[seqfan] Re: Puzzle

[Frank Adams-Watters]

Working backwards with the rule: subtract r unless the current value is 
<= r, otherwise reciprocate, it appears experimentally that most 
rational numbers < 1 work. The smallest counterexample (by denominator) 
is 7/9. (I'm not sure that 7/9 can't be solved, but if it can be solved 
using this rule, it requires more than 10000 steps.)

For example, 3/11 gives us:

1 8/11 5/11 2/11 11/2 115/22 109/22 103/22 97/22 91/22 85/22 79/22 
73/22 67/22 61/22 5/2 49/22 43/22 37/22 31/22 25/22 19/22 13/22 7/22 
1/22 22 239/11 236/11 233/11 230/11 227/11 224/11 221/11 218/11 215/11 
212/11 19 206/11 203/11 200/11 197/11 194/11 191/11 188/11 185/11 
182/11 179/11 16 173/11 170/11 167/11 164/11 161/11 158/11 155/11 
152/11 149/11 146/11 13 140/11 137/11 134/11 131/11 128/11 125/11 
122/11 119/11 116/11 113/11 10 107/11 104/11 101/11 98/11 95/11 92/11 
89/11 86/11 83/11 80/11 7 74/11 71/11 68/11 65/11 62/11 59/11 56/11 
53/11 50/11 47/11 4 41/11 38/11 35/11 32/11 29/11 26/11 23/11 20/11 
17/11 14/11 1

Reverse the order to get the solution.

Franklin T. Adams-Watters

-----Original Message-----
From: M. F. Hasler <oeis at hasler.fr>
To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>
Sent: Tue, Jan 20, 2015 8:49 pm
Subject: [seqfan] Re: Puzzle


Obviously, for r=0 one reaches 1 after the first step.
If r>0, you can reach 1 only via +r, so you must be able to reach 1-r.
For rational r=m/n, you must thus be able to reach (n-m)/n, or 
equivalently
(n-km)/n for any integer k>0, but negative numbers and zero are never
reached so 1 <= k < n/m so m < n.
I.e., r < 1, which includes Franklin's result that r cannot be integer
(except 0).

But I do not agree with his reasoning that r cannot be irrational. 
(Because
the integer coefficients in the rational fraction can be zero.)
For example, r = sqrt(2)/2 seems to work:
+ ÷ + + + + ÷ +
should yield 1 (where ÷ is reciprocal).

Maximilian
Le 20 janv. 2015 20:05, "David Wilson" <davidwwilson at comcast.net> a 
écrit :

> 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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>
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>

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