https://susam.net/fizz-buzz-with-cosines.html

Solving Fizz Buzz with Cosines

By Susam Pal on 20 Nov 2025
\[ \gdef\arraystretch{1.2} \]

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? The words 'Fizz', 'Buzz' and 'FizzBuzz' repeat in a
periodic manner throughout the sequence. What else is periodic?
Trigonometric functions! Perhaps we can use trigonometric functions
to encode all four rules of the sequence in a single closed-form
expression. That is what we are going to explore in this article. By
the end, we will obtain a discrete Fourier series that can take any
integer \( n \) and select the corresponding text to be printed.

Contents 

  * Definitions
      + Symbol Functions
      + Fizz Buzz Sequence
  * Indicator Functions
  * Complex Exponentials
  * Cosines
  * Discrete Fourier Transform
  * Conclusion

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 \( \{ s_0, s_1, s_2, s_3 \} \) for
integers \( n \) by: \begin{align*} s_0(n) &= n, \\ s_1(n) &= \mathtt
{Fizz}, \\ s_2(n) &= \mathtt{Buzz}, \\ s_3(n) &= \mathtt{FizzBuzz}. \
end{align*} 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 \( \mathtt{Fizz}, \) \( \mathtt{Buzz} \) 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. We therefore call \( f(n)
\) the index function.

Indicator Functions 

Here is the index 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:

\( I_5(n) \) \( I_3(n) \) \( f(n) \)
\( 0 \)      \( 0 \)      \( 0 \)
\( 0 \)      \( 1 \)      \( 1 \)
\( 1 \)      \( 0 \)      \( 2 \)
\( 1 \)      \( 1 \)      \( 3 \)

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 \( 3. \)

Graph Graph of \( \frac{1}{3} + \frac{2}{3} \cos \left( \frac{2 \pi
x}{3} \right) \)

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 \( 5. \)

Graph Graph of \( \frac{1}{5} + \frac{2}{5} \cos \left( \frac{2 \pi
x}{5} \right) + \frac{2}{5} \cos \left( \frac{4 \pi x}{5} \right) \)

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.

Graph Graph of \( \frac{11}{15} + \frac{2}{3} \cos \left( \frac{2 \pi
x}{3} \right) + \frac{4}{5} \cos \left( \frac{2 \pi x}{5} \right) + \
frac{4}{5} \cos \left( \frac{4 \pi x}{5} \right) \)

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])

Discrete Fourier Transform 

The keen-eyed might notice that the expression we obtained for \( f
(n) \) is a finite Fourier series. This is not surprising, since the
output of a Fizz Buzz programme depends only on \( n \bmod 15 . \)
Any function on a finite cyclic group can be written exactly as a
finite Fourier expansion. In this section, we recover the same
function \( f(n) \) using the discrete Fourier transform. It is worth
mentioning here that the calculations presented here are quite
tedious to do by hand. Nevertheless, this section offers a glimpse of
how such coefficients are calculated. By the end, we will arrive at
exactly the same \( f(n) \) as before. There is nothing new to
discover here. We simply obtain the result by a different, more
direct and noticeably tedious method. If this doesn't sound
interesting to you, you may safely skip this section.

We know that \( f(n) \) is a periodic function with period \( 15. \)
To apply the discrete Fourier transform, we look at one complete
period using the indices \( n = 1, 2, \dots, 15. \) Over this period,
the values are: \begin{array}{c|ccccccccccccccc} n & 1 & 2 & 3 & 4 &
5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 \\ \hline f(n) & 0 &
0 & 1 & 0 & 2 & 1 & 0 & 0 & 1 & 2 & 0 & 1 & 0 & 0 & 3 \end{array} The
discrete Fourier transform gives us the constants defined by: \[ c_k
= \frac{1}{15} \sum_{n = 1}^{15} f(n) e^{-2 \pi i k n / 15}, \] for \
( k = 1, 2, \dots, 15. \) These constants reconstruct \( f(n) \) via
the inverse transform: \[ f(n) = \sum_{k = 1}^{15} c_k e^{2 \pi i k n
/ 15} \] where \( n \in \mathbb{Z}. \) Let us compute the first
constant: \[ c_1 = \frac{1}{15}\left( \begin{aligned} e^{-2 \pi i \
cdot 3/15} &+ 2 e^{-2 \pi i \cdot 5/15} + e^{-2 \pi i \cdot 6/15} + e
^{-2 \pi i \cdot 9/15} \\ &+ 2 e^{-2 \pi i \cdot 10/15} + e^{-2 \pi i
\cdot 12/15} + 3 e^{-2 \pi i \cdot 15/15} \end{aligned} \right) = 0.
\] Similarly, \[ c_2 = \frac{1}{15}\left( \begin{aligned} e^{-2 \pi i
\cdot 6/15} &+ 2 e^{-2 \pi i \cdot 10/15} + e^{-2 \pi i \cdot 12/15}
+ e^{-2 \pi i \cdot 18/15} \\ &+ 2 e^{-2 \pi i \cdot 20/15} + e^{-2 \
pi i \cdot 24/15} + 3 e^{-2 \pi i \cdot 30/15} \end{aligned} \right)
= 0. \] and \[ c_3 = \frac{1}{15}\left( \begin{aligned} e^{-2 \pi i \
cdot 9/15} &+ 2 e^{-2 \pi i \cdot 15/15} + e^{-2 \pi i \cdot 18/15} +
e^{-2 \pi i \cdot 27/15} \\ &+ 2 e^{-2 \pi i \cdot 30/15} + e^{-2 \pi
i \cdot 36/15} + 3 e^{-2 \pi i \cdot 45/15} \end{aligned} \right) = \
frac{2}{5}. \] Continuing in this manner, we find all the constants:
\begin{align*} c_1 &= c_2 = c_4 = c_7 = c_8 = c_{11} = c_{13} = c_
{14} = 0, \\ c_3 &= c_6 = c_9 = c_{12} = \frac{2}{5}, \\ c_5 &= c_
{10} = \frac{1}{3}, \\ c_{15} &= \frac{11}{15}. \end{align*} Using
the inverse transform, we get \begin{align*} f(n) &= \sum_{k = 1}^
{15} c_k \, e^{2 \pi i k n / 15} \\ &= \frac{11}{15} + \frac{2}{5} \
left( e^{2 \pi i \cdot 3n/15} + e^{2 \pi i \cdot 6n/15} + e^{2 \pi i
\cdot 9n/15} + e^{2 \pi i \cdot 12n/15} \right) \\ &\phantom{=} \
phantom{\frac{11}{15}} + \frac{1}{3} \left( e^{2 \pi i \cdot 5n/15} +
e^{2 \pi i \cdot 10n/15} \right) \\ &= \frac{11}{15} + \frac{2}{5} \
left( e^{6 \pi i n/15} + e^{12 \pi i n/15} + e^{18 \pi i n/15} + e^
{24 \pi i n/15} \right) \\ &\phantom{=} \phantom{\frac{11}{15}} + \
frac{1}{3} \left( e^{10 \pi i n/15} + e^{20 \pi i n/15} \right) \\ &=
\frac{11}{15} + \frac{2}{5} \left( e^{6 \pi i n/15} + e^{12 \pi i n/
15} + e^{-12 \pi i n/15} + e^{-6 \pi i n/15} \right) \\ &\phantom{=}
\phantom{\frac{11}{15}} + \frac{1}{3} \left( e^{10 \pi i n/15} + e^
{-10 \pi i n/15} \right) \\ &= \frac{11}{15} + \frac{2}{5} \left( 2 \
cos \left( \frac{6 \pi n}{15} \right) + 2 \cos \left( \frac{12 \pi n}
{15} \right) \right) \\ &\phantom{=} \phantom{\frac{11}{15}} + \frac
{1}{3} \left( 2 \cos \left( \frac{10 \pi n}{15} \right) \right) \\ &=
\frac{11}{15} + \frac{4}{5} \cos \left( \frac{2 \pi n}{5} \right) + \
frac{4}{5} \cos \left( \frac{4 \pi n}{5} \right) + \frac{2}{3} \cos \
left( \frac{2 \pi n}{3} \right). \end{align*} This gives us exactly
the same function \( f(n) \) we obtained earlier. The difference is
only in how we arrived there. Working out Fourier coefficients by
hand is slow and mechanical. In practice these sums are almost always
computed automatically by numerical software or computer algebra
systems. Still, this exercise shows that our humble Fizz Buzz index
function can be expressed precisely using the machinery of Fourier
analysis.

Conclusion 

To summarise, we have defined the Fizz Buzz sequence as \[ (s_{f(n)}
(n))_{n = 1}^{\infty} \] where \[ f(n) = \frac{11}{15} + \frac{2}{3}
\cos \left( \frac{2 \pi n}{3} \right) + \frac{4}{5} \cos \left( \frac
{2 \pi n}{5} \right) + \frac{4}{5} \cos \left( \frac{4 \pi n}{5} \
right). \] 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:

python3 -c 'from math import cos, pi; [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)))]) for n in range(1, 101)]'

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.

Comments | #python | #programming | #mathematics | #puzzle
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