A242457 - OEIS

%I #17 Aug 18 2014 08:20:39

%S 2,3,3,0,0,0,0,4,0,0,0,4,0,0,0,0,0,0,0,0,0,0,0,5,0,0,0,0,0,0,0,0,0,0,

%T 0,0,0,0,0,5,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,5,0,0,0,0,0,0,0,0,

%U 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0

%N Least number k such that k!/n is prime, or 0 if no such number exists.

%C a(n) <= n + 2 for all n > 0.

%H Jens Kruse Andersen, <a href="/A242457/b242457.txt">Table of n, a(n) for n = 1..10000</a>

%e a(3) = 3 because 3!/3 = 6/3 = 2, which is prime.

%e a(4) = 0 because there is no k such that k!/4 is a prime. For all k > 3, k! has 24 as a divisor, so therefore k!/4 has 6 as a divisor and is therefore certainly composite.

%p N:= 7: # to get all a(n) for n <= N!

%p A:= Array(1..N!):

%p for k from 1 to N do

%p for p in select(isprime,[$2..k]) do

%p     if A[k!/p] = 0 then A[k!/p]:= k fi

%p od

%p od:

%p seq(A[n],n=1..N!); # _Robert Israel_, Aug 18 2014

%o (PARI)

%o a(n)=for(k=1,n+2,s=k!/n;if(floor(s)==s,if(ispseudoprime(s),return(k))))

%o n=1;while(n<100,print1(a(n),", ");n++)

%Y Cf. A242456.

%K nonn

%O 1,1

%A _Derek Orr_, Aug 16 2014