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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 (list; graph; refs; listen; history; text; internal format)
 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 Robert G. Wilson v, Table of n, a(n) for n = 1..100 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 Adjacent sequences:  A173225 A173226 A173227 * A173229 A173230 A173231 KEYWORD nonn,base AUTHOR José María Grau Ribas, Feb 13 2010 STATUS approved

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Last modified September 18 18:55 EDT 2021. Contains 347533 sequences. (Running on oeis4.)