5
$\begingroup$

Suppose we have a necklace with $n$ beads. Each bead is either red or blue. I'd like to ask how to count the number of necklaces $f(n,m,k)$ satisfying the following requirements:

1) There are exactly $m$ red beads; $(0 \leq m \leq n)$.

2) No two adjacent red beads;

3) The number of blue beads with two adjacent red beads is $k$ exactly.

Note that rotation of necklace is counted differently. For example, "blue blue blue red" is different from "red blue blue blue".

Some test cases: $f(4,2,2)=2: \color{red}{1}0\color{red}{1}0, 0\color{red}{1}0\color{red}{1}$

$f(5,2,1)=5: 0\color{red}{1}011, 10\color{red}{1}01, 110\color{red}{1}0, 0110\color{red}{1}, \color{red}{1}0110$, where $0$ represents a red bead and $1$ represents a blue one. I've colored the blue beads satisfying the third requirement.

Note that $a_n=\sum_{m,k}f(n,m,k)$ are Lucas numbers. I've also noticed that for fixed $m$ and $k$, the series $f(n,m,k)$ seems to have a generating function which looks like $\frac{a-bx}{(1-x)^c}$ where $a$, $b$ and $c$ are constants related to $m$ and $k$.

$\endgroup$
1
$\begingroup$

The magic formula $$f(n,m,k)=2{m-1 \choose k-1}{n-2m-1\choose m-k-1}+3{m-1 \choose k}{n-2m \choose m-k}$$

Derivation

Any necklace satisfying the three requirements has $m$ strings of one or more consecutive blue beads bordered by red beads. It must therefore have $m-k$ strings of two or more consecutive blue beads.

Consider those necklaces which do not contain three consecutive blue beads (we will call them minimal necklaces). Clearly $n=3m-k$ for a minimal necklace.

We will consider two cases:

  1. The first bead is red and the last two beads are red then blue OR the last bead is red and the first two beads are blue then red.
  2. Either the first and last bead OR the first two beads OR the last two beads are both blue.

There are $2{m-1 \choose k-1}$ minimal necklaces of type 1 and $3{m-1 \choose k}$ minimal necklaces of type 1.

Finally, we can use the second theorem of stars and bars to calculate the number of ways to add an extra $n-3m+k$ blue balls to a minimal necklace by appending them to strings of two or more consecutive blue balls.

For type 1 necklaces there are $n-2m-1 \choose m-k-1$ ways to do this.

For type 2 necklaces, blue balls can be added at either the start or the end of the necklace so there are $n-2m \choose m-k$ ways to do this.

The Lucas number connection

We will call a necklace good if it does not have adjacent red beads. Since rotations are counted differently, we will assume that each necklace has a special location where beads can be inserted.

$a_n$ is the number of good n-bead necklaces. I can explain why $a_n=a_{n-1}+a_{n-2}$. This is the recurrence relation for Lucas numbers.

Any good necklace of length $n-1$ can be turned into a good necklace of length $n$ by inserting a blue bead.

Any good necklace of length $n-2$ which has one red bead adjacent to its special location can be turned into a good necklace of length $n$ by inserting a blue bead and a red bead, and there is a unique order in which these two beads can be inserted.

Now consider the good necklaces of length $n-2$ which have two blue beads adjacent to their special location. By inserting first a blue bead then a red bead we can create a good necklace of length $n$ which has not already been counted. Note that, if the second bead to be inserted is blue, then we end up adding a blue bead to a length $n-1$ good necklace and creating a necklace that has already been counted.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.