A379944 Smallest number of leading digits of n! that form a prime (or 0 if none exist).

0, 0, 1, 0, 1, 0, 1, 1, 0, 1, 1, 1, 2, 0, 0, 2, 1, 1, 0, 7, 1, 1, 2, 1, 0, 5, 0, 0, 1, 8, 1, 0, 1, 6, 1, 3, 1, 2, 1, 1, 0, 1, 8, 38, 1, 2, 1, 1, 5, 34, 1, 5, 6, 0, 1, 0, 1, 6, 1, 2, 2, 1, 1, 2, 8, 9, 1, 1, 1, 2, 2, 0, 2, 5, 1, 1, 0, 4, 2, 2, 1, 1, 2, 1, 1, 1, 1

Offset: 0

Author

Carson R. Smith, Jan 07 2025

Keywords

• nonn
• base

Comments

It appears that as n gets large, a(n) can become arbitrarily large.

It appears that values of n such that a(n) = 0 exist for arbitrarily large n.

Examples

			For n = 3, 3! = 6, 6 is not prime, a(3) = 0.
For n = 19, 19! = 121645100408832000, 1216451 is the smallest prime, a(19) = 7.
		

Links

• Carson R. Smith, Table of n, a(n) for n = 0..2000

Crossrefs

Cf. A000040, A000142, A046277.

Programs

Mathematica

A379944[n_] := Catch[Do[If[PrimeQ[FromDigits[#[[;; k]]]], Throw[k]],{k,Length[#]}] & [IntegerDigits[n!]]; 0];
Array[A379944, 100, 0] (* Paolo Xausa, Jan 16 2025 *)

PARI

a(n) = my(d=digits(n!)); for (k=1, #d, if (isprime(fromdigits(Vec(d, k))), return(k))); \\ Michel Marcus, Jan 08 2025

Python

import math
from sympy import isprime
def a(n):
    factorial = str(math.factorial(n))
    for d in range(1, len(factorial)+1):
        if isprime(int(factorial[:d])):
            return d
    return 0

Extensions

More terms from Jinyuan Wang, Jan 07 2025