47
$\begingroup$

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.

$\endgroup$
  • 22
    $\begingroup$ $3$ is the $2^{nd}$ prime, along with its mirror, $3$, which is the $2^{nd}$ prime (where $2$ is the mirror of $2$). Moreover, $3$ is also a palindrome in binary. So, clearly, it's the best prime (oh, and $5$ is pretty good too. So is $7$) $\endgroup$ – Milo Brandt Nov 16 '14 at 22:17
  • 4
    $\begingroup$ I think that might be all of them. I tested this over the first $100,000$ primes and $2$, $3$, $5$, $7$, $11$, $37$, and $73$ are all I found, but I can't think of any good argument as to why that would be (nor can I think of any way to refine a brute force search) $\endgroup$ – Milo Brandt Nov 17 '14 at 0:14
  • 3
    $\begingroup$ I actually also just finished testing the first $100,000$ primes. Those including the trivial ones are all that popped up. A proof would be cool! $\endgroup$ – Andrey Kaipov Nov 17 '14 at 0:17
  • 9
    $\begingroup$ 143787341 is the next, but it is a trivial solution (it's a palindrome so not a proper emirp). 11853735811 is the next trivial solution (but there may be a non-trivial one before it). These (with 2,3,5,7,11) make an interesting set: palindromic primes with palindromic prime positions. $\endgroup$ – DanaJ Nov 26 '14 at 21:25
  • 3
    $\begingroup$ There is an article about Sheldon Number in the November 2015 issue of Math Horizons. $\endgroup$ – Patrick Li Nov 11 '15 at 19:01
6
$\begingroup$

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
$\endgroup$
  • $\begingroup$ There are 7 sheldon cooper primes. Neat. A prime number of sheldon primes. $\endgroup$ – Nick Jul 17 '18 at 19:42
1
$\begingroup$

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).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.