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    msdavis26's Avatar
    msdavis26 Posts: 2, Reputation: 1
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    #1

    Jan 2, 2010, 08:17 PM
    What is the number sequence for 7,21,42,126,252,756?
    How do I find the rule for this number sequence? 7,21,42,126,252,756
    msdavis26's Avatar
    msdavis26 Posts: 2, Reputation: 1
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    #2

    Jan 2, 2010, 08:19 PM

    How do I find the rule for number sequence: 7,21,42,126,252,756
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    galactus Posts: 2,271, Reputation: 282
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    #3

    Jan 3, 2010, 09:18 AM

    The numbers are multiplied by 3, then by 2, by 3, then by 2, and so on. The next number would be 1512, then 4536, etc.
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    #4

    Jan 9, 2010, 09:10 AM

    galactus, how do you find the sequence?
    Trial and error? Look and see what works?
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    J_9 Posts: 40,298, Reputation: 5646
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    #5

    Jan 9, 2010, 09:14 AM
    Quote Originally Posted by Stratmando View Post
    galactus, how do you find the sequence?
    Trial and error? Look and see what works?
    Finding the sequence is easy.

    7 x 3 = 21

    21 x 2 + 42

    42 x 3 = 126

    It's common repetition. If you look closely at all of the numbers in the list, you can see a pattern.
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    #6

    Jan 9, 2010, 09:31 AM
    Quote Originally Posted by Stratmando View Post
    galactus, how do you find the sequence?
    Trial and error? Look and see what works?
    No, I did not use trial and error. I just 'seen' it.
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    J_9 Posts: 40,298, Reputation: 5646
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    #7

    Jan 9, 2010, 09:33 AM
    Quote Originally Posted by galactus View Post
    No, I did not use trial and error. I just 'seen' it.
    Me too, and I've been awake for ever. I guess it's just something you get used to.
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    Unknown008 Posts: 8,076, Reputation: 723
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    #8

    Jan 9, 2010, 10:23 AM

    Yep, just seeing '7', '21', and '42', we know they are multiples of 7. So, there could be some multiplication in there.
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    #9

    Jan 9, 2010, 11:37 AM

    How about a sequence like:
    1, 4, 9, 17, 34, 76, 99?
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    #10

    Jan 9, 2010, 11:39 AM
    The above sequence are random numbers, Basically, I'm asking, is their a Formula, or programs that can determine, or do you just see what works?
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    #11

    Jan 9, 2010, 11:58 AM
    Quote Originally Posted by Stratmando View Post
    How about a sequence like:
    1, 4, 9, 17, 34, 76, 99?
    Ok, I have two options, either that is not a sequence, or the formula is very complex.

    If plotted, there are no real pattern. The 'increase' decreases after 76, which is quite a rare sight to see.

    EDIT: Oops, seems that you posted before I did. I checked here, then had to give food to my dog and came back to answer.
    galactus's Avatar
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    #12

    Jan 9, 2010, 01:17 PM
    Quote Originally Posted by Stratmando View Post
    The above sequence are random numbers, Basically, I'm asking, is their a Formula, or programs that can determine, or do you just see what works?
    There is no said formula that determines the pattern of all sequences in general.

    You just see the pattern for each sequence.

    Some are easier than others.

    Did you just make that one up? It is not a sequence with any definitive pattern that I can see. That does not mean there is not one though. I am not that arrogant:):)

    One thing that can often be done is what is called 'finite differences'.

    Using this method one can find a quadratic, cubic, etc that will give the nth term of the sequence, assuming there is one.

    Let's say we have:



    Subtract between the numbers and we see a pattern:



    Keep going:





    See? We ended up with a bunch of 1's.

    Since we have 3 rows before we got to the 1's, the polynomial that represents it will be a cubic.

    I am not going to show all the work, but suffice to say it is:



    Now, we can plug in any number for x and find the number in the sequence.

    Say we want the 10th number in the sequence.



    I used a simple example. We can see these are the triangular numbers and find the next one without a polynomial, but if we wanted a number further up the ladder it is handy.

    Another sequence is called a 'recursion relation'.

    Say we had



    We can do subs and find them by performing each iteration. But, there is a way to solve it in general. Then, we just plug in to find whichever number we need in the sequence.

    We can change it to a quadratic:



    This factors:



    The roots are r=2 and 5

    So, we write

    Use the initial conditions to find b and c:





    We solve the little system:









    There is the formula. If we want the 10th number for instance, we plug in n=10



    Just a small sequence tutorial if you like.
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    #13

    Jan 9, 2010, 03:47 PM
    Strat, see if you can pick up this sequence

    2, 4, 12, 24, 72, 144
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    #14

    Jan 9, 2010, 03:52 PM
    432
    I see it is Times 2, times 3, times 2, times 3...
    I think galactus answered my question, Thanks All.
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    #15

    Jan 9, 2010, 03:53 PM
    So, you see the pattern then?

    The OP was a sequence much the same. We were just able to see the pattern.
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    #16

    Jan 9, 2010, 03:56 PM

    It wasn't I didn't see the solution, I was curious if their was some other way(math wise)

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