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Thread: How can I figure out eqn to this sequence? 8, 10, 14, 22, 38, 70,...

  1. #1

    Question How can I figure out eqn to this sequence? 8, 10, 14, 22, 38, 70,...

    My sequence starts at 4. Double 4 to get 8. 8 minus 3 to get 5. Double 5 to get 10. 10 minus 3 to get 7 double 7 to get 14 and so forth.

    4, 8, 5, 10, 7, 14, 11, 22, 19, 38........

    It can be simplified for my specific needs, as I only care about every other number, starting with 8.

    8, 10, 14, 22, 38, 70........

    I double the difference with every number; since 8 + 2 = 10, 10 + 4 = 14, 14 + 8 = 22, 22 + 16 = 38 and 38 + 32 = 70

    a. How can I figure out the equation for this sequence?

    b. Is there a way to figure out the equation for the original sequence?

  2. #2
    Elite Member
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    Quote Originally Posted by Liftamanjaro View Post
    My sequence starts at 4. Double 4 to get 8. 8 minus 3 to get 5. Double 5 to get 10. 10 minus 3 to get 7 double 7 to get 14 and so forth.

    4, 8, 5, 10, 7, 14, 11, 22, 19, 38........

    It can be simplified for my specific needs, as I only care about every other number, starting with 8.

    8, 10, 14, 22, 38, 70........

    I double the difference with every number; since 8 + 2 = 10, 10 + 4 = 14, 14 + 8 = 22, 22 + 16 = 38 and 38 + 32 = 70

    a. How can I figure out the equation for this sequence?

    b. Is there a way to figure out the equation for the original sequence?
    So we have:

    term 1 = 8 = a1

    term 2 = 10 = 8 + 2 = a2 = a1 + 2 = a1 + 2(2-1)

    term 3 = 14 = 10 + 4 = a3 = a2 + 2*2 = a2 + 2(3 - 1)

    term 4 = 22 = 14 + 8 = a4 = a3 + 2(4 - 1)

    Do you see the pattern.....
    “... mathematics is only the art of saying the same thing in different words” - B. Russell

  3. #3
    Elite Member
    Join Date
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    7,356
    Quote Originally Posted by Liftamanjaro View Post
    My sequence starts at 4. Double 4 to get 8. 8 minus 3 to get 5. Double 5 to get 10. 10 minus 3 to get 7 double 7 to get 14 and so forth.
    4, 8, 5, 10, 7, 14, 11, 22, 19, 38........
    I like to use recursive definitions:
    [TEX]\begin{align*}a_1&=4 \\\text{if }n\ge 2,~a_n&= (2\cdot a_{n-1}[n \mod 2])+(a_{n-1}-3)[n+1 \mod 2] \end{align*}[/TEX]
    Please see HERE.
    “A professor is someone who talks in someone else’s sleep”
    W.H. Auden

  4. #4
    Senior Member
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    For [tex] \ \ n \ge 1, [/tex]

    [tex]a_n \ = \ 2^{(n - 1)/2} \ + \ 3, \ \ \ if \ \ n \ \ odd[/tex]

    [tex]a_n \ = \ \ \ \ \ \ 2^{n/2} \ + \ 6, \ \ \ if \ \ n \ \ even[/tex]

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