Simpler version of the Josephus problem. Circle of $n$ people, every second is removed
let $f(n)$ represent the last one from a circle of $n$ people
I've proved $f(1)=1$; $f(2n)= 2f(n) - 1$; $f(2n+1) = 2f(n) + 1$
Hypothesis: $f(2^m + l)=1+2l$, $2^m > l$
Can the proof be: $2^m + l$ is either $2n$ or $2n+1$?
when its even:
$f(2^m + l) = 2f((2^m + l)/2) - 1 = 2( 1 + l) -1$
Here can I use $f((2^m + l)/2)= 1 + l$? And similarly for when $2^m + l = 2n+1$? And why?
Do I need to have $f(2) = 1$ in the base case as well as $f(1) = 1$?
If viable, is this a type of induction?
And on a similar note. Instead of if $f(n)$, then prove $f(n+1)$; can you do if $f(n)$, then prove $f(2n)$ and $f(2n+1)$? If this is viable, what do you call this type of induction?