A031144 -id:A031144 - cp's OEIS frontend

A031145 Factorials with a record number of zeros.

1, 120, 5040, 479001600, 6402373705728000, 121645100408832000, 2432902008176640000, 1124000727777607680000, 15511210043330985984000000, 304888344611713860501504000000

Offset: 0

N. J. A. Sloane

			From _Alonso del Arte_, May 19 2017: (Start)
Note that 5040 has two zeros, even though only one of them is a trailing zero.
Although 3628800 has one more zero than 362880, it still has as many zeros as 5040, and for that reason it is not in this sequence.
Thus the next term after 5040 is 479001600, which has four zeros. (End)

Michael De Vlieger, Table of n, a(n) for n = 0..59

Cf. A031144 (indices of factorials), A027868.

Function[s, Map[Position[s, #][[1, 1]] &, Union@ FoldList[Max, s]]! ]@ Table[DigitCount[n!, 10, 0], {n, 28}] (* Michael De Vlieger, May 20 2017 *)
DeleteDuplicates[Table[{n!,DigitCount[n!,10,0]},{n,50}],GreaterEqual[#1[[2]],#2[[2]]]&][[;;,1]] (* Harvey P. Dale, Dec 03 2023 *)

More terms from Erich Friedman.

Name clarified by Alonso del Arte, May 19 2017

A375348 a(n) is the mode of the digits of n! not counting trailing zeros (using -1 if multimodal).

Original entry on oeis.org

1, 1, 2, 6, -1, -1, -1, -1, -1, 8, 8, 9, 0, 2, -1, -1, 8, -1, 7, -1, -1, -1, 7, 8, 3, 1, 6, 8, -1, -1, 8, 2, 3, 8, 9, -1, 9, -1, 0, 8, 1, -1, -1, 3, 8, 6, -1, 1, 7, 2, 6, -1, 8, 3, -1, 5, 4, 2, -1, 8, 4, 0, 2, 6, -1, 2, 4, 6, 1, 2, 8, 8, 8, 0, 2, 4, -1, 8, 2, 1, 5, 7, 4, -1, 1, 0

Offset: 0

Stefano Spezia and Robert G. Wilson v, Aug 12 2024

Inspired by A356758.
If we were to count trailing zeros, then would have a(n) = 0 for all n >= 34. Therefore we only consider the decimal digits of A004154(n).
Conjecture: excluding -1, as n -> oo, all digits occur equally often.

			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.

Cf. A004154, A027869, A031144, A034886, A061010, A137579, A137580, A356758.

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

A375575 a(n) is the least frequent digit of n! not counting trailing zeros, or -1 if there is more than one least frequent digit.

Original entry on oeis.org

1, 1, 2, 6, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, 9, -1, -1, -1, -1, -1, -1, -1, -1, -1, -1, 2, -1, -1, 7, 0, 4, -1, -1, -1, -1, -1, -1, 8, -1, -1, -1, 8, -1, -1, 9, -1, -1, 0, 9, 9, -1, -1, -1, 1, -1, -1, 2, -1, -1, 5, 5, 1, 4, 5, 7, -1, 5, -1, 6, 6, 0, -1, 5, 9, 6, -1, 0, 5, 9

Offset: 0

Stefano Spezia and Robert G. Wilson v, Aug 19 2024

Analogous to A375348.
If we were to count trailing zeros, then a(n) would never equal zero, for all n's >= 0. Therefore we only consider the decimal digits of A004154.
Conjecture: excluding -1, as n -> oo, the digits distribution is uniform as in A375348.

			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 two least frequent digits, 2 and 4.
a(14) = 9 because 14! = 87178291200 and, not counting the two trailing 0's, there are two 1's, two 2's, two 7's, two 8's but only one 9.

Cf. A004154, A027869, A031144, A034886, A061010, A137579, A137580, A375348.

f:= proc(n) local L,j;
  L:= convert(n!,base,10);
  for j from 1 while L[j] = 0 do od:
  L:= Statistics:-Tally(L[j...-1]);
  L:= sort(L,(a,b) -> rhs(a) < rhs(b));
  if nops(L) >= 2 and rhs(L[2]) = rhs(L[1]) then -1 else lhs(L[1]) fi
end proc:
map(f, [$0..100]); # Robert Israel, Sep 02 2024

Rarest[lst_] := MinimalBy[ Tally[lst], Last][[All, 1]]; a[n_] := If[ Length[c = Rarest[ IntegerDigits[n!/10^IntegerExponent[n!, 10]] ]] >1, -1, c[[1]]]; Array[a, 80, 0]

from collections import Counter
from sympy import factorial
def A375575(n): return -1 if len(k:=Counter(str(factorial(n)).rstrip('0')).most_common()) > 1 and k[-1][1]==k[-2][1] else int(k[-1][0]) # Chai Wah Wu, Sep 15 2024

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A031145 Factorials with a record number of zeros.

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A375348 a(n) is the mode of the digits of n! not counting trailing zeros (using -1 if multimodal).

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A375575 a(n) is the least frequent digit of n! not counting trailing zeros, or -1 if there is more than one least frequent digit.

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