# Next number in a sequence of randomly chosen numbers?

by jframe
Tags: chosen, number, numbers, randomly, sequence
 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?
HW Helper
P: 3,684
 Quote by jframe what is the next number in the above sequence?
42. There are an infinite number of sequences that, apart from some finite chunk at the beginning, have 42 as all members.
 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?
 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 non-42 finite chunk at the beginning is larger than 12 numbers
 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.
 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?
 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.
HW Helper
P: 3,684
 Quote by jframe is there a formula i can use for 2 6 9 23 17 36 22 24 26 32 41 43?
Sure, the one I suggested above can be coded in Pari as
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.
 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 11th-degree polynomial i have a feeling the answer to my original question is no. thanks for the help, though
 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.