Concatenating the first n semiprimes (in order) to get a semiprime $469101415$…

The concatenation of the first $1,2,3,6,43$, and $61$ semiprimes (in order) is a semiprime (!),

• $4=2 . 2$
• $46=2 . 23$
• $469=7 . 67$
• $469101415=5 . 93820283$
• $4691014152122....121122123129$ (proven semiprime,though no factors are known)
• $46910141521222526....183185187=108525583p$. (p is prime)

After these I don't find anymore such semiprime up to the first $350$ semiprimes. My question: Is there anymore semiprime of such form ?

• Could you elaborate on how you know the last two concatenations are semiprimes? In particular, what algorithm shows that a number has exactly two factors without explicitly finding them? – Barry Cipra Apr 21 '16 at 11:34
• @BarryCipra , p there means a prime. And I say it proven semiprime because I saw it at one site titled _prime curios_at $34$+ page. – Michael AMH Apr 21 '16 at 11:38
• And yes, there's an algorithm for semiprimeness test (!), but I forget the connection, but I believe I also saw it at oeis. Incredibly this algorithm can determine whether a $5000$ digit number is a semiprime or not, without mentioning the factors at all (!!) : ) – Michael AMH Apr 21 '16 at 11:48
• I assume the link you mean is primes.utm.edu/curios/page.php?curio_id=11492 . It would help a lot if you explained how you go about testing for semiprimeness (although I assume most concatenations get ruled out because they're divisible by a couple of small primes). – Barry Cipra Apr 21 '16 at 11:54
• I just checked it manually with my laptop using online prime number calculator, and yes most of them divisible by small primes, especially primes below $1000$. – Michael AMH Apr 21 '16 at 12:18