Talk:Primes of the Form of a Power Less One
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Suppose n is not prime, i.e. n = r s where r, s > 1. Then: $\displaystyle 2^{r s} - 1 = \left({2^r - 1}\right) \sum_{k=0}^{s-1} {2^{r k}}$
Should this be $\displaystyle 2^{r s} - 1 = \left({2^r - 1}\right) \sum_{k=0}^{s-1} {2^{r+k}}$ --Cynic (talk) 04:49, 24 March 2009 (UTC)
Yep, sorry, slip of the finger. --Matt Westwood 06:20, 24 March 2009 (UTC)
Actually no I don't think I do ... $2^{r s} - 1 = \left({2^r - 1}\right) \left({1 + 2^r + 2^{2r} + 2^{3r} + \ldots + 2^{\left({s-1}\right)r}}\right)$. --Matt Westwood 06:54, 25 March 2009 (UTC)
