[seqfan] Re: Puzzle

[Frank Adams-Watters]

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?