%I #20 Sep 23 2014 11:10:23

%S 0,0,0,1,1,1,1,2,2,2,1,1,4,2,1,1,1,2,1,2,1,1,1,1,1,3,3,2,1,2,7,3,1,3,

%T 2,2,8,1,1,7,2,1,1,5,3,2,2,2,3,1,3,8,5,1,1,4,3,1,4,5,3,6,1,2,1,2,1,3,

%U 1,2,2,1,3,1,6,3,1,3,4,2,3,8,4,1,3,34,1

%N Let b_k=3...3 consist of k>=1 3's. Then a(n) is the smallest k such that the concatenation b_k and prime(n) is prime, or a(n)=0 if there is no such prime.

%C Conjecture: for n>=4, a(n)>0.

%C Records >=1: 1,2,4,7,8,34,... correspond to primes 7,19,41,127,157,443,...

%H Peter J. C. Moses, <a href="/A242775/b242775.txt">Table of n, a(n) for n = 1..2000</a>

%e For n<=3, a(n) = 0, because 3..32, 3..33 and 3..35 can never be prime, whatever the number of 3's that are concatenated.

%e For n=4, prime(n)=7, 37 is prime. So a(4)=1.

%o (PARI) a(n) = {if (n<=3, return (0)); p = prime(n); k = 1; while (! isprime(p = eval(concat("3", Str(p)))), k++); k; } \\ _Michel Marcus_, Sep 17 2014

%Y Cf. A232210, A247341, A247342.

%K nonn

%O 1,8

%A _Vladimir Shevelev_, Sep 13 2014

%E More terms from _Peter J. C. Moses_, Sep 14 2014