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%I #23 Aug 20 2019 03:18:45
%S 1,2,7,7,19,22,17,74,29,76,67,49,31,43,95,59,31,41,173,79,94,73,233,
%T 107,73,29,43,201,89,274,191,349,346,199,173,249,89,373,662,197,453,
%U 166,257,865,487,254,149,852,758,389,181,151,699,634,577,542,199,61,278,482,467
%N Least k such that floor[n!/k] is prime.
%C Conjecture: There exists a number k such that for all n > k, a(n) is prime. Motivation: If p is the least prime >n then all the numbers from n to p-1 divide n!. And most of the numbers from p+1 to q also divide n! where q is the least prime > p, etc. and the dividend is composite in almost all cases.
%C I conjecture the opposite: there are infinitely many composites in this sequence. Indeed, heuristics suggest the composites may be of density 1. - _Charles R Greathouse IV_, Apr 07 2013
%H Charles R Greathouse IV, <a href="/A090521/b090521.txt">Table of n, a(n) for n = 2..1000</a>
%H Amarnath Murthy, <a href="http://doi.org/10.5281/zenodo.9289">Smarandache Reciprocal Function and an elementary inequality</a> Smarandache Notions Journal, Vol. 11, 2000.
%e a(5)=7 because the first values of floor(5!/k) for k=1,2,... are 120,60,40,30,24,20,17,15,13,12,... and among these the first prime number is 17, corresponding to k=7.
%p a:= proc(n) local k; for k while not isprime(floor(n!/k)) do od; k end:
%p seq(a(n), n=2..70); # _Emeric Deutsch_, Apr 18 2005
%t lk[n_]:=Module[{f=n!,k=1},While[!PrimeQ[Floor[f/k]],k++];k]; Array[lk,70,2] (* _Harvey P. Dale_, Mar 06 2015 *)
%o (PARI) a(n)=my(N=n!);for(k=if(n>3,n+1,n-1),N\3+2,if(ispseudoprime(N\k),return(k))) \\ _Charles R Greathouse IV_, Apr 04 2013
%Y Cf. A090522.
%K nonn
%O 2,2
%A _Amarnath Murthy_, Dec 07 2003
%E More terms from _Emeric Deutsch_, Apr 18 2005