Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Next number in a sequence of randomly chosen numbers?

  1. Feb 25, 2009 #1
    if one were to randomly choose some numbers, such as the 12 below, is it always possible to determine a pattern and figure out the next number? or is this only possible if we are lucky enough to randomly choose numbers that fit a pattern (for example, if we randomly chose the numbers 1 2 3 4 5 6 7 8 9 10 11 12)

    2 6 9 23 17 36 22 24 26 32 41 43 ?

    what is the next number in the above sequence?
     
  2. jcsd
  3. Feb 25, 2009 #2

    CRGreathouse

    User Avatar
    Science Advisor
    Homework Helper

    42. There are an infinite number of sequences that, apart from some finite chunk at the beginning, have 42 as all members.
     
  4. Feb 25, 2009 #3
    What do you define to be a pattern? For example, if that happens to be the digits in an irrational number, does that count? What if the pattern is 2 6 9 23 17 36 22 24 26 32 41 43 2 6 9 23 17 36 22 24 26 32 41 43 2 6 9 23 17 36 22 24 26 32 41 43 .... etc?
     
  5. Feb 25, 2009 #4
    assuming that the nonsense you are making up is true, it's possible that the non-42 finite chunk at the beginning is larger than 12 numbers
     
  6. Feb 25, 2009 #5
    Yes, so the next number could be 3, what's your point? You have to define something that excludes "nonsense" answers to begin to solve your problem.
     
  7. Feb 25, 2009 #6
    yes, qntty, that would technically count as a pattern, but i'm not content with it

    i'd like to find a formula i can use that outputs these numbers, so that i can figure out what the next number is

    for example, if i randomly chose the numbers 2 4 6 8 10 12 14 16 18 20 22 24, i could use the formula n*2

    is there a formula i can use for 2 6 9 23 17 36 22 24 26 32 41 43?
     
  8. Feb 25, 2009 #7
    A degree n polynomial can be fitted to n+1 points, so yes that's possible. I don't know of any general method to do this for 12 points though.
     
  9. Feb 25, 2009 #8

    CRGreathouse

    User Avatar
    Science Advisor
    Homework Helper

    Sure, the one I suggested above can be coded in Pari as
    Code (Text):
    jframe(n)=if(n<1,0,[2,6,9,23,17,36,22,24,26,32,41,43,42][min(n,13)])
    Also, there's an 11th-degree polynomial that fits those points exactly.
     
  10. Feb 25, 2009 #9
    is there any way to simplify it and get a less complex formula?

    for example, if i randomly chose 1 2 3 4 5 6 7 8 9 10 11 12, i could come up with a less complex formula than an 11th-degree polynomial

    i have a feeling the answer to my original question is no. thanks for the help, though
     
  11. Feb 25, 2009 #10
    Not unless the numbers actually are 1,2,3,...,12 because the degree of a polynomial that fits it will be 11 in most cases. To make it unique perhaps you should say that the expression used to describe the nth term must be a polynomial with a minimal degree, otherwise the next term could still be anything.
     
  12. Feb 25, 2009 #11

    CRGreathouse

    User Avatar
    Science Advisor
    Homework Helper

    Ah. Now you're getting into the interesting realm of Kolmogorov complexity.

    In general, you won't be able to find a way to get a less complex formula to describe your data than the data itself. For some sequences it is possible. The troubles:
    • You must define the environment in which complexity is to be measured (e.g., Pari programs, or closed-form RPN formulas using the following characters: "012n+-*/^!") and the measure of complexity (e.g. characters).
    • Discovering the shortest formula takes exponential time.
     
  13. Feb 25, 2009 #12
    If the numbers are truly random then there are no patterns. Patterns and randomness are mutually exclusive. Even when you pick 1 2 3 4 5 6 7 8 9 10 11 12 if the source of the numbers was random then it's no more likely that the next number will be 13 than any other number.
     
  14. Feb 25, 2009 #13
    thanks, cr, that was helpful. i think i might have to surrender to this problem

    i'm not picking a random 13th number, dale. i'm looking for the simplest pattern/formula i can find in the randomly chosen 12 numbers to determine the 13th number. if i'm lucky enough to pick 1 2 3 4 5 6 7 8 9 10 11 12, then the simplest formula i can find is f(n)=n
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Similar Discussions: Next number in a sequence of randomly chosen numbers?
  1. Number sequence (Replies: 5)

  2. Number Sequences (Replies: 10)

  3. Number Sequence (Replies: 1)

Loading...