A027869 Number of 0's in n!.

0, 0, 0, 0, 0, 1, 1, 2, 2, 1, 2, 2, 4, 4, 2, 4, 4, 4, 5, 6, 7, 7, 8, 5, 6, 9, 8, 9, 10, 7, 9, 7, 10, 8, 11, 9, 10, 12, 16, 12, 9, 15, 13, 13, 12, 13, 16, 11, 14, 14, 19, 18, 18, 17, 18, 18, 17, 20, 17, 19, 19, 26, 20, 21, 20, 20, 23, 22, 25, 21, 20, 25, 23, 35

Offset: 0

Author

N. J. A. Sloane

Keywords

• nonn
• base

Links

• Alois P. Heinz, Table of n, a(n) for n = 0..10000 (first 1001 terms from T. D. Noe)
• Index entries for sequences related to factorial numbers.

Crossrefs

Cf. A000142, A137579, A137580, A034886.

Programs

Mathematica

Table[Count[IntegerDigits[n!], 0], {n, 0, 100}] (* T. D. Noe, Apr 10 2012 *)
DigitCount[Range[0,80]!,10,0] (* Harvey P. Dale, Jul 08 2020 *)

PARI

a(n)=my(d=digits(n!)); sum(i=1,#d,d[i]==0) \\ Charles R Greathouse IV, Jul 06 2017

Python

from math import factorial
def a(n): return str(factorial(n)).count('0')
print([a(n) for n in range(74)]) # Michael S. Branicky, Jan 11 2022

Formula

a(n) = A034886(n) - (A079680(n) + A079714(n) + A079684(n) + A079688(n) + A079690(n) + A079691(n) + A079692(n) + A079693(n) + A079694(n)). - Reinhard Zumkeller, Jan 27 2008

A027868(n) <= a(n). - Reinhard Zumkeller, Jan 27 2008

Conjecture: a(n) ~ (9*A027868(n) + A034886(n))/10. This formula is based on the assumption that the digits other than trailing zeros are uniformly randomly distributed. - Nicolas Bělohoubek, Jan 11 2022