n students standing in line are to be divided into teams and in each team appointed a captain. How many ways to do that? (I need last transformation)

Question

There are $n$ students in the class. They stand in a line in front of the teacher, who is to divide them into any number of non-empty teams (in particular, the sets can be of size $1$ or $n$) and in each set appoint a captain. A team can only consist of students standing consecutively in the line. In how many different ways can the teacher do this?

Arrange a suitable equation or system of recursive equations and determine the general formula.

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I belive that the same question was asked right here: Counting the number of ways to divide into teams - complicated

There's no concrete solution provided so I tried to get my own.

I calculated manually the numbers of sequences for some of the initial values of $n$:

• $a_1 = 1$ (because we can have just $1$ team and $1$ captain of that team)
• $a_2 = 1 + 2 = 3$ (because we can have $2$ teams in which case the captains are obvious or $1$ team in which case we need to choose the captain in one of $2$ ways)
• $a_3 = 1 + 2 \cdot 2 + 3 = 8$ (because we can have $3$ teams in which case the captains are obvious or we can have in $2$ ways $2$ teams of $1 + 2$ students in which case we need to choose the captain in one of $2$ ways or we can have one big team of $3$ students for which we need to choose the captain in one of $3$ ways)
• $a_4 = 1 + 2 \cdot 3 + 2 \cdot 2 + 3 \cdot 2 + 4 = 21$ (because we can have $4$ teams in which case the captain are obvious or we can have in $2$ ways $2$ teams of $1 + 3$ students in which case we need to choose the captain in one of $3$ ways or we can have in one way $2$ teams of $2+2$ students in which case we can choose $2$ captains in $2$ ways each or we can have in $3$ ways $3$ teams of $1 + 1 + 2$ students in which case we can choose captain in $2$ ways or we can have one big team of $4$ students for which we need to choose the captain in one of $4$ ways)

When we add one student in the line at the plece n, we have:

• one big team with n students (so we can choose the captain of that team in one of n ways) and...
• as many teams with captains as in the line of lenght n-1, but we have that one 'new student' alone in his own team and...
• as many teams with captains as in the line of lenght n-2, but we have that one 'new student' in a team of 2 (so we can choose the captain of that team in one of 2 ways) and...
• as many teams with captains as in the line of lenght n-3, but we have that one 'new student' in a team of 3 (so we can choose the captain of that team in one of 3 ways) and so on

Therefore, I thought that the recurrence equation would look like this:

$a_n = n + a_{n-1} + 2a_{n-2} + 3a_{n-3} +...$

As you can see while comparing with the numbers of sequences for some of the initial values of $n$ (those that I calculated manually above), the formula does work.

Now, from the comments below I know that the general formula needs to be of form: $a_n = 3a_{n-1} - a_{n-2}$

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Could somebody explain to me how to get from: $$a_n = n + a_{n-1} + 2a_{n-2} + 3a_{n-3} +...$$ to: $$a_n = 3a_{n-1} - a_{n-2}$$ ?

I would like to see an algebraic way of transformation, not a proof by induction.

Answer 1

A generating function approach. We consider the recurrence relation \begin{align*} a_n&=3a_{n-1}-a_{n-2}\qquad\qquad n\geq 3\tag{1}\\ a_1&=1\\ a_2&=3 \end{align*} and want to show this relation fulfills \begin{align*} \color{blue}{a_n=n+a_{n-1}+2a_{n-2}+3a_{n-3}+\cdots}\tag{2} \end{align*}

Setting $n=2$ in the recurrence relation (1) we find from $a_2=3a_1-a_0$ that $a_0=0$. We consider the generating function \begin{align*} A(z)=\sum_{n=0}^{\infty}a_nz^n=z+3z^2+21z^3+55z^4+\cdots \end{align*}

Using the Ansatzmethode we obtain from (1) \begin{align*} \color{blue}{\sum_{n=2}^{\infty}a_nz^n} &=3\sum_{n=2}^{\infty}a_{n-1}z^n-\sum_{n=2}^{\infty}a_{n-2}z^n\tag{3}\\ &=3z\sum_{n=2}^{\infty}a_{n-1}z^{n-1}-z^2\sum_{n=2}^{\infty}a_{n-2}z^{n-2}\\ &=3z\sum_{n=1}^{\infty}a_{n}z^{n}-z^2\sum_{n=0}^{\infty}a_{n}z^{n}\\ &=3z\left(A(z)-a_0\right)-z^2A(z)\\ &\,\,\color{blue}{=\left(3z-z^2\right)A(z)}\tag{4} \end{align*}

Since the left-hand side in (3) is \begin{align*} \sum_{n=2}^{\infty}a_nz^n=A(z)-a_0-a_1z=A(z)-z \end{align*} we get from (4) \begin{align*} A(z)-z&=\left(3z-z^2\right)A(z)\\ \color{blue}{A(z)}&\color{blue}{=\frac{z}{1-3z+z^2}}\tag{5} \end{align*}

We now have to show that $a_n$ also fulfills the recurrence relation (2).

We consider a sequence $b_n, n\geq 0$ fulfilling (2) with the same starting values $b_0=0, b_1=1, b_2=3$ and show it has the same generating function $A(z)$. We start with \begin{align*} \color{blue}{b_n}&=n+b_{n-1}+2b_{n-2}+3b_{n-3}+\cdots\\ &\color{blue}{=n+\sum_{k=0}^nkb_{n-k}}\tag{6} \end{align*} and write this recurrence relation using generating functions. We get \begin{align*} \color{blue}{B(z)}&=\sum_{n=0}^{\infty}b_nz^n\tag{7}\\ &=\sum_{n=0}^{\infty}nz^n+\sum_{n=0}^{\infty}\left(\sum_{k=0}^nkb_{n-k}\right)z^n\\ &=z\sum_{n=0}^{\infty}nz^{n-1}+\left(\sum_{k=0}^{\infty}kz^k\right)\left(\sum_{l=0}^{\infty}b_lz^l\right)\tag{8}\\ &=z\frac{d}{dz}\left(\frac{1}{1-z}\right)+\left(z\frac{d}{dz}\left(\frac{1}{1-z}\right)\right)B(z)\tag{9}\\ &\,\,\color{blue}{=\frac{z}{(1-z)^2}+\frac{z}{(1-z)^2}B(z)}\tag{10} \end{align*}

Comment:

• In (8) we use the Cauchy product of power series.

• In (9) we have the derivation of the geometric series $\sum_{n=0}^{\infty}z^n=\frac{1}{1-z}$.

We get from (7) and (10) \begin{align*} B(z)&=\frac{z}{(1-z)^2}+\frac{z}{(1-z)^2}B(z)\\ B(z)(1-z)^2&=z+zB(z)\\ B(z)\left(1-2z+z^2\right)&=z+zB(z)\\ \color{blue}{B(z)}&\color{blue}{=\frac{z}{1-3z+z^2}} \end{align*} We conclude since the generating function $B(z)=A(z)$, that $a_n, n\geq 0$ also fulfills the recurrence relation (6) resp. (2) and the claim follows.

Answer 2

It does work. You just need to consider $a_0 = 1$ since there is, so to speak, $1$ way of choosing if there is no student at all.

Answer 3

Could somebody explain to me how to get from: $$a_n = n + a_{n-1} + 2a_{n-2} + 3a_{n-3} +...$$ to: $$a_n = 3a_{n-1} - a_{n-2}$$ ?

The usual trick is to use $a_{n-1}$ and $a_{n-2}$ to cancel the sums.

For each $n\ge 4$ we have

$a_n=n+\sum_{i=1}^{n-1} a_i(n-i)$,

$a_{n-1}=n-1+\sum_{i=1}^{n-2} a_i(n-1-i)$.

So $a_n-a_{n-1}=1+a_{n-1}+\sum_{i=1}^{n-2} a_i$.

Similarly

$a_{n-1}-a_{n-2}=1+a_{n-2}+\sum_{i=1}^{n-3} a_i$.

So $(a_n-a_{n-1})-(a_{n-1}-a_{n-2})=(a_{n-1}+a_{n-2})-a_{n-2}$,

that is $a_n=3a_{n-1}-a_{n-2}$.