A086766 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.

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, 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, 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

Offset: 1

Author

Amarnath Murthy, Sep 10 2003

Keywords

• base
• nonn

Comments

Conjecture: No term is zero. [Warning: This is known to be wrong, see below. - M. F. Hasler, Jan 08 2015]

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

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

From Farideh Firoozbakht, Jan 07 2015: (Start)

The conjecture is not true and there exist many numbers n such that a(n)=0.

Theorem: If m is a positive integer and a(10^m)=r then r+1 divides m+1.

Corollary: If p is a prime number then a(10^(p-1))=0 or (10^(p^2)-1)/(10^p-1) is a prime number.

By using the theorem and its corollary we can prove that for m = 2, 3, ..., 275 a(10^m)=0.

What is the smallest odd prime p, such that (10^(p^2)-1)/(10^p-1) is a prime number (and a(10^(p-1)) could be nonzero)?

What is the smallest integer m > 1 such that a(10^m) is nonzero?

Conjecture: If n is not of the form 10^m then a(n) is nonzero.

M. F. Hasler has checked proofs of the theorem and its corollary.

(End)

Examples

			a(2) = 3, 2221 is a prime but 21 and 221 are composite.
		

Links

• Derek Orr, Values of a(n) > 1000 for n < 1000

Crossrefs

Cf. A087403.

Programs

PARI

a(n)=for(k=1,10^4,if(ispseudoprime((n/(10^#Str(n)-1))*(10^(#Str(n)*k+1)-10)+1),return(k)))
vector(46,n,a(n)) \\ Derek Orr, Oct 02 2014

Extensions

More terms from Ray Chandler, Sep 23 2003