Consider that $c$ is integer that is $\geq \ 3$, also $d$ is non negative integer. How to figure out general formula for congruence: $$ \left( 2^{\large 2\cdot3^{c-3} +\large d}\right)\bmod 3^{c-1} $$ I observed that this equals to: $$ f(c) = \begin{cases} 3^{c-2} + 2^d , & \mbox{if } d\mbox{ is even} \\ 2 \cdot 3^{c-2} + 2^d, & \mbox{if } d\mbox{ is odd} \end{cases}$$ But only when $c$ is big enough for some fixed $d$. Before some $c$, initial congruence and my hypothesized formula aren't equal. For example, $d = 13$ than reminders are: $$ 8, 2, 65, 92, 659, 902, 6005, 1631, 47558, 126290, 362486, 1071074, 3196838, 9574130, 28706006, 86101634, 258288518 ... $$ Starting with $c = 3 $ and formula gives: $$ 8198, 8210, 8246, 8354, 8678, 9650, 12566, 21314, 47558, 126290, 362486, 1071074, 3196838, 9574130, 28706006, 86101634, 258288518 ...$$ Notice that after 8th term (when $c = 10 $), this two sequences are same.So for $ d = 13 $ point of converge is 8th term. For other $d$'s we get sequence: $$0, 0, 0, 1, 1, 3, 3, 4, 4, 5, 5, 6, 6, 8, 8, 9, 9, 10, 10, 11, 11, 13, 13, 14, 14, 15, 15, 17, 17, 18, 18, 19, 19...$$ Beginning with $d=0$. (Missing numbers are $\{2,7,12,16 ...\}$ and more terms are unknown for me)
My question is, can I figure out general formula for initial congruence, that will be correct for all $c$ and $d$. May be it requires Fermat's little theorem?