This is related to one way you can generalize Mersenne primes \(2^n  1\).
If you try to change the \(2\) into \(A+1\) with \(A>1\), we have seen above that the resulting \((A+1)^n1\) can never be prime because it is divisible by \(A\).
Therefore we can consider instead \[\frac{(A+1)^n1}{A}\] which may or may not be prime, just like Mersenne numbers. This is what some people call repunits in base \(A+1\).
For example \(\frac{10^{19}1}{9}\) or 1111111111111111111 is prime.
/JeppeSN
Last fiddled with by JeppeSN on 20180103 at 23:27
