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Sequence Problem

  1. Jun 13, 2007 #1
    I was wondering, is it possible to find an explicit sequence for: 1, 1, 2, 2, 3, 3, 4 ,4 ,5 ,5 ,6 ,6 etc.
    without using the greatest integer function, or a recurrency formula? Any help would be appreciated.
     
  2. jcsd
  3. Jun 13, 2007 #2
    whats wrong with recurrence?
     
  4. Jun 13, 2007 #3
    nothing wrong with recurrence, its just that my professor said its possible to find a_n without using it and I can't figure it out
     
  5. Jun 13, 2007 #4
    are you in calc 2?
     
  6. Jun 13, 2007 #5
    The odd terms are (n+1)/2, n = 1, 3, 5, 7, ....
    The even terms are n/2, n = 2, 4, 6, 8, ....

    All you have to do is to figure out how to alternate between the two functions and that can be achived by using the alternating (-1)^n

    To warm up, figure out how to describe the sequence
    a, b, a, b, a, b, a, b, ....
    using the alternating (-1)^n.

    After you figure that, you can just plug in the two formulas from your original sequence instead of a, b.
     
  7. Jun 13, 2007 #6
    oooooooooooooooooooooooooooooo
     
  8. Jun 13, 2007 #7
    yep, I'm currently in calc 2, and thanks a bunch smallphi!
     
  9. Jun 14, 2007 #8
    both of those series yield even numbers
     
  10. Jun 14, 2007 #9
    Eh? Look again.
     
  11. Jun 14, 2007 #10
    what? for n=1,3,5,7,9 the first sequence yields even numbers.
    for n=2,4,6,8,10 the second sequence yields even numbers?
     
  12. Jun 14, 2007 #11
    Sloan's is your friend

    Many possibles http://www.research.att.com/~njas/sequences/?q=1%2C1%2C2%2C2%2C3%2C3%2C4%2C4%2C5%2C5%2C6%2C6%2C7%2C7%2C8%2C8&language=english&go=Search" [Broken]
     
    Last edited by a moderator: May 2, 2017
  13. Jun 14, 2007 #12

    Office_Shredder

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    For the odd numbers:
    1 -> (1+1)/2 = 1
    3 -> (3+1)/2 = 2
    5 -> (5+1)/2 = 3

    Even numbers:

    2 -> 2/2 = 1
    4 -> 4/2 = 2
    6 -> 6/2 = 3
     
  14. Jun 14, 2007 #13
    Not sure if the floor function is legal for your purposes but if so, this works

    m = {1, 2, 3, 4, ...}

    [tex]a = \lfloor \frac{m}{2} \rfloor + 1[/tex]

    Added later:

    Oops! Sorry didn't see this on first read: "without using the greatest integer function"
     
    Last edited: Jun 14, 2007
  15. Jun 14, 2007 #14
    This works without using the floor or ceil functions

    m = {0, 1, 2, 3, ... }

    [tex]a = \frac{2 * m + (-1) ^ m - 1}{4}[/tex]
     
  16. Jun 17, 2007 #15
    In an earlier post, I came up with this to answer the OP

    m = {0, 1, 2, 3, ... }

    [tex]a = \frac{2 * m + (-1) ^ m - 1}{4}[/tex]

    But I was wondering whether this can be extended to produce sequences with repeating terms of width n? The [tex]-1^m[/tex] trick doesn't seem to help.

    It's trivially easy to do it with floor

    m = {0, 1, 2, ... }
    n = {1, 2, 3, ...}

    [tex]a_m = \lfloor \frac{m}{n} \rfloor[/tex]

    But what about without floor?
     
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