A027869 -id:A027869 - cp's OEIS frontend

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.

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

A332242 Numbers k such that k! has exactly k nonzero decimal digits.

Original entry on oeis.org

1, 48, 49, 53, 54, 57

Offset: 1

Chai Wah Wu, Feb 07 2020

No other terms < 100000.

Conjecture: these 6 terms are the only terms of the sequence, i.e., there are no terms larger than 57.

			48! = 12413915592536072670862289047373375038521486354677760000000000 has 48 nonzero decimal digits, so 48 is a term.

Cf. A000142, A027869, A034886.

q:= n-> nops(subs(0=NULL, convert(n!, base, 10)))=n:
select(q, [$0..100])[];  # Alois P. Heinz, Feb 07 2020

isok(k) = #select(x->(x != 0), digits(k!)) == k; \\ Michel Marcus, Feb 08 2020

A332242_list, i, n = [], 0, 1
while i < 1000:
    s = str(n)
    if len(s) - s.count('0') == i:
        A332242_list.append(i)
    i += 1
    n *= i

{ k : A034886(k) - A027869(k) = k }.

A356739 a(n) is the smallest k such that k! has at least n consecutive zeros immediately after the leading digit in base 10.

Original entry on oeis.org

7, 153, 197, 7399, 24434, 24434, 9242360, 238861211, 238861211

Offset: 1

Christian Perfect, Aug 25 2022

			a(1) = 7, because 7! = 5040 has one zero immediately after the leading digit, and there is no k<7 with that property.

Cf. A000142, A027869.

from itertools import count
t = 1
n = 1
for k in count(1):
    t *= k
    while str(t)[1:1+n] == '0'*n:
        print(n,k)
        n += 1

a(5)-a(6) from Amiram Eldar, Aug 25 2022

a(7)-a(9) from Jinyuan Wang, Aug 25 2022

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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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A332242 Numbers k such that k! has exactly k nonzero decimal digits.

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A356739 a(n) is the smallest k such that k! has at least n consecutive zeros immediately after the leading digit in base 10.

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