A242567 - OEIS

%I #25 Dec 12 2024 02:52:36

%S 1,1,0,1,18,1,712,5031,14,1,18,1,479001586,1719,87178291184,1,3024,1,

%T 40,633,124748,1,86,51847,625793187628,123,20404,1,210,1,

%U 265252859812191058636308479999968,755,263130836933693530167218012159999966

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

%C a(n) = 1 iff n+1 is prime.

%C For n > 2, in order for (n!+k)/(n+k) to be an integer, the smallest integer possible is 2. Thus, a(n) <= n!-2n for all n > 2.

%C Let q = (n!+k)/(n+k). Then, k = (n!-n)/(q-1)-n. So, a(n) = d-n, where d is the smallest divisor (> n) of n!-n. (for all n >= 4) - _Hiroaki Yamanouchi_, Sep 29 2014

%H Hiroaki Yamanouchi, <a href="/A242567/b242567.txt">Table of n, a(n) for n = 1..79</a>

%e (6!+1)/(6+1) = 103 is an integer. Thus a(6) = 1.

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

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

%Y Cf. A006093.

%K nonn

%O 1,5

%A _Derek Orr_, May 17 2014

%E a(13), a(15), a(21), a(25), a(31) and a(33) from _Hiroaki Yamanouchi_, Sep 29 2014