On the growth rate of random Fibonacci sequences

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On the growth rate of random Fibonacci sequences

• Benoît Rittaud (Université Paris-13)
• with Élise Janvresse Thierry de la Rue (CNRS -
  Université de Rouen)

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Random Fibonacci sequences
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Random Fibonacci sequences

• Classical Fibonacci sequence
• Fn Fn1Fn2
• (F0 a, F1 b)

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Random Fibonacci sequences

• Classical Fibonacci sequence
• Fn Fn1Fn2
• (F0 a, F1 b)
• Main property Fn1/Fn goes to ? (1v5)/2, so

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Random Fibonacci sequences

• Classical Fibonacci sequence
• Fn Fn1Fn2
• (F0 a, F1 b)
• Main property Fn1/Fn goes to ? (1v5)/2, so
• (Fn)1/n ? ?

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Random Fibonacci sequences

• Random Fibonacci sequence
• Fn Fn1?Fn2
• (F0 a, F1 b)

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Random 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.

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Random 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?

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Random Fibonacci sequences

• Two points of view

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Random Fibonacci sequences

• Two points of view
• average growth rate E(Fn)1/n ? ?

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Random Fibonacci sequences

• Two points of view
• average growth rate E(Fn)1/n ? ?
• almost-sure growth rate E(Fn1/n)? ?

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Random Fibonacci tree
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Random Fibonacci tree
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Random Fibonacci tree
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Random Fibonacci tree
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Random Fibonacci tree
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Random Fibonacci tree
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Random Fibonacci tree

• Theorem 1
• This tree (denote it by R) has the structure of
  the Fibonacci tree.

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Random Fibonacci tree

• Theorem 2
• All pairs (a, b) of mutually primes integers
  appears exactly once in R with a as parent of b.

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Random 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.

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The average point of view
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The average point of view
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The average point of view
Sn 2Sn1Sn3
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The average point of view

• Since Sn 2Sn1Sn3, the growth rate in R is
  known.

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The 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.

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The 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).

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The almost-sure point of view
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The almost-sure point of view

• Divakar Viswanath (1999)
• E(Fn1/n) ? 1.13198824

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The 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.

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The almost-sure point of view

• A simplification

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The almost-sure point of view

• A simplification
• where ?? is an explicit measure defined
• inductively on Stern-Brocot intervals.

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Generalizations
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Generalizations

• Unbalanced coin ( with probability p)

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Generalizations

• Unbalanced coin ( with probability p)
• OK for both points of view.

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Generalizations

• Unbalanced coin ( with probability p)
• OK for both points of view.
• Crititical p
• 1/3 for almost-sure, 1/4 for average

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Generalizations

• Unbalanced coin ( with probability p)
• Linear case Fn Fn1 ?Fn2

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Generalizations

• Unbalanced coin ( with probability p)
• Linear case Fn Fn1 ?Fn2
• OK again (but irrelevant for average).

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Generalizations

• Unbalanced coin ( with probability p)
• Linear case Fn Fn1 ?Fn2
• ?-sequences Fn ?Fn1 ?Fn2

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Generalizations

• Unbalanced coin ( with probability p)
• Linear case Fn Fn1 ?Fn2
• ?-sequences Fn ?Fn1 ?Fn2
• maybe hard in general, but

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Generalizations

• 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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Generalizations

• 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)

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Generalizations

• 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.