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    jk's Avatar
    jk Posts: 3, Reputation: 1
    New Member
     
    #1

    Dec 2, 2009, 12:09 PM
    1,2,6,? 72
    What are the two numbers missing in the following sequence of numbers : 1,2,6,? 72
    Clough's Avatar
    Clough Posts: 26,677, Reputation: 1649
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    #2

    Dec 2, 2009, 06:26 PM
    Hi, jk!

    Are these homework questions that you've been asking on this site?

    Thanks!
    shihouzhuge's Avatar
    shihouzhuge Posts: 131, Reputation: 6
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    #3

    Dec 2, 2009, 09:58 PM

    Hi,jk and Clough.

    1,2,6,? 72
    The answer might be 1,2,6,12,36,72.

    But I think the question was not asked in a right way. jk,You should give your question more clearly. On the other hand, I think some homeworks should be solved by yourself, and if you are confused at it then ask here.

    Thanks!
    Unknown008's Avatar
    Unknown008 Posts: 8,076, Reputation: 723
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    #4

    Dec 3, 2009, 10:57 AM

    I can't find another possible sequence... :(

    If that was like that: (1, 2), (6, ), (? 72) then I'd have put what shihouzhuge has put.
    shihouzhuge's Avatar
    shihouzhuge Posts: 131, Reputation: 6
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    #5

    Dec 4, 2009, 02:55 AM
    Quote Originally Posted by Unknown008 View Post
    I can't find another possible sequence... :(

    If that was like that: (1, 2), (6, ?), (?, 72) then I'd have put what shihouzhuge has put.
    Unknown008.
    :), I think the regular may be *2,*3,*2,*3...
    Thanks!
    elscarta's Avatar
    elscarta Posts: 118, Reputation: 20
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    #6

    Dec 4, 2009, 06:37 AM

    Tn = T(n-1) + 3T(n-2) +1

    gives

    1, 2, 6, 13, 32, 72
    shihouzhuge's Avatar
    shihouzhuge Posts: 131, Reputation: 6
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    #7

    Dec 4, 2009, 07:12 AM
    Quote Originally Posted by elscarta View Post
    Tn = T(n-1) + 3T(n-2) +1

    gives

    1, 2, 6, 13, 32, 72
    Hi,elscarta.
    I think you give a nice answer, and the question may have many answers.
    Good job!

    Thanks!
    elscarta's Avatar
    elscarta Posts: 118, Reputation: 20
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    #8

    Dec 4, 2009, 07:48 AM

    Mathematically it is possible to find a polynomial of order 5 which gives any numbers that you want for the missing numbers.
    Unknown008's Avatar
    Unknown008 Posts: 8,076, Reputation: 723
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    #9

    Dec 4, 2009, 10:02 AM
    Quote Originally Posted by elscarta View Post
    Tn = T(n-1) + 3T(n-2) +1

    gives

    1, 2, 6, 13, 32, 72
    I would never have thought of that one... You used some program?
    elscarta's Avatar
    elscarta Posts: 118, Reputation: 20
    Junior Member
     
    #10

    Dec 4, 2009, 05:14 PM
    Quote Originally Posted by Unknown008 View Post
    I would never have thought of that one... You used some program?
    Actually it was trial and error.

    Firstly I noticed that 2 x (2 + 1) = 6 so maybe T(n) = 2x(T(n-1) + T(n-2))
    but this gives
    1, 2, 6, 16, 44, 120 which is too big so I reduced the weight of the T(n-1) term
    So I thought 2 + 4 x 1 = 6 so maybe T(n) = T(n-1) + 4x T(n-2) but this gives
    1, 2, 6, 14, 38, 94 which is still too big so I thought that there must be a constant added in not just a combination of the previous terms.
    This lead to 2 + 3 x 1 + 1 = 6 so maybe T(n) = T(n-1) + 3x T(n-2) +1 which turned out to work.
    Unknown008's Avatar
    Unknown008 Posts: 8,076, Reputation: 723
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    #11

    Dec 4, 2009, 10:49 PM

    Wonderful! Genius! :)

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