A242775 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.

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

Offset: 1

Author

Vladimir Shevelev, Sep 13 2014

Keywords

• nonn

Comments

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

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

Examples

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

Links

• Peter J. C. Moses, Table of n, a(n) for n = 1..2000

Crossrefs

Cf. A232210, A247341, A247342.

Programs

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

Extensions

More terms from Peter J. C. Moses, Sep 14 2014