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A173228
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The number of trailing zeros in (10^n)!
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3
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2, 24, 249, 2499, 24999, 249998, 2499999, 24999999, 249999998, 2499999997, 24999999997, 249999999997, 2499999999997, 24999999999998, 249999999999997, 2499999999999996, 24999999999999995, 249999999999999995, 2499999999999999995, 24999999999999999996
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OFFSET
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1,1
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COMMENTS
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For n > 1, the number a(n) of trailing end 0's in (10^n)! is short of (10^n)/4 by A055223(n). - Lekraj Beedassy, Oct 27 2010
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LINKS
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FORMULA
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MATHEMATICA
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a[n_] := Sum[Floor[10^n/5^i], {i, Floor[Log[5, 10^n]]}]; Array[f, 18] (* edited by Robert G. Wilson v, Jul 22 2012 *)
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PROG
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(Python)
from math import log, ceil
def a(n):
return sum(10**n // 5**k for k in range(1, ceil(log(10, 5) * n)))
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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STATUS
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approved
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