# 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.

• $3$ 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 $7$) Commented Nov 16, 2014 at 22:17
• 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) Commented Nov 17, 2014 at 0:14
• 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! Commented Nov 17, 2014 at 0:17
• 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. Commented Nov 26, 2014 at 21:25
• There is an article about Sheldon Number in the November 2015 issue of Math Horizons. Commented Nov 11, 2015 at 19:01

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

• There are 7 sheldon cooper primes. Neat. A prime number of sheldon primes.
– Nick
Commented Jul 17, 2018 at 19:42

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

• 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." Commented Jul 21, 2019 at 15:17
• It is in the October 2019 issue. Commented Oct 6, 2019 at 14:15
• Precise reference to this paper on https://doi.org/10.1080/00029890.2019.1626672. Commented Jul 23, 2022 at 9:19

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).