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Number series

  1. Dec 16, 2008 #1
    This isnt a homework question so i decided to post it here.

    what is n=5 if n1=1663, n2=1527, n3=1126, n4=1096

    Im sure this number series is actually quite easy to get but i cant figure it out... any ideas?
     
  2. jcsd
  3. Dec 17, 2008 #2

    HallsofIvy

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    Do you understand that choosing ANY number for n5 is equally valid? It might be that there are simple formulas that will give those numbers. Is that what you mean?

    If I were looking for a simple formula, I might try subtracting the numbers, finding the difference between two consecutive numbers: 1663- 1527= 136, 1527- 1126= 401, 1126- 1096= 36. That doesn't look particularly promising. Dividing one number by the next doesn't give anything either.

    But we can always find a cubic function that will give any four numbers.

    If we try f(n)= an3+ bn2+ cn+ d, then we want f(1)= a+ b+ c+ d= 1663, f(2)= 8a+ 4b+ 2c+ d= 1527, f(3)= 27a+ 9b+ 3c+ d, f(4)= 64a+ 16b+ 8c+ d= 1096. Solve those four equations for a, b, c, and d to find f(n) inj general, then calculate f(5).

    Again, that is a possibility. There exist an infinite number of perfectly reasonable sequence that have those first four numbers.
     
  4. Dec 17, 2008 #3
    What HallsOfIvy is saying is that knowing the first few elements of a sequence is not sufficiently to determine the entire sequence.

    In math, we consider all sequences to be equally valid possibilities. That means we have to accept it even if a sequence appears to be "unnatural" or even if it was created specially to prove a point (what we would call "being a smartass" in every day language ;-)

    To choose a simplified example, take the sequence 1, 2, 3, .... What is the next number in the sequence? Well, let's look at a few sequences that begin with these three numbers.....

    * The sequence of natural numbers. The next element would be "4".
    * The sequence which repeats the elements 1, 2, and 3 repeatedly. (1, 2, 3, 1, 2, 3, 1, 2, 3, ...). The next element would be "3".
    * The sequence beginning with "1", then continuing on with all the prime numbers. The next element would be "5".

    These are three relatively simple possibilities. But there are an infinity of others.

    So even if you specified the first million elements of your sequence, there's no way to know for sure which sequence you're talking about for sure.
     
  5. Dec 18, 2008 #4

    uart

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    Here's an easy one for you KaneOris.

    Find the missing number "x" is this sequence.

    5, 10, x, 40
     
    Last edited: Dec 18, 2008
  6. Dec 18, 2008 #5
    uart, I say x = 13.

    Reason: 273·2n + 1 is prime for n = 5, 10, 13, 40 and no other numbers between them.


    Ok, so I cheated a bit, but really, x could be anything, which is what Halls and Tac-Tics were saying.
     
  7. Dec 18, 2008 #6

    uart

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    That's a good one Adrian.

    Actually I messed this one up. I thought I had a simple sequence (based on the geometry of polygons) that went "5, 10, 21, 40" but I made a mistake and the last term actually should have been 42 (which kind of messes it up) :blushing:. Now to save face I'll have to come up with another solution (other than the obvious one of 20).

    Ok here it is. The missing term is 15 and my sequence was,

    5*flloor( (n^2 + (n+1)^2)/5 ).

    Which gives the sequence 5, 10, 25, 40, 60, ....
     
  8. Dec 18, 2008 #7
    The point is you can make up a lot of numbers in that place and fit a polynomial to it. In fact if you have n points, then n-1 polynomial will fit it perfectly (as would n, n + 1, ... etc degree polynomial).
     
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