%I #10 Jul 30 2014 20:25:39
%S 0,0,1,1,2,1,2,1,2,0,3,1,2,1,2,0,3,1,2,1,2,0,3,1,2,4,4,0,3,1,2,1,2,0,
%T 3,0,3,1,2,0,4,1,2,1,2,0,3,1,2,0,0,0,3,1,2,0,0,0,3,1,2,1,2,0,3,0,3,1,
%U 2,0,4,1,2,1,2,0,3,0,3,1,2,0,4,1,2,0,0,0,3,1,2,0,0,0,3,0,4,1,2,0,0,1,2,1,2,0,3,1,2,1,2,0,3,1,2,0,0,0,3
%N Least number k > 0 such that n - k! is prime, or 0 if no such k exists.
%C a(n)! < n for all n. Thus a(n) = 0 is definite.
%H Jens Kruse Andersen, <a href="/A245715/b245715.txt">Table of n, a(n) for n = 1..10000</a>
%e 11 - 1! = 10 is not prime.
%e 11 - 2! = 9 is not prime.
%e 11 - 3! = 5 is prime. Thus a(11) = 3.
%o (PARI)
%o a(n)=for(k=1,n,if(ispseudoprime(n-k!),return(k)))
%o vector(150,n,a(n))
%K nonn
%O 1,5
%A _Derek Orr_, Jul 30 2014
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