This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
DigitCount[#,10,9]&/@(Range[0,100]!) (* Harvey P. Dale, Dec 12 2013 *)
a:= n-> numboccur(2, convert(n!, base, 10)): seq(a(n), n=0..101); # Alois P. Heinz, Apr 26 2021
Table[DigitCount[n!,10,2],{n,0,110}] (* Harvey P. Dale, Jun 20 2021 *)
a(n) = #select(x->(x==2), digits(n!)); \\ Michel Marcus, Apr 26 2021
n=10: 10! = 3628800 => a(10) = Max{1,4,5,7,9} = 9;
n=11: 11! = 39916800 => a(11) = Max{2,4,5,7} = 7.
n=10: 10! = 3628800 => a(10) = Min{1,4,5,7,9} = 1;
n=11: 11! = 39916800 => a(11) = Min{2,4,5,7} = 2.
Table[Length[Select[IntegerDigits[n!],Positive]],{n,0,70}]
a(n) = #select(x->(x>0), digits(n!)); \\ Michel Marcus, Aug 26 2022
from math import factorial
def a(n): return len(str(factorial(n)).replace("0", ""))
print([a(n) for n in range(71)]) # Michael S. Branicky, Aug 26 2022
a137581 = a055641 . a004154 -- Reinhard Zumkeller, Apr 01 2015
A137581:= proc(n) uses StringTools; CountCharacterOccurrences(TrimRight(SubstituteAll( convert(n!,string),"0"," "))," ") end proc; # Robert Israel, May 07 2012
f[n_] := f[n] = Block[{a = n!}, While[Mod[a, 10] == 0, a = a/10]; Count[ IntegerDigits@a, 0]]; Table[f@n, {n, 0, 98}] (* Robert G. Wilson v, Jan 28 2008 *)
Table[DigitCount[n!/10^IntegerExponent[n!,10],10,0],{n,0,100}] (* Harvey P. Dale, Apr 10 2023 *)
a(12) = 47916 since 12! = 479001600.
Table[FromDigits[Select[IntegerDigits[n!],Positive]], {n,0,26}]
a(n) = fromdigits(select(x->(x>0), digits(n!))); \\ Michel Marcus, Aug 26 2022
from math import factorial
def a(n): return int(str(factorial(n)).replace("0", ""))
print([a(n) for n in range(27)]) # Michael S. Branicky, Aug 26 2022
a(0) = a(1) = 1 because 0! = 1! = 1 and 1 is the only digit present; a(4) = -1 since 4! = 24 and there are only two digits appearing with the same frequency, 2 and 4. a(14) = -1 because 14! = 87178291200 and, not counting the two trailing 0's, there are two 1's, two 2's, two 7's, and two 8's.
a[n_] := If[Length[c=Commonest[IntegerDigits[n! / 10^IntegerExponent[n!]]]] > 1, -1, c[[1]]]; Array[a, 86, 0]
from collections import Counter from sympy import factorial def A375348(n): return -1 if len(k:=Counter(str(factorial(n)).rstrip('0')).most_common(2)) > 1 and k[0][1]==k[1][1] else int(k[0][0]) # Chai Wah Wu, Sep 15 2024
A000142(a(10)) = 11! = 399168*10^2; A000142(a(11)) = 14! = 871782912*10^2; A000142(a(12)) = 31! = 822283865417792281772556288*10^7.
isok(k) = my(f=k!); while(!(f % 10), f \= 10); #select(x->(x == 0), digits(f)) == 0; \\ Michel Marcus, Jun 28 2023
a(2) = 7 because 7! = 5040 is the first prime factorial followed by 11! = 39916800 to contain exactly 2 0's.
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