MHB Find a closed-form expression for f(n)

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The discussion centers on finding a closed-form expression for the function f(n), defined as f(n) = 1 + n + n(n-1) + n(n-1)(n-2) + ... for n > 1, with f(1) = 1. Examples illustrate the function's values: f(2) = 3, f(3) = 10, and f(4) = 41. A recurrence relation, f(n) = 1 + n*f(n-1), is proposed as a starting point for deriving a formula. However, there is skepticism about the existence of a simple closed form. The conversation emphasizes the challenge of simplifying the expression while exploring its recursive nature.
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Find a closed-form expression for $f(n)$ where

$f(n)=1+n+n(n-1)+n(n-1)(n-2)+...+n(n-1)\cdots2$ for $n>1$

and $f(1)=1$.

(So, $f(2)=1+2=3$

$f(3)=1+3+3\cdot 2=10$

$f(4)=1+4+4\cdot 3+4\cdot 3\cdot 2=41$

and so on.)
 
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It's easy to convert this into a recurrence relation, but I doubt that a simple closed form exists.


$$f(n)=n!\left(1+\frac{1}{2!}+\dots+\frac{1}{n!}\right)$$
 
Evgeny.Makarov said:
It's easy to convert this into a recurrence relation, but I doubt that a simple closed form exists.

I actually started with the recurrence relation $f(n)=1+nf(n-1)$ to arrive at my formula in post #1.
 
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