# Next number in a sequence of randomly chosen numbers?

#### jframe

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?

#### CRGreathouse

Homework Helper
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.

#### qntty

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?

#### jframe

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

#### qntty

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.

#### jframe

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?

#### qntty

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.

#### CRGreathouse

Homework Helper
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
Code:
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.

#### jframe

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

#### qntty

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.

#### CRGreathouse

Homework Helper
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
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.

#### DaleSwanson

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.

#### jframe

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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