Question

How to express a recurrent sequence in a form of algebraic function?

Answer 1

I think i understand what you are looking for. You want a **closed form(include this phrase in your question)** of the given recurrence relation.

That is : a\_n = f(n);  (some function of n)

 Now not every recurrence relation has a closed form. Also here, its a non-linear one. I really don't think it's possible to get a closed form.

Answer 2

You don't need a closed form expression to graph the function. In order to graph the function, you would have to evaluate the function at various values of n, which you can easily do with the recursive definition.  

To get an idea what the closed form expression would look like, we can start by assuming the first term is 1. We will then get the sequence 1, 3, 11, 123, 15132 and so on. At this point, the +2 is negligible and you would get an expression that is approximately something to the power of 2^x.

Answer 3

What you have there is already a function. To evaluate the x^(th) iteration, plug in the (x-1)^(th) result, square it and add 2. Sometimes it *is* that easy!

Here's something you might not have known, though. If you let the constant term be anything (instead of just 2), then what you have is called the Mandelbrot set — hang on, no, the Mandelbrot set is the set of numbers for which that expression remains bounded (doesn't fly off to plus or minus infinity). It's a big thing in the complex number world!

Answer 4

What about this answer on stackexchange?

>Depending on the initial conditions, this sequence may be oeis.org/A102847 or oeis.org/A072191

Answer 5

pretty hard to find closed form. i was thinking try use discrete fourier transform and find dft a(n-1) ^2  by inverse ft convolution (a(n-1) * a(n-1))?

Answer 6

You're taking the square of the previous output here, and adding it by two.

f(x) = ((x-1)\^2 + 2)\^2 + 2

A recurrent sequence is already an algorithm, as is a function.  It's just spraying a fresh coat of paint on it really.