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Old 2019-10-17, 15:03   #2
Dr Sardonicus
 
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Feb 2017
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Quote:
Originally Posted by miket View Post
Let n be an odd positive integer, Let o=ordn2 be the order of 2 modulo n and m the period of 1/n, k is number of distinct odd residues contained in set {2^1,2^2,...,2^{n−1}} modulo n.

If odd part of o,m and k is 1 and k divide n-1, then n is item in the sequence 17, 257, 641, 65537, ….
<snip>
First, I don't understand why n = 3 isn't on the list.

Second, the "period" of 1/n apparently is to the base ten. If this period is a power of 2, then the multiplicative order of 10 (ten) (mod n) must also be a power of 2 (so 5 can not divide n), and the multiplicative order of 5 (mod n) must therefore also be a power of 2. The multiplicative orders may, of course, be different powers of 2.

Third, if n is prime, and the multiplicative order o > 1 of 2 (mod n) is even, then exactly half the representatives of {1, 2, ... 2^(o-1) (mod n)} in the interval [1,n-1] -- that is, o/2 of them -- are odd. I leave the proof as an exercise for the reader.
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