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Math Help - Need help figuring out this sequence...

  1. #1
    snakeyster
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    Need help figuring out this sequence...

    Basicaly, I need help on this sequence, as my homework is due in tomorrow. One question on sequences is:

    1
    3
    11
    67

    I'm supposed to find out what comes next. Can anyone help? Even if you tell me how you got the answer, that would be a great help
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by snakeyster View Post
    Basicaly, I need help on this sequence, as my homework is due in tomorrow. One question on sequences is:

    1
    3
    11
    67

    I'm supposed to find out what comes next. Can anyone help? Even if you tell me how you got the answer, that would be a great help
    Since the online encyclopedia of integer sequences knows this not, I would
    suggest that this is not a mathematicaly defined sequence but is somehow
    connected with the site you got it from

    RonL
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    i think a_n = 7n^3 - 39n^2 + 70n - 37 for n \in \mathbb {N} works

    thus, the fifth term is: 213


    (i found the cubic that hit the points (1,1), (2,3), (3,11) and (4,67) by solving a system of equations)
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  4. #4
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    Hello, snakeyster!

    I need help on this sequence: . 1,\,3,\,11,\,67,\,\cdots
    I see a pattern . . .

    . . \begin{array}{ccccc} a_1 & = & 1 & & \\<br />
a_2 & = & {\color{red}2}(a_1)\,{\color{red}+\,1} & = & 3 \\<br />
a_3 & = & {\color{red}4}(a_2)\,{\color{red}-\,1} & = & 11 \\<br />
a_4 & = & {\color{red}6}(a_3)\,{\color{red}+\,1} & = & 67\end{array}

    I bet the next one is:

    . . \begin{array}{ccccc} a_5 & = & {\color{red}8}(a_4)\,{\color{red}-\,1} & = & 535\end{array}


    The general term seems to be: . a_n\;=\;2(n-1)(a_{n\text{-}1}) + (\text{-}1)^n . for  n \geq 2

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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Soroban View Post
    Hello, snakeyster!

    I see a pattern . . .

    . . \begin{array}{ccccc} a_1 & = & 1 & & \\<br />
a_2 & = & {\color{red}2}(a_1)\,{\color{red}+\,1} & = & 3 \\<br />
a_3 & = & {\color{red}4}(a_2)\,{\color{red}-\,1} & = & 11 \\<br />
a_4 & = & {\color{red}6}(a_3)\,{\color{red}+\,1} & = & 67\end{array}

    I bet the next one is:

    . . \begin{array}{ccccc} a_5 & = & {\color{red}8}(a_4)\,{\color{red}-\,1} & = & 535\end{array}


    The general term seems to be: . a_n\;=\;2(n-1)(a_{n\text{-}1}) + (\text{-}1)^n . for  n \geq 2

    hmm, we got different answers. i guess i made a mistake then. oh well, good job! I don't know how you see these pattens
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  6. #6
    Grand Panjandrum
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    Quote Originally Posted by Jhevon View Post
    hmm, we got different answers. i guess i made a mistake then. oh well, good job! I don't know how you see these pattens
    You made no mistake, one can chose the next term to have an arbitary
    value and find a polynomial that will take all the given values and our arbitary
    next value.

    Your cubic does give the first four terms, and so is as valid a rule for the
    next term as any other, and I suspect it involves no more arbitary constants
    than does Soroban's

    RonL
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  7. #7
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Jhevon View Post
    hmm, we got different answers. i guess i made a mistake then. oh well, good job! I don't know how you see these pattens
    You can find an infinite number of polynomial fits to a finite data set. So the fact that you and Soroban came up with different answers is not a surprise.

    But I'm with you. I'd love to know how he comes up with this stuff!

    -Dan
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    You made no mistake, one can chose the next term to have an arbitary
    value and find a polynomial that will take all the given values and our arbitary
    next value.

    Your cubic does give the first four terms, and so is as valid a rule for the
    next term as any other, and I suspect it involves no more arbitary constants
    than does Soroban's

    RonL
    Quote Originally Posted by topsquark View Post
    You can find an infinite number of polynomial fits to a finite data set. So the fact that you and Soroban came up with different answers is not a surprise.

    But I'm with you. I'd love to know how he comes up with this stuff!

    -Dan
    Pfft! sequences are so weird!
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  9. #9
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    HellO!

    RonL and Dan are, of course, absolutely correct.

    A sequence of numbers is meaningless unless there is a stipulation,
    . . a promise that that the sequence "continues in a similar manner",
    . . that there is indeed a "reasonable" pattern to the terms.

    I've posted the following before, but I enjoy showing it off.


    Find the next term of the sequence: . \bf{1,\,3,\,5,\,7,\,\cdots}

    Answer: 8


    Justification
    I was using this function: . f(n) \;=\;-\frac{1}{24}\left(n^4 - 10n^3 + 35n^2 - 98x + 48\right)

    "That's unfair!" you say?
    Okay, here's a better explanation . . .

    The sequence: . 1,\,3,\,5,\,7,\,8,\,\cdots is the sequence of natural numbers
    . . whose English names contain the letter "e".

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  10. #10
    Grand Panjandrum
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    Quote Originally Posted by Soroban View Post
    HellO!

    RonL and Dan are, of course, absolutely correct.

    A sequence of numbers is meaningless unless there is a stipulation,
    . . a promise that that the sequence "continues in a similar manner",
    . . that there is indeed a "reasonable" pattern to the terms.

    I've posted the following before, but I enjoy showing it off.


    Find the next term of the sequence: . \bf{1,\,3,\,5,\,7,\,\cdots}

    Answer: 8


    Justification
    I was using this function: . f(n) \;=\;-\frac{1}{24}\left(n^4 - 10n^3 + 35n^2 - 98x + 48\right)

    "That's unfair!" you say?
    Okay, here's a better explanation . . .

    The sequence: . 1,\,3,\,5,\,7,\,8,\,\cdots is the sequence of natural numbers
    . . whose English names contain the letter "e".
    My favourite is: find the next term in 1, 2, 4, 8, 16.

    The answer that I am thinking of is 31. See here in particular scroll down to sequence A000127.

    RonL
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