I don't think there is an easy way to do this. Clearly there is no characteristic polynomial as the coefficients are variable. I plugged it into wolfram alpha after working on it to no avail and its general form is $a_n=2e\Gamma(n,1)-1$ where $a_1=1$.

My original strategy was to substitute $b_n=a_{n+1}-c_1a_n$ for some constant $c$ such that everything cancels out neatly, but it got really messy and I ended up having $c_1=\dfrac{-n+\sqrt{n^2-4}}{2}$ (which is almost certainly wrong) then I stopped.

Divide through by $(n-1)!$ to get $\frac{a_n}{(n-1)!} = \frac{a_{n-2}}{(n-2)!} + \frac{n}{(n-1)!}$. Then if we let $b_n = \frac{a_n}{(n-1)!}$, we should have $b_1 = a_1$ and $b_n = b_{n-1} + \frac{n}{(n-1)!}$, so $b_n = a_1 + \sum_{k=2}^{n} \frac{k}{(k-1)!}$. Then $a_n = a_1(n-1)! + \sum_{k=2}^n \frac{k(n-1)!}{(k-1)!}$. Perhaps that sum has a nice simple closed form?

Is there a nice way to express the summation in a closed form? Summations in general do not count as a closed form. $\displaystyle\sum_{i=1}^ni$ is not a closed form, but $\dfrac{n(n+1)}{2}$ is.

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## Comments

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TopNewestYou have to do plug and chug

This is called plug :

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

This is called chug

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

Continue doing this until you arrive at a base case....You can see the general form after a few chugs. which you can prove by induction....

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I don't think there is an easy way to do this. Clearly there is no characteristic polynomial as the coefficients are variable. I plugged it into wolfram alpha after working on it to no avail and its general form is $a_n=2e\Gamma(n,1)-1$ where $a_1=1$.

$\Gamma(a,x)$ is the Incomplete Gamma Function which I have no clue of its definition.

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My original strategy was to substitute $b_n=a_{n+1}-c_1a_n$ for some constant $c$ such that everything cancels out neatly, but it got really messy and I ended up having $c_1=\dfrac{-n+\sqrt{n^2-4}}{2}$ (which is almost certainly wrong) then I stopped.

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Yes me too! I got that too, but it doesn't work the same way as typical characteristic eqn with variables...

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Divide through by $(n-1)!$ to get $\frac{a_n}{(n-1)!} = \frac{a_{n-2}}{(n-2)!} + \frac{n}{(n-1)!}$. Then if we let $b_n = \frac{a_n}{(n-1)!}$, we should have $b_1 = a_1$ and $b_n = b_{n-1} + \frac{n}{(n-1)!}$, so $b_n = a_1 + \sum_{k=2}^{n} \frac{k}{(k-1)!}$. Then $a_n = a_1(n-1)! + \sum_{k=2}^n \frac{k(n-1)!}{(k-1)!}$. Perhaps that sum has a nice simple closed form?

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Is there a nice way to express the summation in a closed form? Summations in general do not count as a closed form. $\displaystyle\sum_{i=1}^ni$ is not a closed form, but $\dfrac{n(n+1)}{2}$ is.

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No, I don't think so. Although $\sum \frac{(k-2)}{(k-1)!}$ is telescoping, this still leaves $\sum \frac2{(k-1)!}$, twice the partial sum for $e$.

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I would use generating functions.

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a1=1,a2=3,a3=9,a4=31

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