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:

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

Using this procedure, what is the average number of iterations required to find the rounded-square-root of a $N$-digit number ($10^{N - 1} \le n \lt 10^{N}$)?
Give your answer rounded to $10$ decimal places.