Find$\{a_{n}\}$such that $a_{n+1}=\frac{a_{n}^{2}+3}{a_{n}+1},\,a_{1}=1$

Question

I met the following problem when I explained some math problems to a high school student.Find$\{a_{n}\}$such that $$a_{n+1}=\frac{a_{n}^{2}+3}{a_{n}+1},\,a_{1}=1.$$ Rewrite $$a_{n+1}=\frac{a_{n}^{2}+3}{a_{n}+1}$$ as $$(a_{n}+A)(a_{n}-a_{n+1}+B)=0$$,but I don't have $A$ and $B$ that satisfy the conditions.

$3=\frac{3^2+3}{3+1}$, but $a_{1}=1$.

$a_{n+1}-3=\frac{a_{n}(a_{n}-3)}{a_{n}+1}$ doesn't seem work.

Could anyone help me, thanks a lot.

Answer 1

Not an answer, but it doesn't fit in a comment. The first few values of $a_n$ are: $$ 1\\ 2\\ 7/3\\ 38/15\\ 2119/795\\ 3193118/1158315\\ 14221083479599/5040330115395\\ 139226999175098461748609438/48541941503963464907166315\\ 26453117554225500188222788581576818186109055874713519/9114668934704209925639212726593462750208755315360195 $$ I doubt there's a compact formula for $a_n.$

Answer 2

I think that you may actually want the equation $\ (a_{n}+A)(a_{n}-a_{n+1}+B)=C \ $ in which case the solution is $\ (a_{n}+1)(a_{n}-a_{n+1}-1)=-4. \ $ By definition we have $\ (a_n+1)a_{n+1} = a_n^2+3. \ $ Now $\ (a_n+1)(a_n-1) = a_n^2-1 \ $ and subtracting the two equations gives the solution you want.

Note that if $\ a_n \to 3\ $ then $\ (a_n + 1) \to 4\ $ and $\ (a_{n}-a_{n+1}-1) \to -1\ $ which is consistent with $\ (a_{n}+1)(a_{n}-a_{n+1}-1)=-4.$

You wrote:

$a_{n+1}-3=\frac{a_{n}(a_{n}-3)}{a_{n}+1}$ doesn't seem work.

but it does work. It implies that as $\ a_n\to 3\ $ the ratio $\ \frac{a_n}{a_n+1} \to \frac34<1\ $ and the sequence $\ a_n\ $ converges linearly to $3.$