The function $f(n)$ satisfies $f(1) = 1$ and $f(2n + 1) = f(n) + 1$ for $n \ge 0.$  Find $f(15).$
I need to find $f(15),$ which means I need to plug in $2n + 1 = 15$ into the recursive formula. Solving for $n,$ I get $n = 7,$ so $f(15) = f(7) + 1.$ Now I need to find $f(7),$ which means I need to plug in $2n + 1 = 7$ into the recursive formula. Solving for $n,$ I get $n = 3,$ so $f(7) = f(3) + 1.$ Now I need to find $f(3),$ which means I need to plug in $2n + 1 = 3$ into the recursive formula. Solving for $n,$ I get $n = 1,$ so $f(3) = f(1) + 1.$ Finally, I know that $f(1) = 1,$ so $f(3) = 1 + 1 = 2.$ Therefore, $f(7) = 2 + 1 = 3,$ and $f(15) = 3 + 1 = 4.$ 