The function $f(n)$ satisfies $f(1) = 1$ and $f(2n + 1) = f(n) + 1$ for $n \ge 0.$ Find $f(15).$

The recursion f(2n + 1) = f(n) + 1 tells us that f(1), f(2), f(3), ... satisfies an arithmetic progression with common difference 2.  The terms f(1) = 1 and f(15) are separated by 14 terms, so f(15) = 2*14 + 1 = 29.

Thanks for trying but the answer is 24

f(3) = f(2*1+1)=f(1)+1=2

f(7)=f(2*3+1)=f(3)+1=3

f(15)=f(2*7+1)=f(7)+1=4