%I
%S 1,3,1,1,11,1,1,2,2,1,9,3,1,5,1,3,15,1,1,2,1,60,3,1,1,2,1,1,5,5,1,2,1,
%T 6,12,3,12,3,5,1,2,1,1,5,3,1,0,2,1,9,2,1,6,1,6,18,1,3,45,1,6,3,1,1,2,
%U 1,0,3,1,1,2,3,4,8,1,1,6,2,36,96,1,1,5,304,6,2,6,1,2,2,1,2,5,1,6,5,1,2,1,0
%N a(n) = smallest r where (concatenation of n, r times with itself)*10 + 1 is a prime given by A087403(n), or 0 if no such number exists.
%C Conjecture: No term is zero. [Warning: This is known to be wrong, see below.  _M. F. Hasler_, Jan 08 2015]
%C a(47), a(67), a(100), a(107), a(114) are zero or larger than 1000.  _Ray Chandler_, Sep 23 2003; edited by _M. F. Hasler_, Jan 08 2015
%C a(47) > 10000 or 0. a(67) > 10000 or 0. a(100) > 10000 or 0. a(107) = 2478. a(114) = 1164. See link for more details.  _Derek Orr_, Oct 02 2014
%C From _Farideh Firoozbakht_, Jan 07 2015: (Start)
%C The conjecture is not true and there exist many numbers n such that a(n)=0.
%C Theorem: If m is a positive integer and a(10^m)=r then r+1 divides m+1.
%C Corollary: If p is a prime number then a(10^(p1))=0 or (10^(p^2)1)/(10^p1) is a prime number.
%C By using the theorem and its corollary we can prove that for m = 2, 3, ..., 275 a(10^m)=0.
%C What is the smallest odd prime p, such that (10^(p^2)1)/(10^p1) is a prime number (and a(10^(p1)) could be nonzero)?
%C What is the smallest integer m > 1 such that a(10^m) is nonzero?
%C Conjecture: If n is not of the form 10^m then a(n) is nonzero.
%C _M. F. Hasler_ has checked proofs of the theorem and its corollary.
%C (End)
%H Derek Orr, <a href="/A086766/a086766.txt">Values of a(n) > 1000 for n < 1000</a>
%e a(2) = 3, 2221 is a prime but 21 and 221 are composite.
%o (PARI)
%o a(n)=for(k=1,10^4,if(ispseudoprime((n/(10^#Str(n)1))*(10^(#Str(n)*k+1)10)+1),return(k)))
%o vector(46,n,a(n)) \\ _Derek Orr_, Oct 02 2014
%Y Cf. A087403.
%K base,nonn
%O 1,2
%A _Amarnath Murthy_, Sep 10 2003
%E More terms from _Ray Chandler_, Sep 23 2003
