How to check if an integer can be represented with all set bits for any given base? By all set bits, I mean all the set bits are placed consecutively together. e.g., for base 2, we have 1, 3, 7, 15, 31, 63, i.e., any integer $x$ such that $x = 2^m - 1$ for some integer $m$.
In binary, $1$ is $1$ and $3$ is $11$ and $7$ is $111$ and $15$ is $1111$ and so on.
I want to be able to do this with any base. Based on wolfram, I am inducing that the formula for checking an integer $x$ and some base $b$ and some non-negative exponent $m$ is
$$
x = \frac{1}{b - 1}\left(b^m - 1 \right)
$$
How can you intuitively derive this formula?

After I wrote out $x = b^0 + b^1 + b^2 + \ldots + b^m$, it became obvious to me how to derive this expression.
Let $f(m) = x = b^0 + b^1 + b^2 + \ldots + b^m$, then we have
$$
f(m) / b = 1/b + b^0 + b^1 + \ldots + b^{m - 1} \\
= 1/b + f(m) - b^m \\
\implies f(m) = 1 + bf(m) - b^{m + 1} \\
\therefore f(m) = \frac{b^{m + 1} - 1}{b - 1}
$$
 A: Your formula for $x$ is a geometric sum of $m$ digits of all $1$s,
$$
{\overbrace{111\cdots1}^m}_b = b^0+b^1+\cdots + b^{m-1} = \frac{b^m-1}{b-1} = \frac{10_b^m-1}{10_b-1}$$
To apply the usual proof of geometric sum specifically to $x$ in base $b$,
$$\begin{array}{crcrl}
&10_b x &= &{\overbrace{111\ldots 1}^m0}_b\\
-& x &= &{\overbrace{11\ldots 11}^m}_b\\
\hline
&(10_b-1)x &= &1{\overbrace{00\ldots00}^m}_b & -1\\
&&= &10^m_b&-1\\
\hline
&x &= &\dfrac{10^m_b-1}{10_b-1}
\end{array}$$
A: I don't know if you'll find this intuitive, but it may, at least, make the result more memorable. First note that your number $x$ with $m$ bits "set" is $x = b^{m-1} + b^{m - 2} + \ldots + b + 1$. Then lay out the calculation of $(b- 1)x$ like this:
$$
\begin{align*}
(b - 1)(b^{m-1} + b^{m - 2} + \ldots + b + 1)
  &= \begin{array}[t]{ccccccc}
        b^m &+b^{m-1} &+ &\ldots &+ b^2 &+ b\\
               &-b^{m-1} &- &\ldots &- b^2 &- b &- 1
      \end{array}\\
  &= b^m - 1
\end{align*}
$$
Now divide both sides by $b - 1$. This is how I remember the formula for the sum of a geometric progression.
