(Shouldn’t you have i_2=k-1 ?)
Lower bounds for k (Crandall):
Using \log_23 and its convergents \frac{p_n}{q_n}, a well know inequality states that for any pair of integer S_k, k with k<q_n we have \lvert S_k - k\log_23 \rvert>\lvert p_n - q_n\log_23 \rvert. We also have the classical continued fraction inequality \lvert p_n - q_n\log_23 \rvert>\frac{1}{q_n+q_{n+1}}
From here you have \lvert 2^{S_k} - 3^k\rvert>3^k\lvert S_k\log2 - k\log3 \rvert>3^k\log2\lvert p_n - q_n\log_23 \rvert>\frac{3^k\log2}{q_n+q_{n+1}}
which plugged into a_0<\frac{k3^{k-1}}{2^{S^k}-3^k} from here gives X_0<a_0<\frac{k(q_n+q_{n+1})}{3\log2} or k>\frac{X_0\cdot 3\log2}{q_n+q_{n+1}} where X_0=2^{71}. The other option is k>q_n so that k>\min\{q_n,\frac{X_0\cdot 3\log2}{q_n+q_{n+1}}\}
To find the right q_n just search for the minimum one satisfying X_0<\frac{q_n(q_n+q_{n+1})}{3\log2} and verify that q_{n-1}<\frac{X_0\cdot 3\log2}{q_n+q_{n+1}} (if not, pick q_{n-1})
Here, with X_0=2^{71} we have q_{n}=65470613321 and k>2.4\cdot 10^{10}