Eccoci.

We define the rounded-square-root of a positive integer $n$ as the square root of $n$ rounded to the nearest integer.

The following procedure (essentially Heron's method adapted to integer arithmetic) finds the rounded-square-root of $n$:

Let $d$ be the number of digits of the number $n$.
If $d$ is odd, set $x_0 = 2 \times 10^{(d-1)/2}$.
If $d$ is even, set $x_0 = 7 \times 10^{(d-2)/2}$.
Repeat:

$$x_{k+1} = \Biggl\lfloor{\dfrac{x_k + \lceil{n / x_k}\rceil}{2} }\Biggr\rfloor$$

until $x_{k+1} = x_k$.

As an example, let us find the rounded-square-root of $n = 4321$.
$n$ has $4$ digits, so $x_0 = 7 \times 10^{(4-2)/2} = 70$.
$$x_1 = \Biggl\lfloor{\dfrac{70 + \lceil{4321 / 70}\rceil}{2} }\Biggr\rfloor = 66$$ $$x_2 = \Biggl\lfloor{\dfrac{66 + \lceil{4321 / 66}\rceil}{2} }\Biggr\rfloor = 66$$ Since $x_2 = x_1$, we stop here.
So, after just two iterations, we have found that the rounded-square-root of $4321$ is $66$ (the actual square root is $65.7343137\cdots$).

The number of iterations required when using this method is surprisingly low.
For example, we can find the rounded-square-root of a $5$-digit integer ($10\,000 \le n \le 99\,999$) with an average of $3.2102888889$ iterations (the average value was rounded to $10$ decimal places).

Using the procedure described above, what is the average number of iterations required to find the rounded-square-root of a $14$-digit number ($10^{13} \le n \lt 10^{14}$)?
Give your answer rounded to $10$ decimal places.

Note: The symbols $\lfloor x \rfloor$ and $\lceil x \rceil$ represent the floor functionthe largest integer not greater than $x$ and ceiling functionthe smallest integer not less than $x$ respectively.