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

Author

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

Keywords

• base
• easy
• sign

Comments

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.

Examples

			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.
		

Crossrefs

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

Programs

Maple

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

Mathematica

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]

Python

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