
#1
Feb2509, 09:24 PM

P: 5

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
Feb2509, 09:35 PM

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#3
Feb2509, 09:37 PM

P: 290

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?




#4
Feb2509, 09:45 PM

P: 5

next number in a sequence of randomly chosen numbers?
assuming that the nonsense you are making up is true, it's possible that the non42 finite chunk at the beginning is larger than 12 numbers




#5
Feb2509, 09:48 PM

P: 290

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.




#6
Feb2509, 09:51 PM

P: 5

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? 



#7
Feb2509, 09:57 PM

P: 290

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.




#8
Feb2509, 09:59 PM

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#9
Feb2509, 10:10 PM

P: 5

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 11thdegree polynomial i have a feeling the answer to my original question is no. thanks for the help, though 



#10
Feb2509, 10:11 PM

P: 290

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.




#11
Feb2509, 10:16 PM

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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:




#12
Feb2509, 11:11 PM

P: 351

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.




#13
Feb2509, 11:15 PM

P: 5

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 


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