Equivalently, primes of the form 4*10^n + 9*R_n, where R_n is the repunit (A002275) of length n.
If m is in the sequence then m appears at the end of m^3, in fact if n>1 and m=5*10^n1 then m appears at the end of m^3.  Farideh Firoozbakht, Nov 10 2005
If n is in the sequence then 4n is a term of A067206. Namely the digits of 4n end in phi(4n)  the proof is easy.  Farideh Firoozbakht, Dec 30 2006
The next term  a(7)  has 211 digits.  Harvey P. Dale, Feb 20 2016
