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(All dots here means concatenation.)

Let $s= ab$ be a semiprime number, then I call s a "prime" semiprime if all these following conditions are satisfied:

  1. The reversal of $s$ is a prime
  2. The concatenation of $a$ and $b$ in any order(i.e. $a{.}b$ and $b{.}a$) are primes.
  3. And these following six are all also primes $s{.}a{.}b$, $s{.}b{.}a$, $a{.}s{.}b$, $b{.}s{.}a$, $a{.}b{.}s$, $b{.}a{.}s$.

I've search numbers up to $10000$, but I couldn't find a single semiprime that satisfies all these conditions, so what is the smallest "prime" semiprime?

Michael Galuza
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1 Answers1

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There are likely to be infinitely many, and you're not likely to find any.

As a rough estimate, the reversal of $s$ and the concatenations of $a$ and $b$ are of the order of $s$, and the concatenations of $s$, $a$ and $b$ are of the order of $s^2$. So you want $s$ to be a semiprime (density $\log\log s\,/\,(2\log s)$), three numbers of the order of $s$ to be primes (density $1/\log s$) and six numbers of the order of $s^2$ to be primes (density $1/\log s^2=1/(2\log s)$). The density of such numbers is roughly

$$ \frac{\log\log s}{2\log s}\left(\frac1{\log s}\right)^3\left(\frac1{2\log s}\right)^6=\frac{\log\log s}{128(\log s)^{10}}\;. $$

The integral of that density from where you ended your search, $10^4$, up to a reasonable search limit of $10^{13}$ is, according to Wolfram|Alpha, about $7\cdot10^{-4}$. Nevertheless, the integral over this density to infinity diverges, so if you keep looking long enough, you should eventually find one. Expect about two of them with $18$ digits.

joriki
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