How amount of Emirp Primes depends on the base of numeral system

Question

There was a problem on searching for primes which, if their decimal notation is reverted, yield another primes, like 37 => 73 or 13 => 31. Some of colleagues pointed to the term Emirp invented for such values.

But since being an Emirp depends on numeral system I'm curious, what kind of corelation could exist here. I thought "intuitively" that if the base of numeral system is prime, then there will be more emirps.

I've tried to count emirps among first 100000 of primes with different bases:

base    emirps
   2     11117
   3     10105
   4      8426
   5     10929
   6      3921
   7      9809
   8      4694
   9      6998
  10      5985
  11     11976

So it looks like my suggestion is correct. However I have no idea what more precise law may lie below this fact.

So the question is - could anyone hint on what qualities / properties of numbers may affect the frequency of encountering emirps. May the phi(N) be somehow related?

P.S. Numbers in my example could be somewhat rough due to method of counting, however I believe they reflect the tendency.

Answer 1

If $n$ has many divisors, the last digit of the base-$n$ representation of a prime will be significantly restricted. It follows that the first digit of an emirp will also be significantly restricted. (For example, in base $10$, no prime that starts with $2$, $4$, $5$, $6$, or $8$ can be an emirp.)

More precisely, a base-$n$ emirp has $\phi(n)$ possible first digits, where $\phi(n)$ is Euler's totient function, and so a randomly selected prime will have roughly a $\phi(n)/(n-1)$ chance of not being excluded from emirpality (ytilamirp?) simply by virtue of its first digit. If this is all that's going on, the third row of this extended version of your table should be considerably smoother than the second:

base    emirps     emirps * (n-1) / (phi(n))
   2     11117                         11117
   3     10105                         10105
   4      8426                         12639
   5     10929                         10929
   6      3921                          9802
   7      9809                          9809
   8      4694                         10953
   9      6998                          9330
  10      5985                         10773
  11     11976                         11976

...which indeed it seems to be, if I haven't made any mistakes.

Answer 2

Here is one general property of all prime numbers except for $2$ and $3$: $p\equiv\pm1\pmod6$.

So perhaps you will find the lowest amount of emirps on a base which is a multiple of $6$.

Answer 3

Probably the largest effects will be $\varphi(b)$ (only numbers starting or ending with digits coprime to $b$ can be emirps > $b$) and the prime divisors of $b-1$ (since the digit sum can't be divisible by any of these).