A173228 The number of trailing zeros in (10^n)!

2, 24, 249, 2499, 24999, 249998, 2499999, 24999999, 249999998, 2499999997, 24999999997, 249999999997, 2499999999997, 24999999999998, 249999999999997, 2499999999999996, 24999999999999995, 249999999999999995, 2499999999999999995, 24999999999999999996

Offset: 1

Author

José María Grau Ribas, Feb 13 2010

Keywords

• nonn
• base

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

Programs

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 *)

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

Formula

a(n) = Sum_{k>=1} floor(10^n/5^k). - Stephen G Cappella, Dec 13 2017