> f(n)=(10^(6*n)+1)/(10^(2*n)+1)

In fairness, I should highlight the smiley,

> f(1)= 9901 is prime

> f(2)= 99990001 is prime

> f(3)= 999999000001 is prime

> f(4)= 9999999900000001 is prime

> Puzzle: Find the next prime of this form :-)

lest folk fall into the trap of trying values

of n with prime divisors greater than 3.

Feeding pfgw the following ABC2 file,

I found nothing new:

ABC2 Phi(2^$a*3^$b,10)

a: from 2 to 6

b: from 1 to 5

so the heuristics are truly grim!

David