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^n-1 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