A356758 -id:A356758 - cp's OEIS frontend

A356757 Omit zero digits from factorial numbers.

1, 1, 2, 6, 24, 12, 72, 54, 432, 36288, 36288, 399168, 47916, 622728, 871782912, 137674368, 2922789888, 35568742896, 64237375728, 121645148832, 243292817664, 5199421717944, 11247277776768, 258521673888497664, 624484173323943936, 15511214333985984, 4329146112665635584

Offset: 0

Stefano Spezia, Aug 26 2022

			a(12) = 47916 since 12! = 479001600.

Index entries for sequences related to factorial numbers

Cf. A000142, A004719.
Cf. A004154, A243657, A321475.
Cf. A027869 (number of omitted zero digits), A356758 (number of nonzero digits).

Table[FromDigits[Select[IntegerDigits[n!],Positive]], {n,0,26}]

a(n) = fromdigits(select(x->(x>0), digits(n!))); \\ Michel Marcus, Aug 26 2022

from math import factorial
def a(n): return int(str(factorial(n)).replace("0", ""))
print([a(n) for n in range(27)]) # Michael S. Branicky, Aug 26 2022

a(n) = A004719(A000142(n)).

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

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A356757 Omit zero digits from factorial numbers.

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