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A173228
The number of trailing zeros in (10^n)!
3
2, 24, 249, 2499, 24999, 249998, 2499999, 24999999, 249999998, 2499999997, 24999999997, 249999999997, 2499999999997, 24999999999998, 249999999999997, 2499999999999996, 24999999999999995, 249999999999999995, 2499999999999999995, 24999999999999999996
OFFSET
1,1
COMMENTS
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
LINKS
David S. Hart, James E. Marengo, Darren A. Narayan, and David S. Ross, On the number of trailing zeros in n!, College Math. J., 39(2) (2008) 139-145.
A. M. Oller-Marcen, A new look at the trailing zeros of n!, arXiv:0906.4868 [math.NT], 2009.
A. M. Oller-Marcen, J. Maria Grau, On the Base-b Expansion of the Number of Trailing Zeros of b^k!, J. Int. Seq. 14 (2011) 11.6.8
FORMULA
a(n) = Sum_{k>=1} floor(10^n/5^k). - Stephen G Cappella, Dec 13 2017
MATHEMATICA
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 *)
PROG
(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)))
# Stephen G Cappella, Dec 13 2017
CROSSREFS
Sequence in context: A019520 A300400 A187584 * A061190 A300399 A221082
KEYWORD
nonn,base
AUTHOR
STATUS
approved