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

what is the next number in the above sequence?

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

what is the next number in the above sequence?

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42. There are an infinite number of sequences that, apart from some finite chunk at the beginning, have 42 as all members.what is the next number in the above sequence?

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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?

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Sure, the one I suggested above can be coded in Pari asis there a formula i can use for 2 6 9 23 17 36 22 24 26 32 41 43?

Code:

`jframe(n)=if(n<1,0,[2,6,9,23,17,36,22,24,26,32,41,43,42][min(n,13)])`

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

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.

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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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