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

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

Author

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

Keywords

• base
• easy
• sign

Comments

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.

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 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.
		

Crossrefs

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

Programs

Mathematica

a[n_] := If[Length[c=Commonest[IntegerDigits[n! / 10^IntegerExponent[n!]]]] > 1, -1, c[[1]]]; Array[a, 86, 0]

Python

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