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Thread: Leftmost digit of a randomly chosen Fibonacci number

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    Leftmost digit of a randomly chosen Fibonacci number

    .


    Randomly choose one Fibonacci number.

    Will its leftmost digit more likely belong to the set {1, 4, 8, 9} or to the set {2, 3, 5, 6, 7}?
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    Re: Leftmost digit of a randomly chosen Fibonacci number

    Quote Originally Posted by greg1313 View Post
    .


    Randomly choose one Fibonacci number.

    Will its leftmost digit more likely belong to the set {1, 4, 8, 9} or to the set {2, 3, 5, 6, 7}?
    IIRC, doesn't the leftmost digit have repetition? Something like it repeats every 67 numbers?
    Last edited by SlipEternal; Jun 6th 2018 at 11:11 AM.
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    Re: Leftmost digit of a randomly chosen Fibonacci number

    Quote Originally Posted by SlipEternal View Post
    IIRC, doesn't the leftmost digit have repetition? Something like it repeats every 67 numbers?
    I didn't know about that alleged information. I went to this list of the first 300 Fibonacci numbers just
    to quickly look at the leading 1's starting at n = 2 for F(2).

    The first 300 Fibonacci numbers, factored

    n
    -------

    2
    2 + 67 = 69
    69 + 67 = 136
    136 + 67 = 203
    203 + 67 = 270


    F(n) corresponding to these have leftmost digits of 1.


    - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -


    Possible open hint: Consider the distribution in Benford's Law.
    Last edited by greg1313; Jun 6th 2018 at 12:10 PM.
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    Re: Leftmost digit of a randomly chosen Fibonacci number

    Quote Originally Posted by greg1313 View Post
    I didn't know about that alleged information. I went to this list of the first 300 Fibonacci numbers just
    to quickly look at the leading 1's starting at n = 2 for F(2).

    The first 300 Fibonacci numbers, factored

    n
    -------

    2
    2 + 67 = 69
    69 + 67 = 136
    136 + 67 = 203
    203 + 67 = 270


    These have leftmost digits of 1.


    - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -


    Possible open hint: Consider the distribution in Benford's Law.
    If you look at the leftmost digit of $F(n+67)$, apparently, it is extremely similar to the leftmost digit of $F(n)$. It is the same something like 97% of the time. Although, IIRC, that was a heuristic approach that discovered that, and I am not sure if it has been proven yet. It was true for the first thousand or so terms.

    This graph does a good job of showing how similar they are:

    http://www.wolframalpha.com/input/?i...g%5B10%5D))%5D
    Last edited by SlipEternal; Jun 6th 2018 at 12:45 PM.
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    Re: Leftmost digit of a randomly chosen Fibonacci number

    $\phi^{67}\approx 10^{14} $ is why the digits are so close.

    Anyway, the split is almost 50/50 based on Bedford's law.
    Thanks from greg1313
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    Re: Leftmost digit of a randomly chosen Fibonacci number

    Quote Originally Posted by SlipEternal View Post

    Anyway, the split is almost 50/50 based on Bedford's [sic] law.
    Yes, one can look at the distribution of Benford's Law at this site: https://en.wikipedia.org/wiki/Benford%27s_law
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