Kaprekar number

Let $n$ be a $k$-digit integer in base $b$. Then $n$ is said to be a Kaprekar number in base $b$ if $n^2$ has the following property: when you add the number formed by its right hand digits to that formed by its left hand digits, you get $n$.

Or to put it algebraically, an integer $n$ such that in a given base $b$ has

$$n^2=\sum _{i=0}^{k-1}{d}_i{b}^i$$

(where $d_x$ are digits, with $d_0$ the least significant digit and $d_{k-1}$ the most significant) such that

$$\sum _{i=\frac{k}{2}+1}^k{d}_i{b}^{i-\frac{k}{2}-1}+\sum _{i=1}^{\frac{k}{2}}{d}_i{b}^{i-1}=n$$

if $k$ is even or

$$\sum _{i=\lceil \frac{k}{2}\rceil }^k{d}_i{b}^{i-\lfloor \frac{k}{2}\rfloor -1}+\sum _{i=1}^{\frac{k}{2}}{d}_i{b}^{i-1}=n$$

if $k$ is odd.

$b^x-1$ for a natural $x$ is always a Kaprekar number in base $b$.