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Thread: Sequences

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    Sequences

    Here are the first six numbers in three number sequences. In each case, give me the seventh number in the sequence...

    1) 1, 7, 20, 44, 81, 135
    2) 2499, 2484, 2419, 2244, 1875, 1204
    3) 135, 234, 371, 552, 783, 1070
    Waggers
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    Re: Sequences

    1) <span style="background-color: #FFFF00; color: #FFFF00; font-weight: bold">208</span hide>

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    Re: Sequences

    3) <span style="background-color: #FFFF00; color: #FFFF00; font-weight: bold">1419</span hide>

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    Re: Sequences

    2) <span style="background-color: #FFFF00; color: #FFFF00; font-weight: bold">99</span hide>

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    Re: Sequences

    <span style="background-color: #FFFF00; color: #FFFF00; font-weight: bold">Polynomial sequences are relatively easy to crack:
    - List the numbers.
    - List their differences.
    - List the differences of the differences.
    - Etc., until you arrive at constant values.
    - You can then easily go back up, constructing the next item in each list.</span hide>

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    Re: Sequences

    I'm interested in your solving technique - could you elaborate?

    <span style="background-color: #FFFF00; color: #FFFF00; font-weight: bold">Taking the first puzzle as an example, you said to list the differences of the values and then the differences of the diffeences, etc...
    So the puzzle has 6 givens: 1,7,20,44,81,135.
    The differences are: 6,13,24,37,54.
    Differences of differences are: 7,11,13,17.
    Then 4,2,4</span hide>

    Where now?

    <img src=/S/cheers.gif border=0 alt=cheers width=30 height=16>
    - Ricky

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    Re: Sequences

    <span style="background-color: #FFFF00; color: #FFFF00; font-weight: bold">I think Hans then assumes that the 4,2,4 is a recurring pattern, making the next number 2.
    You then reverse the process, giving the next difference of the differences as 19,
    the next difference as 73
    and so the next number in the sequence is 208</span hide>

    Of course, using this method would be harder if I'd only given 3 or 4 numbers in the original sequence.
    Waggers
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    Re: Sequences

    <span style="background-color: #FFFF00; color: #FFFF00; font-weight: bold">Actually, the first one cannot be n^2 - (n-1) as Dave claims.
    In this one, I assumed that 7,11,13,17 was a series of prime numbers, so that the next one would be 19.</span hide>

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    Re: Sequences

    Hans is correct, of course. The actual formula I used for the first series is:
    <span style="background-color: #FFFF00; color: #FFFF00; font-weight: bold">f(n) = n^3 - f(n-1) (for n>1 and assuming that f(0)=0)

    The link to prime numbers is interesting though, and works for f(8) as well. However, the sequences diverge at f(9).
    Using HansV's method, f(9) would be 429, but my f(9) is 425.

    So not only did I publish the wrong answer, but there's more than one correct answer anyway! Sorry...</span hide>
    Waggers
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    Re: Sequences

    <img src=/S/shrug.gif border=0 alt=shrug width=39 height=15>
    Ok, so it was too easy...
    Well done, <img src=/S/hansv.gif border=0 alt=HansV width=27 height=26>

    <span style="background-color: #FFFF00; color: #FFFF00; font-weight: bold">
    Sequence 1: f(n) = (n^2) - (n-1) - THIS IS WRONG! see <post#=505,520>post 505,520</post: > for correct formula
    Sequence 2: f(n) = 2500 - (n^4)
    Sequence 3: f(n) = n + (n+2)^2 + (n+4)^3
    </span hide>
    Waggers
    If at first you do succeed, you&#39;ve probably missed something.

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