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Lucas Numbers and Generating Functions

  1. Oct 30, 2005 #1
    Here is my problem and my attempt at the answer. Any help or advice is highly appreciated.
    With the famous sequence of Lucas numbers 1, 3, 4, 7, 11, 18... (Where each number is the sum of the last two and the first two are defined as 1 and 3.) use generating functions to find an explicit formula for the Lucas function.
    Attempted Solution
    We have
    where Fj denotes the jth Fibonacci number and n is going to infinity. Then we add that to
    Where F-1 = -1 and F0 = 0

    And that should get us a function of Lucas numbers right?:confused:
    Last edited: Oct 31, 2005
  2. jcsd
  3. Oct 30, 2005 #2


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    The Lucas numbers satisfy the relation

    [tex]L_{n+1} = L_n + L_{n-1}[/tex]

    Just set [itex]L_n = a^n[/itex] and solve for a. Your generating function will be a linear combination of the two solutions. Apply your initial conditions ( [itex]L_1[/itex] and [itex]L_2[/itex]) to determine the two arbitrary constants and you're done! :)
  4. Oct 31, 2005 #3
    I should've been more specific, but we have to use the Fibonacci numbers to generate the Lucas numbers in this manner.

    EDIT: I changed it up a little bit as well.
    Last edited: Oct 31, 2005
  5. Oct 31, 2005 #4


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    In that case, it should be apparent that [itex]L_n = F_{n+1} + F_{n-1}[/itex]
  6. Nov 1, 2005 #5
    Ok so I believe that matches what I was intending on getting at. Thank you again.
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