Minimal prime numbers (no shorter subsequence of digits is prime) I need help with minimal prime numbers, defined as primes such that no shorter subsequence of digits is prime.
I calculated the prime numbers I need, but can not find a way to find minimal prime numbers. On the Wikipedia page stated that there are exactly 26 minimal primes. But how are those numbers calculated? Can you give me a formula and/or an example?
 A: Things about primes and digits almost always amount to brute force. That's because base $10$ (or any base) is essentially an arbitrary system for representing numbers. 
Here, you have that every digit must be from $0,1,4,6,8,9$. Then you start eliminating:


*

*The number can't have $3+$ digits with have two or more digits $1$ because $11$ is prime (except when the number is $11$. 

*The number can't be a $3+$-digit number with a $1$ before a $9$, because $19$ isn't prime 

*...


It's really a dull argument. 
For example, if a $3+$ digit number contains a $1$, then the $1$ has to be the last digit (since the last digit must be $1$ or $9$, and a $1$ cannot come before a $9$.)
Since $41$ and $61$ and $2,3,5,7$ are prime, the only other digits that can occur are $9$ and $0$. Since $991$ is prime, there can't be more $9$s than $1$ for any other minimal prime, and since $9001$ is prime, there can't be more than one nine and two zeros. So that's all the minimal primes containing a digit $1$.
It might be work trying a simpler case - base $6$, say.
$2,3,5$ are still minimal primes in base $6$.
Any prime that does not contain those digits ends in $1$.
There is only one $2$-digit base $6$ prime not containing $2,3,5: 11_6=7$.
For three digits or mre, it must end in $1$ and contain only $0,4$ before. $401_6,441_6$ are both composite. So there are no $3$-digit minimal primes.
Next we try $4$-digits: $4401_6=1009$ and $4441_6$ are prime.
There is only two $5$-digit number we can check now: $40441_6=5353$ is not prime.
$40041_6=5209$ is prime.
$111_8=73_10, $401_8=257_{10}, $ is prime. $
