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Math Help - Number pattern.

  1. #1
    Member integral's Avatar
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    Number pattern.

    1,1,2,4,16,128,4096

    What are the next 3 terms.
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  2. #2
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    the next 2 are 1048576 and 8589934592 I will let someone else reveal the third
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  3. #3
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    Next 3:

    18014398509481984, 309485009821345068724781056, 11150372599265311570767859136324180752990208
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  4. #4
    Member integral's Avatar
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    I would have thought that would have taken some time. It would have for me .

    Please tell me how you can find patterns so quick? Or is it just, your that good at pattern recognition? Thanks
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  5. #5
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    Quote Originally Posted by integral View Post

    Please tell me how you can find patterns so quick?
    This one was quite easy. This is what to do. Write the numbers down and start looking at any relationships between consecutive numbers.

    In this case look at the ratios between each consecutive numbers. For the next term consider the relationship between the previous two ratios.

    1 {\color{red}\times 1 = } , 1 {\color{red}\times 2 = } , 2 {\color{red}\times 2 = } , 4 {\color{red}\times 4 = } , 16 {\color{red}\times 8 = } , 128 {\color{red} \times 32 = } ,\dots
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  6. #6
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    Quote Originally Posted by integral View Post
    I would have thought that would have taken some time. It would have for me .

    Please tell me how you can find patterns so quick? Or is it just, your that good at pattern recognition? Thanks
    The powers of two jumped out at me pretty quickly.

    But just look at each term divided by the preceding term and you should see the pattern quickly too.

    Edit: Forgot to refresh the page before posting, so my "hint" is pretty unnecessary after pickslides's post.
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  7. #7
    Member integral's Avatar
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    Quote Originally Posted by pickslides View Post
    This one was quite easy. This is what to do. Write the numbers down and start looking at any relationships between consecutive numbers.

    In this case look at the ratios between each consecutive numbers. For the next term consider the relationship between the previous two ratios.

    1 {\color{red}\times 1 = } , 1 {\color{red}\times 2 = } , 2 {\color{red}\times 2 = } , 4 {\color{red}\times 4 = } , 16 {\color{red}\times 8 = } , 128 {\color{red} \times 32 = } ,\dots
    Ohrly? Interesting.

    I was playing around to find a number pattern so I started with pascals triangle.
    <br />
\binom{n}{k}

    then I added up each row, but not all rows. I chose a row correlating to a financing number subtracted by 1.

    so

    <br />
\binom{(F_{n}+F_{n-1})-1}{k}

    which gives

    \binom{1-1}{k}
    \binom{1-1}{k}
    \binom{2-1}{k}
    \binom{3-1}{k}

    ect.

    But I wanted the sum so.


    \sum^{F_n-1}_{k}\binom{F_n-1}{k}
    \left [ \sum^{0}_{k}\binom{0}{k} \right ]=1
    \left [ \sum^{0}_{k}\binom{0}{k} \right ]=1
    \left [\sum^{1}_{k}\binom{1}{k}\right ]=2
    \left [ \sum^{2}_{k}\binom{2}{k}\right ]=4
    \left [ \sum^{4}_{k}\binom{4}{k}\right ]=16
    \left [ \sum^{7}_{k}\binom{7}{k}\right ]=128


    While your method works yes (although I did not think of it)


    It does not explain the first term, only all after.

    1,1,2,4,16

    So, you're answers work but are technically incorrect.

    <br />
\int
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  8. #8
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by integral View Post
    So, you're answers work but are technically incorrect.
    I think the answers are just as correct as yours.

    The pattern stated in pickslides's post requires the first three values to be given, just like how the Fibonacci sequence requires the first two values, as base cases for recursion.
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  9. #9
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    Quote Originally Posted by integral View Post
    It does not explain the first term, only all after.
    No need to; the 1st term here is really the "2"; the two 1's are to set up;
    same as with fibonacci sequence's set up with 0 and 1.
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  10. #10
    Member integral's Avatar
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    I suppose, indeed.

    Interesting non the less
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  11. #11
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    Look at the power of the pattern

     0,0,1,2,4,7,12 ....

    If we add one in each term

    we have

    1 , 1 , 2,3,5,8,13 ...


    Fibonacci sequence
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  12. #12
    Member integral's Avatar
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    Quote Originally Posted by simplependulum View Post
    Look at the power of the pattern

     0,0,1,2,4,7,12 ....

    If we add one in each term

    we have

    1 , 1 , 2,3,5,8,13 ...


    Fibonacci sequence
    What is the pattern of? (the first one not the fibonacci.)
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  13. #13
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    Quote Originally Posted by integral View Post
    What is the pattern of? (the first one not the fibonacci.)
    Powers of 2.

     2^0 = 1
     2^0 = 1
     2^1 = 2
     2^2 = 4
     2^4 = 16
     2^7 = 128

    etc.
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  14. #14
    Member integral's Avatar
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    I see, so any sequence you place at the top of your binomial coefficient will be two to the power of that sequence?

    awesome!
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