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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.

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    $\begingroup$ $3$ 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, 2014 at 22:17
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    $\begingroup$ 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, 2014 at 0:14
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    $\begingroup$ 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, 2014 at 0:17
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    $\begingroup$ 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, 2014 at 21:25
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    $\begingroup$ There is an article about Sheldon Number in the November 2015 issue of Math Horizons. $\endgroup$
    – JACKY88
    Nov 11, 2015 at 19:01

3 Answers 3

Reset to default
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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
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  • $\begingroup$ There are 7 sheldon cooper primes. Neat. A prime number of sheldon primes. $\endgroup$
    – Nick
    Jul 17, 2018 at 19:42
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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

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  • $\begingroup$ It was not in the February issue, and it looks like it hasn't actually been published yet. Pomerance's web page says "to appear." $\endgroup$
    – RobPratt
    Jul 21, 2019 at 15:17
  • $\begingroup$ It is in the October 2019 issue. $\endgroup$
    – RobPratt
    Oct 6, 2019 at 14:15
  • $\begingroup$ Precise reference to this paper on https://doi.org/10.1080/00029890.2019.1626672. $\endgroup$
    – Jeppe Stig Nielsen
    Jul 23, 2022 at 9:19
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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).

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