How to find lattice coefficients of all pole IIR filter with filter coefficient $k_m$ is 1?
Watch after 30mins... above question is addressed.

Consider:
   $$ 
          A_{4} (z) = 1 + h_1 z^{-1} + h_2 z^{-2} + h_1 z^{-3} + z^ {-4}$$ 
$$    A_4 (z) = A_3 (z) + k_4 * B_3 (z)$$
$$     A_4 (z) = A_3 (z) + k_4 * z^ {-4} (A_3 (z^ {-1})) $$



Try to break $A_4(z)$ into two equal halves:



$$ A_4 (z) = (1 + h_1 z^ {-1} + \frac{h_2}{2} z^ {-2} ) + z^ {-4} (1 + h_1 z^{1} + \frac{h_2}{2} z^{2} )$$

$$ A_3 (z) = A_2 (z) : k_3 = 0 , k_2 = \frac{h_2}{2} $$

$$A_1 (z ) = \frac{A_2 (z) - k_2  B_2 (z)}{1- k_2 ^ {2}} $$



Solving:
   $$ k_1 = \frac{h_1}{ 1 + \frac{h_2}{2}}$$



For above question : $$k_4 = 1$$
                         $$k_3 = 0$$
                         $$k_2 = \frac{\frac{5}{8}}{2} = \frac{5}{16}$$
                         $$k_1 = \frac{\frac{2}{3}}{1+ \frac{5}{16}} = \frac{32}{63} $$