Closed form sought for $a_1 = a_2 = 1, a_n = 1 + \frac{2}{n} \sum_{i=1}^{n-2} a_i $ where $n>2$ I've been working through a problem that I've got as far as getting a recursive answer to. I was hoping to turn this into more of a "closed form" answer, but haven't really gotten anywhere. I'm hoping that someone can help with this, though anything would be greatly appreciated.

The recursive answer I have is a sequence of real numbers given by $$\begin{gather}
a_1 = a_2 = 1 \\
a_n = 1 + \frac{2}{n} \sum_{i=1}^{n-2} a_i \qquad  (n > 2)
\end{gather}$$
The first few non-trivial members of this sequence are

*

*$a_3 = \frac{5}{3}$

*$a_4 = 2$

*$a_5 = \frac{37}{15}$

*$a_6 = \frac{26}{9}$

*$a_7 = \frac{349}{105}$
I've tried to express these in terms of $a_1$ and $a_2$ and and constants and arrived at

*

*$a_3 = 1 + \frac{2}{3} a_1$

*$a_4 = 1 + \frac{2}{4} a_1 + \frac{2}{4} a_2$

*$a_5 = (1 + \frac{2}{5}) + (\frac{2}{5} + \frac{2^2}{3\cdot5} ) a_1 + \frac{2}{5} a_2$

*$a_6 = (1 + \frac{2}{6} + \frac{2}{6}) + (\frac{2}{6} + \frac{2^2}{3 \cdot 6} + \frac{2^2}{4 \cdot 6}) a_1 + (\frac{2}{6} + \frac{2^2}{4 \cdot 6}) a_2$

*$a_7 = (1 + \frac{2}{7} + \frac{2}{7} + \frac{2}{7} + \frac{2^2}{5 \cdot 7}) + (\frac{2}{7} + \frac{2^2}{3\cdot7} + \frac{2^2}{4 \cdot 7} + \frac{2^2}{5 \cdot 7} + \frac{2^3}{3 \cdot 5 \cdot 7}) a_1 + (\frac{2}{7} + \frac{2^2}{4 \cdot 7} + \frac{2^2}{5 \cdot 7}) a_2$
I am not seeing a pattern developing here.
I also rearranged the above noting that $a_1 = a_2 = 1$ and got

*

*$a_3 = 1 + \frac{2}{3} $

*$a_4 = 1 + 2 (\frac{2}{4})$

*$a_5 = 1 + 3 (\frac{2}{5}) + \frac{2^2}{3\cdot5}$

*$a_6 = 1 + 4 (\frac{2}{6}) + \frac{2^2}{3 \cdot 6} + 2 (\frac{2^2}{4 \cdot 6})$

*$a_7 = 1 + 5 (\frac{2}{7}) + \frac{2^2}{3 \cdot 7} + 2 (\frac{2^2}{4 \cdot 7}) + 3 (\frac{2^2}{5 \cdot 7}) + \frac{2^3}{3 \cdot 5 \cdot 7}$
Here I do notice a couple of things

*

*The expression for $a_n$ begins with "$1 + (n-2) \frac{2}{n}$".

*The remaining terms of the expression look like "$k \dfrac{2^{i+1}}{b_1 \cdots b_{i} \cdot n}$" where each $b_j$ is between $3$ and $n-2$ and consecutive numbers cannot appear among them. The $k$ seems to be determined by the smallest number among the $b_j$, but this is more of a guess than anything right now.

These observations are not really helping me much at all.

After @Gary's answer, using the generating function
$$G(x) = \frac{{1 - e^{ - 2x} }}{{2(x-1)^2}}= \sum_{n = 1}^\infty  {a_n x^n }$$ define
$b_n=n! \,a_n$ which gives the sequence
$$\{1,2,10,48,296,2080,16752,151424,1519744,16766208\}$$ which is $A037256$   in $OEIS$ (have a look here).
It does not seem to present any particularity. In year 2002, Vaclav Kotesovec proposed tha approximation
$$b_n \sim \frac{1}{2} \left(1-\frac{1}{e^2}\right) n\, n!$$ which would make
$$a_n\sim \frac{1}{2} \left(1-\frac{1}{e^2}\right) n$$ which shows a relative error smaller than $1$% for $n> 68$.
Edit
As @Gary commented, I missed the constant term, which according to his comment should be
$$a_n\sim \frac{1}{2} \left(1-\frac{1}{e^2}\right) (n+1)$$
