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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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