Problems & Puzzles: Puzzles

 

 

Problems & Puzzles: Puzzles

Puzzle 1267 How many emirps are there?


On March 19, 2029, Simon Cavegn wrote:

How many emirps are there?


State whether you are excluding or including palindromes.


Q1. Find the number of emirps up to the highest possible n.

  Q2. Find a formula that matches the data. Compare it to n/(ln(n)*ln(n)).

Can a part of the formula be attributed to:

- excluding or including palindromes
- for p > 5 the first digit cannot be 5
- other reverse digit correlations


Q3. Can you create a possibly provable conjecture, or maybe even a proof, that there are infinitely many emirps?


 




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From 8 to 15, May, 2026, contributions came from Emmanuel Vantieghem, Carlos Rivera


Emmanuel wrote:
The number of emirps with  n  digits can be found in  https://oeis.org/A152014 :
{0, 8, 28, 204, 1406, 9538, 70474, 535578, 4192024, 33619380, 274890230, 2294771254}
(for  n = 1  to  12).
From this, we find the number of emirps < 10^n :
{0, 8, 36, 240, 1646, 11184, 81658, 617236, 4809260, 38428640, 313318870, 2608090124}.
The number of primes < 10^n  is:
{4, 25, 168, 1229, 9592, 78498, 664579, 5761455, 50847534, 455052511, 4118054813, 37607912018}.
The quotient of these numbers is (we took  n  > 1) :
{0.32, 0.16667, 0.16599, 0.14658, 0.12151, 0.10604, 0.092959, 0.082443, 0.073880, 0.066752, 0.061018}
Here is a plot 



 

It is clear that we have too little data to formulate a decent conjecture.
But if I was forced to make a guess, I would say that the curve could tend to a limit.  And I would choose a limit value which depends on 10, our number base.
A "good" one could be  1 / (10 ln 10).~ 0.043429448.
That would lead me to the conjecture :
     Em[x] ~ x / (10 (ln 10) ( ln x)).
This is widely far from  n / (log(n)^2.
But I would not bet on it ! 


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Carlos wrote:

Here are my counts asked in Q1 & my comparison to the formula proposed in Q2

Note: By definition an emirp exclude the palprimes:

"An emirp (pronounced /ˈiːmərp/ or /ˈɛmərp/, an anadrome of prime) is a prime number that results in a different prime when its decimal digits (digits in base 10) are reversed.[1] This definition excludes the related palindromic primes...." From https://en.wikipedia.org/wiki/Emirp

I'm not sure if I'm making a correct interpretation of the Simon's phrase in Q2. "Compare it to n/(ln(n)*ln(n))". But here I go with my turn


n Emirps Accum. Of Emirps Palprimes x=10^(n+1) F=x/(ln(x)*ln(x)) Acc. of emirps/F
0 0 0 4 10                         1.89                              -  
1 8 8 1 100                         4.72                          1.70
2 28 36 15 1000                       20.96                          1.72
3 204 240 0 10000                    117.88                          2.04
4 1406 1646 93 100000                    754.45                          2.18
5 9538 11184 0 1000000                 5,239.21                          2.13
6 70474 81658 668 10000000              38,492.18                          2.12
7 535578 617236 0 100000000            294,705.78                          2.09
8 4192024 4809260 5172 1000000000        2,328,539.47                          2.07
9 33619380 38428640 0 1E+10      18,861,169.70                          2.04
 n=x means the range [10^x ,  10^(x+1)]
Computed by my  "Cuenta_de_emirps.py"


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