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

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

Author

William Boyles, May 31 2022

Keywords

• nonn
• easy
• base

Examples

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

Links

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

Crossrefs

Cf. A000079, A027868.

Programs

Haskell

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

Mathematica

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

PARI

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

Python

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

Python

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

Formula

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