; while ((d = digits(k^3)) && (((k % 10) == 0) || (sum(i=1, #d, d[i] == 0) != n)), k++); k; } \\ Michel Marcus, Apr 30 2014
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CROSSREFS
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Cf. A134845.
Sequence in context: A235660 A044118 A044499 * A009953 A009939 A009935
Adjacent sequences: A241486 A241487 A241488 * A241490 A241491 A241492
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KEYWORD
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nonn,base
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AUTHOR
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Derek Orr, Apr 23 2014
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STATUS
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approved
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