login
Least number k >= 0 such that (n!+k)/(n+k) is prime.
2

%I #16 Sep 29 2014 11:26:05

%S 0,1,110,1,5026,10070,362862,1,39916778,34,6227020774,25152407,

%T 1307674367970,50917,355687428095966,256443711659,121645100408831962,

%U 1286,51090942171709439958,111014413076599,25852016738884976639954,51704033477769953279974

%N Least number k >= 0 such that (n!+k)/(n+k) is prime.

%C a(n) <= n!-2n for all n. See A242567.

%C a(68) = 1526549.

%C Since 2 is prime, we see that (n!+k)/(n+k) = 2 when k = n!-2n, which is an integer. Thus, a(n) will always be nonzero. However, it is uncertain whether there are smaller k-values besides n!-2n.

%H Hiroaki Yamanouchi, <a href="/A242568/b242568.txt">Table of n, a(n) for n = 3..60</a>

%e (4!+1)/(4+1) = 5 is prime. Thus, a(4) = 1.

%o (PARI) a(n)=for(k=1,5*10^6,s=(n!+k)/(n+k);if(floor(s)==s,if(ispseudoprime(s),return(k))));

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

%Y Cf. A242567.

%K nonn

%O 3,3

%A _Derek Orr_, May 17 2014

%E a(11)-a(24) from _Hiroaki Yamanouchi_, Sep 29 2014