For a fixed positive number β, consider the recurrence relation
where the sign in the sum is chosen at random for each n independently with equal probabilities for "+" and "−". This is a generalization of the random Fibonacci sequence to values of β ≠ 1.
It can be proven that for any choice of β, the limit
β* ≈ 0.70258 is defined as the threshold value for which
- σ(β) < 1 for 0 < β < β*,
so solutions to this recurrence decay exponentially as n → ∞, and
- σ(β) > 1 for β > β*,
so they grow exponentially. (In both cases, with probability 1.)
Regarding values of σ, we have:
- σ(1) = 1.13198824... (Viswanath's constant), and
- σ(β*) = 1 (by definition).
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