OFFSET
1,2
COMMENTS
There cannot be any primes of this form when k is even, because all such numbers must be divisible by 11. A number is divisible by 11 if the difference between the sum of its odd digits and the sum of its even digits is 0 or divisible by 11. When k is even, the difference is always 0. - Dmitry Kamenetsky, Jul 12 2008
The next term, if one exists, is >100000. - Robert Price, Mar 24 2011
See Kamada link - primecount.txt for terms, primesize.txt for discovery details including probable or proved primes - search on "10021".
LINKS
EXAMPLE
For k=3 we have 10^3+21 = 1000+21 = 1021, which is prime.
MATHEMATICA
q=21; s=""; For[ a=q, a<=q, s="10^n+"<>ToString[ a ]<>":"; n=0; For[ i=1, i< 10^3, If[ PrimeQ[ 10^i+a ], n=1; s=s<>ToString[ i ]<>", " ]; i++ ]; If[ n>0, Print[ s ] ]; a++ ] (* Vladimir Joseph Stephan Orlovsky, May 06 2008 *)
PROG
(PARI) for(n=1, 1e4, if(ispseudoprime(10^n+21), print1(n", "))) \\ Charles R Greathouse IV, Jul 20 2011
CROSSREFS
KEYWORD
more,nonn
AUTHOR
Julien Peter Benney (jpbenney(AT)ftml.net), Jun 01 2005
EXTENSIONS
a(6)=77 inserted by Vladimir Joseph Stephan Orlovsky, May 06 2008
a(13)=8007 from Dmitry Kamenetsky, Jul 12 2008
a(14)=25671 from Robert Price, Nov 08 2010
Edited by Ray Chandler, Dec 24 2010
a(15)=48003 from Robert Price, Dec 31 2010
a(16)=55811 from Robert Price, Jan 09 2011
a(17)=94983 from Robert Price, Mar 24 2011
STATUS
approved