Fermat's Library | The Sheldon Conjecture annotated/explained version.

Here's the video of the 73rd episode when Sheldon mentions his favo...

Except for 11, all palindromic primes have an odd number of digits,...

Here's a some Matlab code if you want to find prime mirrors: ```...

A result of Rosser and Schoenfeld is that $$ \pi (x) \geq \fra...

In February 2019, Carl Pomerance and Chris Spicer were able to prov...

12 November 2015 : : Math Horizons : : www.maa.org/mathhorizons

he Big Bang Theory is a CBS sitcom cre-

ated in 2007. It follows the escapades of ﬁve

friends: Sheldon Cooper, physicist; Leonard

Hofstadter, physicist; Howard Wolowitz,

aerospace engineer; Raj Koothrappali, as-

trophysicist; and Penny, a waitress and aspiring actress

who lives across the hall. In the 73rd episode of the

show, we hear the following conversation [1].

Sheldon: What is the best number? By the way,

there’s only one correct answer.

Raj: Five million, three hundred eighteen thou-

sand, eight?

Sheldon: Wrong. The best number is 73. You’re

probably wondering why.

Leonard: No.

Howard: Uh-uh.

Raj: We’re good.

Sheldon: Seventy-three is the 21st prime number.

Its mirror, 37, is the 12th, and its mirror, 21, is the

product of multiplying, hang on to your hats, seven

and three. Eh? Eh? Did I lie?

Leonard: We get it. Seventy-three is the Chuck

Norris of numbers.

Sheldon: Chuck Norris wishes. In binary, 73 is a

palindrome, 1,001,001 which backwards is 1,001,001,

exactly the same. All Chuck Norris backwards gets

you is Sirron Kcuhc.

Sheldon is correct in each of his statements regarding

73. We will call a number that satisﬁes these properties

(which we deﬁne explicitly below) a Sheldon prime.

Implicit in Sheldon’s assertion that 73 is the best num-

ber is that 73 is the only Sheldon prime. We call this

the Sheldon conjecture.

Although we cannot prove the conjecture, we show

that a counterexample, if one exists, must be very large.

The Mirror Property

We begin by deﬁning some notation that will help

us simplify this problem. First, let p(n) be the nth

prime number. For example,

and

The mirror of a natural number x, which we denote

m(x), is the number obtained by reversing the order

of the digits of x. So

and

Some numbers satisfy the very special property that

These numbers are their own mirrors and

are called palindromic numbers. Palindromic primes,

such as 101 or 373, are rare and have been studied in

their own right [5]. There are inﬁnitely many pal-

indromic numbers, but it is an open question as to

whether there are inﬁnitely many palindromic primes.

As of November 2014, the largest known palindromic

prime was 474,501 digits long [6].

With these deﬁnitions in hand, we are ready to state

the ﬁrst property of a Sheldon prime.

Deﬁnition. The nth prime number r satisﬁes the

mirror property if m(r) is the m(n)th prime number.

That is, m(p(n))=p(m(n)).

There are actually two statements in the above prop-

erty. The ﬁrst is that m(r) is a prime number (which is

not guaranteed). The second is that m(r) is the m(n)th

prime number (deﬁnitely not guaranteed). We will look

at each part of this property.

Palindromic primes satisfy the ﬁrst part of the

property, but they are not the only ones. For example,

389 and

are both prime. Although there

are several obvious conditions that would preclude a

prime from satisfying this property, such as starting

with a 2, 4, 5, 6, or 8, primes that satisfy this ﬁrst

half are common. Of the ﬁrst 10 million primes, 14.99

SHELDON

CONJECTURE

the

Jessie Byrnes, Chris Spicer, and Alyssa Turnquist

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percent of them have prime mirrors.

The second half of the mirror property states that

m(r) is the m(n)th prime number. This requirement

is much stronger than the ﬁrst. Primes such as 2, 3,

5, 7, and 11 satisfy this property because both n and

p(n) are palindromes. Also included in this list is

73, the 21st prime, and 37, the 12th. Looking at the

ﬁrst 10 million primes again, we ﬁnd only one prime

greater than 73 that satisﬁes the mirror property:

In this example, as was

the case with the “trivially” palindromic one-digit

primes and 11, both n and p(n) are palindromes.

The Product Property

For any natural number x, let denote the

product of the digits of x. For instance,

and We are now ready

to state the second property of a Sheldon prime.

Deﬁnition. The nth prime number r satisﬁes the

product property if

That is,

First, notice that if the prime r contains a zero as

one of its digits, then it will not satisfy the product

property because This fact signiﬁcantly

restricts the number of primes that can satisfy the

product property. Suppose q is a k-digit prime. Clearly

q cannot begin with a 0 and it cannot end with a 0 (or

else it is not prime). Assuming that for the remaining

digits the numbers 0–9 are equally likely, the

probability that q does not contain a 0 is

There is a relatively high probability that a small

prime will not contain a 0 digit. For larger primes, the

probability is nearly zero. For one that is 512 digits

long, the probability is approximately

The quantity is the product of the digits of x.

Because each digit of x is one of

2, 3, 5, and 7 are the only possible prime factors of

In number theory, integers that have a small

number of factors are known as smooth numbers.

Formally, integers whose prime factors are all less than

or equal to y are called y-smooth [4]. For instance, 72

is a 3-smooth number because

and 125 is

5-smooth but not 3-smooth.

is a 7-smooth number for all x. In particu-

lar, if n has a prime factor greater than 7, then p(n)

Photos courtesy CBS

From left: Leonard, Sheldon, Raj, and Howard. Sheldon implies that 73 is the only number with the properties he lists.

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14 November 2015 : : Math Horizons : : www.maa.org/mathhorizons

cannot satisfy the product property. This greatly

reduces the number of integers in the running to be

Sheldon primes.

We know that the 21st prime, 73, satisﬁes the

product property, and so does the 7th prime, 17.

The only other instance in the ﬁrst 10 million

primes is the 181,440th prime, 2,475,989 (because

Sheldon Primes

We may now deﬁne a Sheldon prime.

Deﬁnition. A natural number that satisﬁes the

mirror property and the product property is called a

Sheldon prime.

Suppose p(n) is a Sheldon prime. The product prop-

erty implies that for some

But the mirror property enables us say more about the

prime factorization of n.

For most natural numbers, n, n and m(n) have the

same number of digits. But each 2-5 pair in the prime

factorization of n yields a zero at the end of n and de-

creases the number of digits of m(n) by one. We claim

that if 100 divides n, then m(n) is too small for p(n)

to satisfy the mirror property. To justify this claim, we

need to know more about the relative sizes of n and

p(n).

The famous prime number theorem, proven by

Jacques Hadamard around 1896, states that there are

approximately

prime numbers less than x.

A corollary of the prime number theorem states that

p(n) is approximately For instance, this for-

mula states that the millionth prime is approximately

In actuality,

This formula gives good approximations, but not

good enough for our needs. We need a more precise

statement about the size of p(n). In [2] and [3], we are

provided the following upper and lower bounds of p(n).

For

For our purposes, we care less about the value of

p(n) than about the number of digits of p(n). To

compute the number of digits of a number, we need

a basic result involving logarithms. In particular,

is the number of digits of a natural num-

ber x. Recall that is the ﬂoor function; is the

greatest integer less than or equal to x. For instance,

is the number of

digits of 294.

Now suppose 100 divides a natural number n. Then

m(n) has at least two fewer digits than n. On the other

hand, p(n) and m(p(n)) have the same number of dig-

its because p(n) is a prime, which cannot end in a 0.

To show that the numb er of digits of p(m(n)) is too

small to correspond to the number of digits of m(p(n)), we

will use the upper and lower bounds given above. Using

the upper-bound formula for p(m(n)) and the fact that

we have

Using the properties of the logarithm and the ﬂoor func-

tion, we ﬁnd the number of digits of p(m(n)) is at most

Using the lower-bound formula for p(n), we have

Photos courtesy CBS

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

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www.maa.org/mathhorizons : : Math Horizons : : November 2015 15

Thus, the number of digits of p(n) is at least

So, the number of digits of p(n), and hence m(p(n))

because they have the same number of digits, is at

least one more than the number of digits of p(m(n)).

Thus

Consequently, for p(n) to be a Sheldon prime, n must

be a 7-smooth number not divisible by 100. Only 3,039

shows that the product property can be satisﬁed by

larger numbers.

Is there a Sheldon prime other than 73? These au-

thors believe it unlikely. Can you prove the conjecture?

Or can you ﬁnd a counterexample? 

Further Reading

[1] “The Alien Parasite Hypothesis,” The Big Bang

Theory, season 4, episode 10, http://bit.ly/1HYFGkT.

[2] E. Bach and J. Shallit, Algorithmic Number

Theory 1, MIT Press (1996) 233.

[3] P. Dusart, The kth prime is greater than

for , Mathematics of

Computation 68 no. 225 (1999) 411–415.

[4] A. Granville, Smooth numbers: computational

number theory and beyond, Algorithmic Number

Theory 44 (2008) 267–316.

[5] T. P. Hill, The signiﬁcant-digit phenomenon,

Amer. Math. Monthly 102 no. 4 (1995) 322–327.

[6] P. Ribenboim, The Little Book of Bigger Primes,

Springer (2004).

Jessie Byrnes received a math and business degree from

Morningside College and is now pursuing a master’s

degree in mathematics at the University of South

Dakota.

Email: jmb018@morningside.edu

Chris Spicer is an associate professor of mathemat-

ics at Morningside College. He is an avid Big Bang

Theory watcher and fan of all math in popular culture.

Email: spicer@morningside.edu

Alyssa Turnquist is a senior math major at Morning-

side College. She enjoys puzzles, traveling, and playing

the piano. She plans to attend graduate school after she

graduates.

Email: alt007@morningside.edu

http://dx.doi.org/10.4169/mathhorizons.23.2.12

Is there a Sheldon

prime other than 73?

of the ﬁrst billion numbers have this form. In other

words, this criterion alone rules out 99.9997 percent of

the ﬁrst billion primes as possible Sheldon primes.

Sheldon-Norris Primes

The astute reader will notice that we have not ad-

dressed the Chuck Norris property of 73—that it is

a palindrome in binary. After much discussion, the

authors decided that Sheldon mentions this property

solely as icing on the cake, rather than a pillar of his

mathematical argument.

However, a quick check of the ﬁrst 10 million primes

shows, not surprisingly, this is yet another restrictive

property, with only 0.0019 percent of those primes

being binary palindromic. If the chance of ﬁnding

another Sheldon prime is small, the chance of ﬁnding

another Sheldon-Norris prime is even smaller.

So where does all this analysis lead us? Sheldon’s

claim focused on two properties of the number 73.

One of them, that 73 is the 21st prime and its mirror

37 is the 12th, is very restrictive. However, we found

an example of a large prime that satisﬁes the mirror

property,

The second property, the product property, was that

Again, this property is incredibly

restrictive, requiring the n under consideration to be a

7-smooth integer not divisible by 100. In addition, the

prime cannot contain any zeroes as digits. However,

our nontrivial example of

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Discussion

In February 2019, Carl Pomerance and Chris Spicer were able to prove that 73 is in fact the only Sheldon Prime. For more on their proof you can click the link below. [![](https://i.imgur.com/KNzbMjB.png)](https://math.dartmouth.edu/~carlp/sheldon022119.pdf) Here's the video of the 73rd episode when Sheldon mentions his favorite prime [![](https://i.imgur.com/sQl6qj4.png)](https://www.youtube.com/watch?v=RyFr279K9TE) Except for 11, all palindromic primes have an odd number of digits, because the [[divisibility test]] for 11 tells us that every palindromic number with an even number of digits is a multiple of 11. It is not known if there are infinitely many palindromic primes in base 10. The largest known as of July 2019 is (474,501 digits): $$10^{474500} + 999 × 10^{237249} + 1$$ It was found in 2014 by Serge Batalov. On the other hand, it is known that, for any base, almost all palindromic numbers are composite. Banks et al. (2004) proved that almost all palindromes (in any base) are composite, with the precise statement being $$ P(x) \sim O \left (\frac{N(x) \ln \ln \ln x}{\ln \ln x} \right) $$ where $P(x)$ is the number of palindromic primes $\leq$ x and $N(x)$ is the number of palindromic numbers $\leq$ x. A result of Rosser and Schoenfeld is that $$ \pi (x) \geq \frac{x}{\log x}, x \geq 17 $$ This actually allows us to prove that no Sheldon Prime exceeds $10^{45}$! Say $p$ has $k$ digits with the leading digit $a$. Then $n$, which is equal to the product of the digits of $p$, is at most $a×9^{k−1}$. Using the result of Rosser and Schoenfeld, for $p$ ≥ 17, we have $$ n = \pi (p) = \frac{p}{\log p} $$ But $p ≥ a×10^{k−1}$ since $p$ is $k$ digits long. Thus if $p$ has the product property, then the following inequality must be satisfied: $$ a \times 9^{k-1} > \frac{a \times 10^{k-1}}{ \log(a \times 10^{k-1} ) } $$ which implies that $$ \log a + \log (10^{k-1}) > (\frac{10}{9})^{k-1} $$ Since the left side grows linearly in $k$ and the right side grows exponentially, it is clear that it fails for all large values of $k$. Further, if it fails for $a=9$, then it also fails for smaller values of $a$. A small computation and mathematical induction allow us to see that it fails for all $k≥46$. Here's a some Matlab code if you want to find prime mirrors: ``` 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 ``` Alternatively, you can run [this script online](https://www.khanacademy.org/computer-programming/run-testssheldon-cooper-primes/5927601331011584) if you want.

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