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
T
he Big Bang Theory is a CBS sitcom cre-
ated in 2007. It follows the escapades of ļ¬ve
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,
thereā€™s only one correct answer.
Raj: Five million, three hundred eighteen thou-
sand, eight?
Sheldon: Wrong. The best number is 73. Youā€™re
probably wondering why.
Leonard: No.
Howard: Uh-uh.
Raj: Weā€™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 satisļ¬es these properties
(which we deļ¬ne explicitly below) a Sheldon prime.
Implicit in Sheldonā€™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 deļ¬ning 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 inļ¬nitely many pal-
indromic numbers, but it is an open question as to
whether there are inļ¬nitely many palindromic primes.
As of November 2014, the largest known palindromic
prime was 474,501 digits long [6].
With these deļ¬nitions in hand, we are ready to state
the ļ¬rst property of a Sheldon prime.
Deļ¬nition. The nth prime number r satisļ¬es 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 ļ¬rst 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 (deļ¬nitely not guaranteed). We will look
at each part of this property.
Palindromic primes satisfy the ļ¬rst 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 ļ¬rst
half are common. Of the ļ¬rst 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 ļ¬rst. 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
ļ¬rst 10 million primes again, we ļ¬nd only one prime
greater than 73 that satisļ¬es the mirror property:
In this example, as was
the case with the ā€œtriviallyā€ 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.
Deļ¬nition. The nth prime number r satisļ¬es 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 signiļ¬cantly
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 0ā€“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.
So
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, satisļ¬es the
product property, and so does the 7th prime, 17.
The only other instance in the ļ¬rst 10 million
primes is the 181,440th prime, 2,475,989 (because
Sheldon Primes
We may now deļ¬ne a Sheldon prime.
Deļ¬nition. A natural number that satisļ¬es 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 ļ¬‚oor 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 ļ¬‚oor func-
tion, we ļ¬nd the number of digits of p(m(n)) is at most
Using the lower-bound formula for p(n), we have
Photos courtesy CBS
ī€¶ī‹īˆīī‡ī’ī‘ī‚·ī–ī€ƒī†ī’īīīˆī‘ī—ī–ī€ƒī„ī…ī’ī˜ī—ī€ƒī€¦ī‹ī˜ī†īŽī€ƒī€±ī’ī•ī•īŒī–ī€ƒ
ī„ī•īˆī€ƒīŒī†īŒī‘īŠī€ƒī’ī‘ī€ƒī—ī‹īˆī€ƒī†ī„īŽīˆī€‘
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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 satisļ¬ed 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 ļ¬nd a counterexample? ī‘
Further Reading
[1] ā€œThe Alien Parasite Hypothesis,ā€ 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) 411ā€“415.
[4] A. Granville, Smooth numbers: computational
number theory and beyond, Algorithmic Number
Theory 44 (2008) 267ā€“316.
[5] T. P. Hill, The signiļ¬cant-digit phenomenon,
Amer. Math. Monthly 102 no. 4 (1995) 322ā€“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 masterā€™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 ļ¬rst billion numbers have this form. In other
words, this criterion alone rules out 99.9997 percent of
the ļ¬rst 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 73ā€”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 ļ¬rst 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 ļ¬nding
another Sheldon prime is small, the chance of ļ¬nding
another Sheldon-Norris prime is even smaller.
So where does all this analysis lead us? Sheldonā€™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 satisļ¬es 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): $$10^{474500} + 999 Ɨ 10^{237249} + 1$$ 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 $aƗ9^{kāˆ’1}$. Using the result of Rosser and Schoenfeld, for $p$ ā‰„ 17, we have $$ n = \pi (p) = \frac{p}{\log p} $$ But $p ā‰„ aƗ10^{kāˆ’1}$ 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 $kā‰„46$. 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.