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:

$$x_{k+1}=\lfloor {\displaystyle \frac{x_k+\lceil n/x_k\rceil }{2}}\rfloor$$

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

As an example, let us find the rounded-square-root of $n=4321$.
$n$ has $4$ digits, so $x_0=7\times {10}^{(4-2)/2}=70$.
$$x_1=\lfloor {\displaystyle \frac{70+\lceil 4321/70\rceil }{2}}\rfloor =66$$ $$x_2=\lfloor {\displaystyle \frac{66+\lceil 4321/66\rceil }{2}}\rfloor =66$$ 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 ($10000\le n\le 99999$) 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 (${10}^{13}\le n<{10}^{14}$)?
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.