Number of walks on triangular latice Let $a_n$ denote the number of self-avoiding walks of length $n \in \mathbb{N}$ on the triangular lattice, it is the two-dimensional lattice where unit triangles are packed together. I want to show that 
$$2^n \leq 
a_n ≤ 6 \cdot 5^{n−1}, ~~ n \in \mathbb{N}.$$
I am looking for a rigorous way of doing this instead of an intuitive explanation.
I think you forgot to specify that the walks all start at some origin; otherwise there are countably infinitely many of them.

As in your other question, since every vertex has $6$ neighbours, $6\cdot5^{n-1}$ is the number of walks that don't immediately return to the same place, and thus an upper bound for the number of self-avoiding walks.

For the lower bound, choose, say, two directions to the upper right and to the lower right; then in each step you can use one of them, and you'll always be going right and therefore never returning.
I think you forgot to specify that the walks all start at some origin; otherwise there are countably infinitely many of them.

As in your other question, since every vertex has $6$ neighbours, $6\cdot5^{n-1}$ is the number of walks that don't immediately return to the same place, and thus an upper bound for the number of self-avoiding walks.

For the lower bound, choose, say, two directions to the upper right and to the lower right; then in each step you can use one of them, and you'll always be going right and therefore never returning.