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 ?

  • $\begingroup$ 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? $\endgroup$ – Barry Cipra Apr 21 '16 at 11:34
  • $\begingroup$ @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. $\endgroup$ – Michael AMH Apr 21 '16 at 11:38
  • $\begingroup$ 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 (!!) : ) $\endgroup$ – Michael AMH Apr 21 '16 at 11:48
  • $\begingroup$ 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). $\endgroup$ – Barry Cipra Apr 21 '16 at 11:54
  • $\begingroup$ 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 $1000$. $\endgroup$ – Michael AMH Apr 21 '16 at 12:18

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