I am studying through the josephus problem in concrete math , Here is the equation of binary form $$f(1) = α ;$$ $$f(2n + j) = 2f(n) + β_j ,$$ $$\text{ for } j = 0, 1 \text{ and } n \geq 1$$ this part I understand , now the unfolding of the recurrence
\begin{align*}f(b_m b_{m−1}...b_1 b_0)_2 &= 2f(b_m b_{m−1} . . . b_1)_2+ β_{b_0}\\&=4f(b_mb_{m−1}...b_2)_2 + 2β_{b_1} + β_{b_0}\\&= 2^mf((b_m)_2) +2^{m−1}β_{b_{m−1}} + · · · +2β_{b_1}+β_{b_0} \\&=2^mα + 2^{m−1}β_{b_{m−1}} + · · · + 2β_{b_1} + β_{b_0}\end{align*}
I mean what just happened here ?where did this $β_{b_0}$ came from ?There was just two constants $α$ , $β$ and now there are many. Can someone explain pease?
You have $j \in \{0, 1\},$ so you have $\beta_0, \beta_1.$ From the fact that $b_m b_{m−1}\dots b_1 b_0$ is a binary representation of $2n + j$ we have that to get $n$ from $2n + j$ we need to shift bits right and LSB (least significant bit) goes to subscript of $\beta.$ E.g.
$$f(101) = 2f(10) + \beta_1.$$