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Old 2018-01-03, 23:26   #9
JeppeSN
 
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"Jeppe"
Jan 2016
Denmark

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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 \((A+1)^n-1\) can never be prime because it is divisible by \(A\).

Therefore we can consider instead \[\frac{(A+1)^n-1}{A}\] 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 2018-01-03 at 23:27
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