Closed form of the recursion $D_{n+1} = n(D_n + D_{n-1})$

Question

I got this problem from chapter five of Titu Andreescu A path to combinatorics for undergraduates. I'm asked to give a closed form to the given recursion given that $D_1 = 0$ and $D_2 = 1$. And I know that this closed form is $D_n = n!\sum_{k = 0}^n \frac{(-1)^k}{k!}$ which can be proven by using inclusion exclusion, however I'm interested in a solution based purely on recursions given that in the same chapter the following theorem is provided:

Let $z \neq 0$ a complex number. Then $f(n) = z^n$ satisfies $f(n)-a_1f(n-1)- \dots - a_kf(n-k) = 0$ for real constants $a_1 \dots a_k$ if and only if $z$ is root of $x^k- a_1x^{k-1}- \dots -a_k = 0$.

I know that this theorem cannot be used to solve this because in the recursion I gave the coefficients are not constant, but that's the kind of strategy I was looking to use. Any help would be appreciated.

Answer 1

The sequence $x_n:=\frac{D_n}{n!}$ satisfies $x_1=0$, $x_2=\frac12$, and $$x_{n+1}-x_n=-\frac1{n+1}(x_n-x_{n-1}),$$ i.e. $x_1=0$ and $$x_n-x_{n-1}=\frac{(-1)^n}{n!},$$ i.e. $$x_n =\sum_{k = 0}^n \frac{(-1)^k}{k!}.$$

Answer 2

By the way, there is a related theorem which says that if you have a sequence which satisfies a recurrence where the coefficients are polynomials in $n$, then the EGF satisfies a differential equation where the coefficients are polynomials in $x$.

To see how this all connects, consider the two operations: $$ \frac{d}{dx} \sum a_n \frac{x^n}{n!} = \sum a_{n+1} \frac{x^n}{n!} \text{ and } x \sum a_n \frac{x^n}{n!} = \sum n a_{n - 1} \frac{x^n}{n!} $$ See if you just have the derivative operator then you can shift the coefficients and so a recurrence like $f_{n + 2} = f_{n + 1} + f_{n}$ translates to a differential equation $y'' = y' + y$ (each shift corresponds to a derivative).

If we have coefficients which are in terms of $n$, then we can create these by multiplying by $x$. So in your case $$ D_{n + 1} = nD_n + nD_{n-1} \text{ translates to } y' = xy' + xy. $$

This is a homogeneous linear differential equation: $$ \frac{y'}{y} = \frac{x}{1-x} \implies \log y = -x - \log(1-x) \implies y = \frac{e^{-x}}{1-x}. $$

And this is of course the EGF for the derangements problem. You can now use another common identity which just comes from multiplying $e^{-x}$ by $\frac1{1 - x}$:

\begin{align} \frac{1}{1 - x} e^{-x} &= \sum_i x^i \cdot \sum_j \frac{(-1)^j x^j}{j!} \\ &= \sum_n \left( \sum_{i + j = n} \frac{(-1)^j}{j!} \right) x^n \\ &= \sum_n \left( n! \sum_{i + j = n} \frac{(-1)^j}{j!} \right) \frac{x^n}{n!}. \end{align}

This gives us the formula $$ D_n = n! \sum_{j = 0}^n \frac{(-1)^j}{j!}. $$

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This method is reversible too! If $y = \frac{e^{-x}}{1 - x}$ then (using the product rule on $e^{-x} \cdot \frac{1}{1 - x}$)

$$ y' = -\frac{e^{-x}}{1 - x} + \frac{e^{-x}}{(1 - x)^2} = -y + \frac{y}{1 - x}. $$

And that rearranges back to $y' = xy' + xy$ which corresponds to $D_{n + 1} = nD_n + nD_{n - 1}$.