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Smallest prime p such that floor(n!/p) is prime, or 0 if no such prime exists.
1

%I #16 Aug 04 2018 19:46:54

%S 0,0,2,7,7,19,29,17,107,29,151,67,101,31,43,163,59,31,41,173,79,167,

%T 73,233,107,73,29,43,1259,89,317,191,349,541,199,173,577,89,373,997,

%U 197,773,1093,257,1733,487,349,149,1511,2621,389,181,151

%N Smallest prime p such that floor(n!/p) is prime, or 0 if no such prime exists.

%C Conjecture: There are no zeros for n > 2.

%C This conjecture is correct. For m > 1, there is always a prime between m and 2*m. Taking m = n!/4, this gives us a prime p such that floor(n!/p) = 2 or 3. - _Franklin T. Adams-Watters_, Jul 28 2011

%t Do[p = 1; While[ !PrimeQ[Floor[n!/Prime[p]]], p++ ]; Print[Prime[p]], {n, 3, 30}] (* _Ryan Propper_, Jun 23 2005 *)

%Y Cf. A090524.

%K nonn

%O 1,3

%A _Amarnath Murthy_, Dec 07 2003

%E More terms from _Ryan Propper_, Jun 23 2005

%E More terms from _Stefan Steinerberger_, Jun 07 2007