[seqfan] Re: Puzzle

[Zak Seidov]

 r=3/5:
{1, 8/5, 11/5, 14/5, 17/5, 4, 23/5, 26/5, 
   29/5, 32/5, 7, 38/5, 41/5, 44/5, 47/5, 
   10, 1/10, 7/10, 13/10, 19/10, 5/2, 2/5, 1}


Tue, 20 Jan 2015 20:46:07 -0500 от Frank Adams-Watters <franktaw at netscape.net>:
>The best I've got so far is that any a/b where a divides b-1 has this 
>property. Add a/b to 1 b*(b-1)/a times, and you get b. Reciprocate to 
>1/b, and add a/b another (b-1)/a times; this gets you to 1.
>
>Showing that an r does not have this property is harder. One can show 
>that if r does not have the property, neither does r*k for any integer 
>k > 0; a step adding r*k can be transformed into k steps adding r. I 
>think I can show that 1 does not have the property; it would follow 
>that no integer n > 0 has the property. Every number in such a sequence 
>is a rational function of r (with integer coefficients), so no 
>irrational r can have the property.
>
>Franklin T. Adams-Watters
>
>-----Original Message-----
>From: David Wilson < davidwwilson at comcast.net >
>
>Choose any real r >= 0.
>
>Starting with n = 1, on the first step add r, on subsequent steps 
>either add
>r or take the reciprocal as you choose.
>
>For example, if r = 1/4, we can generate the sequence
>
>1, 5/4, 3/2, 7/4, 2, 1/2, 3/4, 1.
>
>For which r is it possible to return to 1 as does this sequence?
>
>
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