Answer: A

Small Hint:

Values greater than $1$ come from even indices; values between $0$ and $1$ come from odd indices

Big Hint:

Reverse the recursion by subtracting $1$ from values above $1$ and taking reciprocals of values below $1$

Solution:

Let $N(r)$ be the index at which the value $r$ occurs. Reversing the recursion gives $N(r)=2N(r-1)$ for $r\gt1,$ and $N(r)=N(\frac{1}{r})+1$ for $0\lt r\lt1.$

Starting with $N(1)=1,$ repeated use gives $N(2)=2,$ $N(\frac{1}{2})=3,$ $N(\frac{3}{2})=6,$ and $N(\frac{2}{3})=7.$ Continuing gives $N(\frac{5}{3})=14,$ $N(\frac{8}{3})=28,$ $N(\frac{3}{8})=29,$ and $N(\frac{11}{8})=58.$

Next, $N(\frac{8}{11})=59,$ $N(\frac{19}{11})=118,$ $N(\frac{11}{19})=119,$ $N(\frac{30}{19})=238,$ $N(\frac{49}{19})=476,$ and $N(\frac{68}{19})=952.$ Finally, $N(\frac{87}{19})=1904$ and $N(\frac{19}{87})=1905.$ Thus $n=1905,$ whose digit sum is $1+9+0+5=15.$

Therefore the correct answer is A.