%I #55 Feb 11 2022 17:17:54
%S 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,
%T 8,11,9,10,12,16,12,9,15,13,13,12,13,16,11,14,14,19,18,18,17,18,18,17,
%U 20,17,19,19,26,20,21,20,20,23,22,25,21,20,25,23,35
%N Number of 0's in n!.
%H T. D. Noe, <a href="/A027869/b027869.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Fa#factorial">Index entries for sequences related to factorial numbers</a>.
%F 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
%F A027868(n) <= a(n). - _Reinhard Zumkeller_, Jan 27 2008
%F 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
%t Table[Count[IntegerDigits[n!], 0], {n, 0, 100}] (* _T. D. Noe_, Apr 10 2012 *)
%t DigitCount[Range[0,80]!,10,0] (* _Harvey P. Dale_, Jul 08 2020 *)
%o (PARI) a(n)=my(d=digits(n!)); sum(i=1,#d,d[i]==0) \\ _Charles R Greathouse IV_, Jul 06 2017
%o (Python)
%o from math import factorial
%o def a(n): return str(factorial(n)).count('0')
%o print([a(n) for n in range(74)]) # _Michael S. Branicky_, Jan 11 2022
%Y Cf. A000142, A137579, A137580, A034886.
%K nonn,base
%O 0,8
%A _N. J. A. Sloane_