|
| |
|
|
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.
|
|
4
|
|
|
|
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
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
1,2
|
|
|
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
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)
|
|
|
LINKS
|
|
|
|
EXAMPLE
|
a(2) = 3, 2221 is a prime but 21 and 221 are composite.
|
|
|
PROG
|
(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
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
base,nonn
|
|
|
AUTHOR
|
|
|
|
EXTENSIONS
|
|
|
|
STATUS
|
approved
|
| |
|
|
