Recurrence equation similar to a geometric progression I have the following recurrence relation:
$$T(i) = \sqrt{T(i-1) \left(T(i+1) + k\right)},$$
with $k \geq 0$, a fixed constant.
I know that when $k=0$, we have:
$$T(i) = \sqrt{T(i-1) T(i+1)},$$
which solution is $T(i) = e^{ai+b}$ (geometric progression), but I do not know if there is a solution for the general case.


*

*If there is no closed-form solution to it, as it seems to be the case, is there an easy way to compute $T(i),\;1 < i < n$ given $T(1) = A$ and $T(n) = B$ ?

*More generally, is there a systematic, elegant way to (numerically) compute recurrences when the initial conditions are not "contiguous" ?

Remind that $T(i)=\sqrt{T(i-1)T(i+1)}$ can find its general solution is just a luck:
$T(i)=\sqrt{T(i-1)T(i+1)}$
$(T(i))^2=T(i-1)T(i+1)$
Let $T(i)=e^{U(i)}$ ,
Then $T(i-1)=e^{U(i-1)}$ , $T(i+1)=e^{U(i+1)}$
$\therefore(e^{U(i)})^2=e^{U(i-1)}e^{U(i+1)}$
$e^{2U(i)}=e^{U(i-1)+U(i+1)}$
$2U(i)=U(i-1)+U(i+1)$
$U(i+1)-2U(i)+U(i-1)=0$
$U(i)=\Theta_1(i)i+\Theta_2(i)$, where $\Theta_1(i)$ and $\Theta_2(i)$ are arbitrary periodic functions with unit period
$\therefore T(i)=e^{\Theta_1(i)i+\Theta_2(i)}$, where $\Theta_1(i)$ and $\Theta_2(i)$ are arbitrary periodic functions with unit period
It does not mean that $T(i)=\sqrt{T(i-1)\left(T(i+1)+k\right)}$ is as lucky as $T(i)=\sqrt{T(i-1)T(i+1)}$ .
