Spektrum problem: Beetroot and Radish primes

The set of Beetroots can be generated recursively by length of the base-10 expansion. First one proves a simple lemma (which basically derives from the definition):

Lemma 1. Let $B_{N}$ denote the set of Beetroots of length $N$ . Then for all $N>1$ a positive natural $n$ is in $B_{N}$ if and only if $n\in\mathbb{P}$ and $n^{-},n^{+}\in B_{N-1}\cup\{1\}$ , where $n^{-},n^{+}$ are the two $N-1$ -digit subsequences of $n$ .

Lemma 2. If $B_{N}=\emptyset$ for some $N>1$ , then $B_{m}=\emptyset$ for all $m\geq N$ . Proof. It suffices to show that $B_{N+1}=\emptyset$ , then repeating the argument the desired result follows by induction. Suppose a Beetroot $n$ exists of length $N+1$ and a. Then by Lemma 1, $n^{-},n^{+}\in B_{N}\cup\{1\}=\emptyset\cup\{1\}=\{1\}$ , so $n^{-},n^{+}=1$ . Now, technically one of these could be of the form $00\ldots 01$ , as a subsequence of $n$ . But not both of them. So at least on subsequence of length $N$ is of the form $1$ . But this is impossible, as $N>1$ . QED.

Appealing to Lemma 1, it is now easy to generate the Beetroots. Furthermore, it is easy to see when to stop.

$B_{1}=\{1,2,3,5,7\}$

$B_{2}=\{11,12,\ldots,77\}\cap\mathbb{P}=\{11,13,17,23,31,37,53,71,73\}$

$B_{3}=\{111,113,\ldots,737\}\cap\mathbb{P}=\{113,131,137,173,311,313,317,373\}$

$B_{4}=\{1131,1137,\ldots,3173\}\cap\mathbb{P}=\{1373,3137\}$

$B_{5}=\{31373\}\cap\mathbb{P}=\emptyset$ .

By Lemma 2 it follows that there are no Beetroots of length $\geq 5$ . Hence, by the above computations, it thus follows that the largest Beetroot is $\fbox{\$b=3137\$}$ .

To compute the Radishes is easier, as we only need the smallest. The digits of a Radish must lie in $\{0,4,6,8,9\}$ . There are by definition no 1-digit Radishes. The only possibly 2-digit Radishes are

$\{40,44,46,\ldots,98,99\}\cap\mathbb{P}=\{89\}$

Hence we need not compute any further. The smallest Radish is $\fbox{\$r=89\$}$ .

The desired answer is thus \fbox{\$b^{\frown}r=313789\$} .