I recently stumbled across Automorphic Numbers (definition and examples).
In simple words, a number $n$ is said to be automorphic if last $d(n)$ digits of $n^2$ are $n$ itself (where $d(n)$ is the number of digits of n)
Example: $n$ = 890625 is automorphic since $n^2$ = 793212890625 where last $d(n)$ = 6 digits of $n^2$ are the same as $n$.
I understand that there is recurrence relation to calculate an automorphic number with $2d(n)$ digits given an automorphic number of $d(n)$ digits; given by,
$n'$ = $(3n^2 - 2n^3)$ $mod$ $10^{2d(n)}$ (Proof)
Now that I understood this, I started playing with automorphic numbers and tried to find out what I call as Reverse Automorphic Numbers. As you might have guessed by now,
A reverse automorphic number $k$ is a number such that last $d(n)$ digits of $k^2$ is the reverse of $k$.
Example: $k = 65766$ is reverse-automorphic since $k^2 = 43251$66756 where last $d(n) = 6$ digits of $k^2$ are the reverse of $k$.
I tried finding out the recurrence relation for reverse-automorphic numbers on the same lines as that of automoprhic numbers but could not proceed. How can I find out a recurrence relation for such reverse-automorphic numbers ?
Thanks in advance!