A055223 -id:A055223 - cp's OEIS frontend

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

José María Grau Ribas, Feb 13 2010

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

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

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 *)

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

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

A354463 a(n) is the number of trailing zeros in (2^n)!.

Original entry on oeis.org

0, 0, 0, 1, 3, 7, 14, 31, 63, 126, 253, 509, 1021, 2045, 4094, 8189, 16380, 32763, 65531, 131067, 262140, 524285, 1048571, 2097146, 4194297, 8388603, 16777208, 33554424, 67108858, 134217720, 268435446, 536870902, 1073741816, 2147483642, 4294967289, 8589934584, 17179869176, 34359738358, 68719476729

Offset: 0

William Boyles, May 31 2022

			For n = 4, (2^4)! = 20922789888000, which has a(4) = 3 trailing zeros.

William Boyles, Table of n, a(n) for n = 0..650

Cf. A000079, A027868.

seq n = aux (2 ^ n) 0
  where
    aux x acc
      | x < 5 = acc
      | otherwise = aux y (acc + y)
      where
        y = x `div` 5

a[n_]:=IntegerExponent[(2^n)!]; Array[a,38,0] (* Stefano Spezia, Jun 01 2022 *)

a(n) = val(1<David A. Corneth, Jun 01 2022

from sympy import factorial, multiplicity
def a(n): return multiplicity(5, factorial(2**n, evaluate=False))
print([a(n) for n in range(39)]) # Michael S. Branicky, Jun 01 2022

def A354463(n):
    c, m = 0, 2**n
    while m >= 5:
        m //= 5
        c += m
    return c # Chai Wah Wu, Jun 02 2022

a(n) = A027868(A000079(n)). - Michel Marcus, Jun 01 2022

a(n) = 2^(n-2) - A055223(n) for n >= 2. - John Keith, Jun 06 2022

A174807 Floor(10^n/4) - A173228(n).

Original entry on oeis.org

0, 1, 1, 1, 1, 2, 1, 1, 2, 3, 3, 3, 3, 2, 3, 4, 5, 5, 5, 4, 3, 5, 6, 7, 5, 8, 8, 6, 8, 10, 10, 8, 6, 7, 8, 8, 10, 7, 9, 9, 10, 11, 10, 9, 10, 9, 11, 11, 11, 11, 12, 13, 13, 12, 14, 10, 14, 17, 15, 13, 13, 12, 15, 14, 16, 15, 12, 14, 15, 15, 16, 15, 15, 15, 16, 13, 12, 16, 17, 14, 20, 20, 20

Offset: 1

José María Grau Ribas, Mar 29 2010

a(n) = A055223(n) if n>1.

Antonio M. Oller-Marcen, José María Grau, On the base b expansion of the number of trailing zeros of b^k!, arXiv:1002.4175 [math.NT], 2010, and J. Int. Seq. 14 (2011) 11.6.8

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

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A354463 a(n) is the number of trailing zeros in (2^n)!.

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A174807 Floor(10^n/4) - A173228(n).

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