Ouch, HTML formatting....:( On Thursday 25 April 2002 01:46, you wrote: > <html><div style='background-color:'><DIV>Hi,</DIV> > <DIV> </DIV> > <DIV> Mersenne primes are of the form 2^p-1. > The usual generalization is primes of the form ((k^p)-1)/(k-1), that is > repprimes in base k. It is a well known result that when </DIV> <DIV>2^p-1 > is composite every prime factor has to be of the form 2np+1 for some n.
Only when p is prime. Neither of the prime factors of 2^4-1 are divisible by 8n+1. In fact 2^n-1 is _always_ divisible by 3 when n is even, yet 3 = 2kp+1 only when k & p = 1. > Does there exist a similar restriction for divisors > of repdigits? Most results for Mersenne numbers generalize > easily but this one doesn't seem to. Any thoughts?</DIV> <DIV> </DIV> Oh, doesn't it? 3 is a special case, so ignore repunits with length 3. 11111 (base 3) = 121 = 11 x 11 = (2.1.5+1)(2.1.5+1) [This case is VERY interesting. It is widely believed that the factorization of Mersenne numbers is square free, although AFAIK there is neither proof nor counterexample - if so then generalized repunits are obviously different in at least this respect!] 1111111 (base 3) = 1093 is prime 11111111111 (base 3) = 88573 = 23 x 3851 = (2.1.11+1)(2.175.11+1) 1111111111111 (base 3) = 797161 is prime 11111 (base 5) = 781 = 11 x 71 = (2.1.5+1)(2.7.5+1) 1111111 (base 5) = 19531 is prime 11111111111 (base 5) = 12207031 is prime 11111 (base 7) = 2801 is prime 1111111 (base 7) = 137257 = 29 x 4733 = (2.2.7+1)(2.338.7+1) 11111 (decimal) = 41 x 271 = (2.4.5+1)(2.27.5+1) 1111111 (decimal) = 239 x 4649 = (2.17.7+1)(2.332.7+1) ... Regards Brian Beesley _________________________________________________________________________ Unsubscribe & list info -- http://www.ndatech.com/mersenne/signup.htm Mersenne Prime FAQ -- http://www.tasam.com/~lrwiman/FAQ-mers
