Sheldon Cooper Primes On the $73^{\text{rd}}$ episode of the Big Bang Theory, Dr. Sheldon Cooper, an astrophysicist portrayed by Jim Parsons $(1973 - \stackrel{\text{hopefully}}{2073})$ revealed his favorite number to be the sexy prime $73$

Sheldon : 
  "The best number is $73$. 
  Why? 
  $73$ is the $21^{\text{st}}$ prime number. 
  Its mirror, $37$, is the $12^{\text{th}}$ 
  and its mirror, $21$, is the product of multiplying $7$ and $3$
  ... and in binary $73$ is a palindrome, $1001001$, which backwards is $1001001$."
Leonard : "$73$ is the Chuck Norris of numbers!"
Sheldon : "Chuck Norris wishes... all Chuck Norris backwards gets you is Sirron Kcuhc!"' 

My question is basically this: Are there any more Sheldon Cooper primes?
But how do I define a Sheldon Cooper Prime? Sheldon emphasizes three aspects of 73


*

*It is an emirp with added mirror properties (ie, the prime's mirror is also a prime with position number mirrored)

*A concatenation of the factors of the position number of the prime yields the prime.

*Binary representation of the prime is a palindrome
I think having all three properties exist simultaneously in a number is difficult.
So, a prime satisfying the first property is good enough.
So, I define a Sheldon Cooper Prime as an emirp with added mirror properties.
Good Luck finding them :D
Edit: Please find primes with position numbers $>9$.
$2,3,..$ are far too trivial.
 A: Up to 10,000,000 $\;\;$ (currently running until 100,000,000)


*

*Emirp with added mirror properties (as defined above): $$2, \;\;\; 3, \;\;\; 5, \;\;\; 7, \;\;\; 11, \;\;\; 37, \;\;\; \text{and}\;\;\; 73.$$

*$+$ Mirror different from original prime:$$37, \;\;\; \text{and}\;\;\;  73.$$

*$+$ Binary representation of the prime is a palindrome: $$73.$$

*$+$ A concatenation of the factors of the position number of the prime yields the prime: $$73.$$

Matlab Code
clc
clear

for i = 1:10000000

    % Prime:
    if (isprime(i))
        cont = 1;
    else
        cont = 0;
    end

    % 1. It is an emirp with added mirror properties: 
    if (cont == 1)

        mirror_i = str2double(fliplr(num2str(i)));

        if (isprime(mirror_i))
            cont = 1;
        else
            cont = 0;            
        end

    end

    if (cont == 1)

        p_i  = length(primes(i));

        p_mi = length(primes(mirror_i));

        mirror_p_i = str2double(fliplr(num2str(p_i)));

        if (mirror_p_i == p_mi)
            cont = 1;
            disp(' ')
            disp(' ')
            disp(['------------->>  ',num2str(i)])
            disp(['Satisfies Condition 1:  ',num2str([mirror_i,p_i,p_mi])])
        else
            cont = 0;            
        end

    end

     % 2. Mirror different from original prime:
    if (cont == 1)

        if (i == mirror_i)
            cont = 0;
        else
            cont = 1;
            disp('Satisfies Condition 2')
        end

    end

    % 3. Binary representation of the prime is a palindrome:
    if (cont == 1)

        bin = dec2bin(i);
        mirror_bin = fliplr(num2str(bin));

        if (bin == mirror_bin)
            cont = 1;
            disp(['Satisfies Condition 3:  ',num2str(str2double(bin))])
        else
            cont = 0;
        end

    end

    % 4. A concatenation of the factors of the position number of the prime
    % yields the prime:
    if (cont == 1)

        if (prod(sscanf(num2str(i),'%1d')) == p_i)
            disp('Satisfies Condition 4')
        end

    end

end

A: The proof of the "Sheldon Conjecture" was published a few months ago in the February edition of the American Mathematical Monthly.
https://math.dartmouth.edu/~carlp/sheldon02132019.pdf
A: I created a [script] you can play with here to test this out. Note that the answer depends on your numerical base -- among all bases I've tried, 10 seems to be the only base in which there's a Sheldon Cooper prime. 
Base 16 seems promising, however -- it has a large number of "special emirps", and actually provides primes with the appropriate product of digits, which very few bases provide. 
Can someone try base = 16, convbase = 2 (and perhaps other bases in multiple tabs) with a large uppercap (e.g. 10,000,000) using fastcount = false? It would take ~15 hours for an upper cap of 10 million -- or just 90 minutes for an uppercap of 1 million -- but I can't leave my laptop on for so long (the fan is malfunctioning).
