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

0,3

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.

Links

Table of n, a(n) for n=0..85.

Example

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.

Mathematica

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

Prog

(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

Crossrefs

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

Sequence in context: A321172 A375575 A127508 * A244814 A090185 A059813

Adjacent sequences: A375345 A375346 A375347 * A375349 A375350 A375351

Keyword

base,easy,sign

Author

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

Status

approved