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RANDOM FIBONACCI SEQUENCES AND THE NUMBER 1.13198824...

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  • #Math
Divakar Viswanath
@DivakarViswanath
(Author)
www.math.lsa.umich.edu
Read on www.math.lsa.umich.edu
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Abstract. For the familiar Fibonacci sequence (defined by f1 = f2 = 1, and fn = fn−1 + fn−2 for n > 2), fn increases exponentially with n at a 5)/2 = 1.61803398 . . . . But for a si... Show More

Abstract. For the familiar Fibonacci sequence (defined by f1 = f2 = 1,
and fn = fn−1 + fn−2 for n > 2), fn increases exponentially with n at a
5)/2 = 1.61803398 . . . . But for a simple modification with both additions and subtractions — the random Fibonacci sequences defined by t1 = t2 = 1, and for n > 2, tn = ±tn−1 ±tn−2, where each ± sign is independent and either + or − with probability 1/2 — it is not
even obvious if |tn| should increase with n. Our main result is that 􏰎
n |tn|→1.13198824... as n→∞
with probability 1. Finding the number 1.13198824 . . . involves the theory of random matrix products, Stern-Brocot division of the real line, a fractal measure, a computer calculation, and a rounding error analysis to validate the computer calculation.

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Greg Egan @gregeganSF · Jun 28, 2022
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[1/2] Make a Fibonacci-like sequence with an equal probability of choosing either sign in the recurrence: f_{n+2} = f_n ± f_{n+1} This is explored in a great paper by Viswanath (thanks to @_onionesque). [link] Here’s the distribution of the nth root of |f_n|
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