Title: On the growth rate of random Fibonacci sequences
1On the growth rate of random Fibonacci sequences
- Benoît Rittaud (Université Paris-13)
- with Élise Janvresse Thierry de la Rue (CNRS -
Université de Rouen)
2Random Fibonacci sequences
3Random Fibonacci sequences
- Classical Fibonacci sequence
- Fn Fn1Fn2
- (F0 a, F1 b)
4Random Fibonacci sequences
- Classical Fibonacci sequence
- Fn Fn1Fn2
- (F0 a, F1 b)
- Main property Fn1/Fn goes to ? (1v5)/2, so
5Random Fibonacci sequences
- Classical Fibonacci sequence
- Fn Fn1Fn2
- (F0 a, F1 b)
- Main property Fn1/Fn goes to ? (1v5)/2, so
- (Fn)1/n ? ?
6Random Fibonacci sequences
- Random Fibonacci sequence
- Fn Fn1?Fn2
- (F0 a, F1 b)
7Random Fibonacci sequences
- Random Fibonacci sequence
- Fn Fn1?Fn2
- (F0 a, F1 b)
- where the ? sign is obtained by tossing a
- (balanced) coin for each n.
8Random Fibonacci sequences
- Random Fibonacci sequence
- Fn Fn1?Fn2
- (F0 a, F1 b)
- where the ? sign is obtained by tossing a
- (balanced) coin for each n.
- Question what about the growth rate?
9Random Fibonacci sequences
10Random Fibonacci sequences
- Two points of view
- average growth rate E(Fn)1/n ? ?
11Random Fibonacci sequences
- Two points of view
- average growth rate E(Fn)1/n ? ?
- almost-sure growth rate E(Fn1/n)? ?
12Random Fibonacci tree
13Random Fibonacci tree
14Random Fibonacci tree
15Random Fibonacci tree
16Random Fibonacci tree
17Random Fibonacci tree
18Random Fibonacci tree
- Theorem 1
- This tree (denote it by R) has the structure of
the Fibonacci tree.
19Random Fibonacci tree
- Theorem 2
- All pairs (a, b) of mutually primes integers
appears exactly once in R with a as parent of b.
20Random Fibonacci tree
- Theorem 3
- The walk from (1, 1) to
- (a, b) in R can be obtained by considering the
continued fraction expansion of a/b.
21The average point of view
22The average point of view
23The average point of view
Sn 2Sn1Sn3
24The average point of view
- Since Sn 2Sn1Sn3, the growth rate in R is
known.
25The average point of view
- Since Sn 2Sn1Sn3, the growth rate in R is
known. - Consider the full tree as a union of (infinite)
copies of R to evaluate the average growth rate
of a random Fibonacci sequence.
26The average point of view
- Since Sn 2Sn1Sn3, the growth rate in R is
known. - Consider the full tree as a union of (infinite)
copies of R to evaluate the average growth rate
of a random Fibonacci sequence. - The result is ?1 1.20556943, where
- ?3 2?21 (? gt 1).
27The almost-sure point of view
28The almost-sure point of view
- Divakar Viswanath (1999)
- E(Fn1/n) ? 1.13198824
29The almost-sure point of view
- Divakar Viswanath (2000)
- E(Fn1/n) ? ? 1.13198824
- obtained by the computation of a (quite
- tiresome) integral given by a formula of
- Furstenberg.
30The almost-sure point of view
31The almost-sure point of view
- A simplification
- where ?? is an explicit measure defined
- inductively on Stern-Brocot intervals.
32Generalizations
33Generalizations
- Unbalanced coin ( with probability p)
34Generalizations
- Unbalanced coin ( with probability p)
- OK for both points of view.
35Generalizations
- Unbalanced coin ( with probability p)
- OK for both points of view.
- Crititical p
- 1/3 for almost-sure, 1/4 for average
36Generalizations
- Unbalanced coin ( with probability p)
- Linear case Fn Fn1 ?Fn2
37Generalizations
- Unbalanced coin ( with probability p)
- Linear case Fn Fn1 ?Fn2
- OK again (but irrelevant for average).
38Generalizations
- Unbalanced coin ( with probability p)
- Linear case Fn Fn1 ?Fn2
- ?-sequences Fn ?Fn1 ?Fn2
39Generalizations
- Unbalanced coin ( with probability p)
- Linear case Fn Fn1 ?Fn2
- ?-sequences Fn ?Fn1 ?Fn2
- maybe hard in general, but
40Generalizations
- Unbalanced coin ( with probability p)
- Linear case Fn Fn1 ?Fn2
- ?-sequences Fn ?Fn1 ?Fn2
- maybe hard in general, but
- lets try with ? v2
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42Generalizations
- Unbalanced coin ( with probability p)
- Linear case Fn Fn1 ?Fn2
- ?-sequences Fn ?Fn1 ?Fn2
- more generally,
- the method works for any ? of the form
- ? 2cos(p/k) (k integer)
43Generalizations
- Unbalanced coin ( with probability p)
- Linear case Fn Fn1 ?Fn2
- ?-sequences Fn ?Fn1 ?Fn2
- more generally,
- the method works for any ? of the form
- 2cos(p/k) (k integer)
- These are Rosen continued fraction.