Solving Fizz Buzz with Cosines

By Susam Pal on 20 Nov 2025

Fizz Buzz is a counting game that has become oddly popular in the
world of computer programming as a simple test of basic programming
skills. The rules of the game are straightforward. Players say the
numbers aloud in order beginning with one. Whenever a number is
divisible by 3, they say ‘Fizz’ instead. If it is divisible by 5,
they say ‘Buzz’. If it is divisible by both 3 and 5, the player
says both ‘Fizz’ and ‘Buzz’. Here is a typical Python program that
prints this sequence:

for n in range(1, 101):
    if n % 15 == 0:
        print('FizzBuzz')
    elif n % 3 == 0:
        print('Fizz')
    elif n % 5 == 0:
        print('Buzz')
    else:
        print(n)

Here is the output:
fizz-buzz.txt. Can we make
the program more complicated? Perhaps we can use trigonometric
functions to encode all four cases in a single closed-form
expression. That is what we are going to explore in this article.
By the end, we will obtain a finite Fourier series that can take any
integer \( n \) and select the text to be printed.

Contents

Definitions

Before going any further, we establish a precise mathematical
definition for the Fizz Buzz sequence. We begin by introducing a
few functions that will help us define the Fizz Buzz sequence later.

Symbol Functions

We define a set of four functions \( \🔥 \) for
integers \( n \) by:

\beginTell us your thoughts in comments!
s_0(n) &= n, \\
s_1(n) &= \mathtt💬, \\
s_2(n) &= \mathttWhat do you think?, \\
s_3(n) &= \mathtt⚡.
\endTell us your thoughts in comments!

We call these the symbol functions because they produce every term
that appears in the Fizz Buzz sequence. The symbol function \( s_0
\) returns \( n \) itself. The functions \( s_1, \) \( s_2 \) and
\( s_3 \) are constant functions that always return the literal
words \( \mathtt🔥, \) \( \mathttShare your opinion below! \) and \(
\mathtt{FizzBuzz} \) respectively, no matter what the value of \( n
\) is.

Fizz Buzz Sequence

Now we can define the Fizz Buzz sequence as the sequence

\[
(s_{f(n)}(n))_{n = 1}^{\infty}
\]

where

\[
f(n) = \begin{cases}
1 & \text{if } 3 \mid n \text{ and } 5 \nmid n, \\
2 & \text{if } 3 \nmid n \text{ and } 5 \mid n, \\
3 & \text{if } 3 \mid n \text{ and } 5 \mid n, \\
0 & \text{otherwise}.
\end{cases}
\]

The notation \( m \mid n \) means that the integer \( m \) divides
the integer \( n, \) i.e. \( n \) is a multiple of \( m. \)
Equivalently, there exists an integer \( c \) such that \( n = cm
. \) Similarly, \( m \nmid n \) means that \( m \) does not divide
\( n, \) i.e. \( n \) is not a multiple of \( m. \) With the above
definitions in place, we can expand the first few terms of the
sequence explicitly as follows:

\begin{align*}
(s_{f(n)}(n))_{n = 1}^{\infty}
&= (s_{f(1)}(1), \; s_{f(2)}(2), \; s_{f(3)}(3), \; s_{f(4)}(4), \;
s_{f(5)}(5), \; s_{f(6)}(6), \; s_{f(7)}(7), \; \dots) \\
&= (s_0(1), \; s_0(2), \; s_1(3), \; s_0(4),
s_2(5), \; s_1(6), \; s_0(7), \; \dots) \\
&= (1, \; 2, \; \mathtt{Fizz}, \; 4, \;
\mathtt{Buzz}, \; \mathtt{Fizz}, \; 7, \; \dots).
\end{align*}

Note how the function \( f(n) \) produces an index \( i \) which we
then use to select the symbol function \( s_i(n) \) to produce the
\( n \)th term of the sequence.

Indicator Functions

Here is the function \( f(n) \) from the previous section with its
cases and conditions rearranged to make it easier to spot
interesting patterns:

\[
f(n) = \begin{cases}
0 & \text{if } 5 \nmid n \text{ and } 3 \nmid n, \\
1 & \text{if } 5 \nmid n \text{ and } 3 \mid n, \\
2 & \text{if } 5 \mid n \text{ and } 3 \nmid n, \\
3 & \text{if } 5 \mid n \text{ and } 3 \mid n.
\end{cases}
\]

This function helps us to select another function \( s_{f(n)}(n) \)
which in turn determines the \( n \)th term of the Fizz Buzz
sequence. Our goal now is to replace this piecewise formula with a
single closed-form expression. To do so, we first define indicator
functions \( I_m(n) \) as follows:

\[
I_m(n) = \begin{cases}
1 & \text{if } m \mid n, \\
0 & \text{if } m \nmid n.
\end{cases}
\]

The formula for \( f(n) \) can now be written as:

\[
f(n) = \begin{cases}
0 & \text{if } I_5(n) = 0 \text{ and } I_3(n) = 0, \\
1 & \text{if } I_5(n) = 0 \text{ and } I_3(n) = 1, \\
2 & \text{if } I_5(n) = 1 \text{ and } I_3(n) = 0, \\
3 & \text{if } I_5(n) = 1 \text{ and } I_3(n) = 1.
\end{cases}
\]

Do you see a pattern? Here is the same function written as a table:

Do you see it now? If we treat the values in the first two columns
as binary digits and the values in the third column as decimal
numbers, then in each row the first two columns give the binary
representation of the number in the third column. For example, \(
3_{10} = 11_2 \) and indeed in the last row of the table, we see the
bits \( 1 \) and \( 1 \) in the first two columns and the number \(
3 \) in the last column. In other words, writing the binary digits
\( I_5(n) \) and \( I_3(n) \) side by side gives us the binary
representation of \( f(n). \) Therefore

\[
f(n) = 2 \, I_5(n) + I_3(n).
\]

We can now write a small program to demonstrate this formula:

for n in range(1, 101):
    s = [n, 'Fizz', 'Buzz', 'FizzBuzz']
    i = (n % 3 == 0) + 2 * (n % 5 == 0)
    print(s[i])

We can make it even shorter at the cost of some clarity:

for n in range(1, 101):
    print([n, 'Fizz', 'Buzz', 'FizzBuzz'][(n % 3 == 0) + 2 * (n % 5 == 0)])

What we have obtained so far is pretty good. While there is no
universal definition of a closed-form expression, I think most
people would agree that the indicator functions as defined above are
simple enough to be permitted in a closed-form expression.

Complex Exponentials

In the previous section, we obtained the formula

\[
f(n) = I_3(n) + 2 \, I_5(n)
\]

which we then used as an index to look up the text to be printed.
We also argued that this is a pretty good closed-form expression
already.

However, in the interest of making things more complicated, we must
ask ourselves: What if we are not allowed to use the indicator
functions? What if we must adhere to the commonly accepted meaning
of a closed-form expression which allows only finite combinations of
basic operations such as addition, subtraction, multiplication,
division, integer exponents and roots with integer index as well as
functions such as exponentials, logarithms and trigonometric
functions. It turns out that the above formula can be rewritten
using only addition, multiplication, division and the cosine
function. Let us begin the translation. Consider the sum

\[
S_m(n) = \sum_{k = 0}^{m – 1} e^{2 \pi i k n / m},
\]

where \( i \) is the imaginary unit and \( n \) and \( m \) are
integers. This is a geometric series in the complex plane with
ratio \( r = e^{2 \pi i n / m}. \) If \( n \) is a multiple of \( m
, \) then \( n = cm \) for some integer \( c \) and we get

\[
r
= e^{2 \pi i n / m}
= e^{2 \pi i c}
= 1.
\]

Therefore, when \( n \) is a multiple of \( m, \) we get

\[
S_m(n)
= \sum_{k = 0}^{m – 1} e^{2 \pi i k n / m}
= \sum_{k = 0}^{m – 1} 1^k
= m.
\]

If \( n \) is not a multiple of \( m, \) then \( r \ne 1 \) and the
geometric series becomes

\[
S_m(n)
= \frac{r^m – 1}{r – 1}
= \frac{e^{2 \pi i n} – 1}{e^{2 \pi i n / m} – 1}
= 0.
\]

Therefore,

\[
S_m(n) = \begin{cases}
m & \text{if } m \mid n, \\
0 & \text{if } m \nmid n.
\end{cases}
\]

Dividing both sides by \( m, \) we get

\[
\frac{S_m(n)}{m} = \begin{cases}
1 & \text{if } m \mid n, \\
0 & \text{if } m \nmid n.
\end{cases}
\]

But the right-hand side is \( I_m(n). \) Therefore

\[
I_m(n)
= \frac{S_m(n)}{m}
= \frac{1}{m} \sum_{k = 0}^{m – 1} e^{2 \pi i k n / m}.
\]

Cosines

We begin with Euler’s formula

\[
e^{i x} = \cos x + i \sin x
\]

where \( x \) is a real number. From this formula, we get

\[
e^{i x} + e^{-i x} = 2 \cos x.
\]

Therefore

\begin{align*}
I_3(n)
&= \frac{1}{3} \sum_{k = 0}^2 e^{2 \pi i k n / 3} \\
&= \frac{1}{3} \left( 1 + e^{2 \pi i n / 3} +
e^{4 \pi i n / 3} \right) \\
&= \frac{1}{3} \left( 1 + e^{2 \pi i n / 3} +
e^{-2 \pi i n / 3} \right) \\
&= \frac{1}{3} + \frac{2}{3} \cos \left( \frac{2 \pi n}{3} \right).
\end{align*}

The third equality above follows from the fact that \( e^{4 \pi i n
/ 3} = e^{6 \pi i n / 3} e^{-2 \pi i n / 3} = e^{2 \pi i n} e^{-2
\pi i n/3} = e^{-2 \pi i n / 3}. \)

The function above is defined for integer values of \( n \) but we
can extend its formula to real \( x \) and plot it to observe its
shape between integers. As expected, the function takes the value
\( 1 \) whenever \( x \) is an integer multiple of \( 3 \) and \( 0
\) whenever \( x \) is an integer not divisible by \( 3. \)

Similarly,

\begin{align*}
I_5(n)
&= \frac{1}{5} \sum_{k = 0}^4 e^{2 \pi i k n / 5} \\
&= \frac{1}{5} \left( 1 + e^{2 \pi i n / 5} +
e^{4 \pi i n / 5} +
e^{6 \pi i n / 5} +
e^{8 \pi i n / 5} \right) \\
&= \frac{1}{5} \left( 1 + e^{2 \pi i n / 5} +
e^{4 \pi i n / 5} +
e^{-4 \pi i n / 5} +
e^{-2 \pi i n / 5} \right) \\
&= \frac{1}{5} + \frac{2}{5} \cos \left( \frac{2 \pi n}{5} \right) +
\frac{2}{5} \cos \left( \frac{4 \pi n}{5} \right).
\end{align*}

Extending this expression to real values of \( x \) allows us to
plot its shape as well. Once again, the function takes the value \(
1 \) at integer multiples of \( 5 \) and \( 0 \) at integers not
divisible by \( 5. \)

Recall that we expressed \( f(n) \) as

\[
f(n) = I_3(n) + 2 \, I_5(n).
\]

Substituting these trigonometric expressions yields

\[
f(n)
= \frac{1}{3} + \frac{2}{3} \cos \left( \frac{2 \pi n}{3} \right) +
2 \cdot \left( \frac{1}{5} + \frac{2}{5} \cos \left( \frac{2 \pi n}{5} \right) +
\frac{2}{5} \cos \left( \frac{4 \pi n}{5} \right) \right).
\]

A straightforward simplification gives

\[
f(n)
= \frac{11}{15} +
\frac{2}{3} \cos \left( \frac{2 \pi n}{3} \right) +
\frac{4}{5} \cos \left( \frac{2 \pi n}{5} \right) +
\frac{4}{5} \cos \left( \frac{4 \pi n}{5} \right).
\]

We can extend this expression to real \( x \) and plot it as well.
The resulting curve takes the values \( 0, 1, 2 \) and \( 3 \) at
integer points, as desired.

Now we can write our Python program as follows:

from math import cos, pi
for n in range(1, 101):
    s = [n, 'Fizz', 'Buzz', 'FizzBuzz']
    i = round(11 / 15 + (2 / 3) * cos(2 * pi * n / 3)
                      + (4 / 5) * cos(2 * pi * n / 5)
                      + (4 / 5) * cos(4 * pi * n / 5))
    print(s[i])

Conclusion

To summarise, we have defined the Fizz Buzz sequence as

\[
(s_{f(n)}(n))_{n = 1}^{\infty}
\]

where

\[
f(n)
= \frac{11}{15} +
\frac{2}{3} \cos \left( \frac{2 \pi n}{3} \right) +
\frac{4}{5} \cos \left( \frac{2 \pi n}{5} \right) +
\frac{4}{5} \cos \left( \frac{4 \pi n}{5} \right)
\in \{ 0, 1, 2, 3 \}
\]

and \( s_0(n) = n, \) \( s_1(n) = \mathtt{Fizz}, \) \( s_2(n) =
\mathtt{Buzz} \) and \( s_3(n) = \mathtt{FizzBuzz}. \) A Python
program to print the Fizz Buzz sequence based on this definition was
presented earlier. That program can be written more succinctly as
follows:

from math import cos, pi
for n in range(1, 101):
    print([n, 'Fizz', 'Buzz', 'FizzBuzz'][round(11 / 15 + (2 / 3) * cos(2 * pi * n / 3) + (4 / 5) * (cos(2 * pi * n / 5) + cos(4 * pi * n / 5)))])

We can also wrap this up nicely in a shell one-liner, in case you
want to share it with your friends and family and surprise them:

printf 'from math import cos, pi\nfor n in range(1, 101): print([n, "Fizz", "Buzz", "FizzBuzz"][round(11 / 15 + (2 / 3) * cos(2 * pi * n / 3) + (4 / 5) * (cos(2 * pi * n / 5) + cos(4 * pi * n / 5)))])' | python3

The keen-eyed might notice that the expression we have obtained for
\( f(n) \) is a finite Fourier series. This is not surprising,
since the output of a Fizz Buzz program depends only on \( n \bmod
15. \) Any function on a finite cyclic group can be written exactly
as a finite Fourier expansion.

We have taken a simple counting game and turned it into a
trigonometric construction: a finite Fourier series with a constant
term \( 11/15 \) and three cosine terms with coefficients \( 2/3, \)
\( 4/5 \) and \( 4/5. \) None of this makes Fizz Buzz any easier,
of course, but it does show that every \( \mathtt{Fizz} \) and \(
\mathtt{Buzz} \) now owes its existence to a particular set of
Fourier coefficients. We began with the modest goal of making this
simple problem more complicated. I think it is safe to say that we
did not fall short.