Let $T_1$ be a triangle with sides $3011,3012,3013$. For $n\ge1$, if $T_n=\triangle ABC$ and $D,E,F$ are the points of tangency of the incircle of $\triangle ABC$ to the sides $AB,BC,AC$, respectively, then $T_{n+1}$ is a triangle with side lengths $AD,BE,CF$, if it exists. What is the perimeter of the last triangle in the sequence?

1. With $b=3012$ (the middle side), $T_n$ has sides $\frac b{2^{n-1}},\frac b{2^{n-1}}\pm1$, valid as a triangle while $b>2^n$.
2. Largest such $n$: $2^{11}=2048<3012<4096=2^{12}$, so $b>2^n$ holds exactly for $n\le11$, and the last triangle is $T_{11}$.
3. Perimeter of the last triangle $=\dfrac{3b}{2^{11-1}}=\dfrac{3b}{2^{10}}=\dfrac{9036}{1024}=\boxed{\dfrac{2259}{256}}$.