Direct way to calculate nth term of cyclically repeating function?

[ktoz]

"cyclic functions" may not be the correct term, but I don't know what else to call them. (only have basic high school math training, no calculus) Here's what I'm looking for.

Given the following series of functions

f(a) = b (step 1)
f(b) = c (step 2)
f(c) = d (step 3)
...
f(n) = a (step n)

Is there a general way to rework these so that you can directly calculate the value at steps 1,2,3 etc rather than having to iterate through all the steps from a to x, something like this?

f(0) = a
f(1) = b
f(2) = c
f(3) = d
...
f(n) = a

This problem arose from a computer program I'm working on and it would be much more efficient to directly calculate the n'th term rather than having to iterate through them all.

Thanks for any help

 

[Tide]

How about using the mod (modulo) function?

 

[ktoz]

Example?

How about using the mod (modulo) function?

Here's a little more background. Using the Mandelbrot generating function Z1 = Z0^2 + c yeilds either:

Escape set - values not in set
Chaotic set - values don't escape but also don't settle into a repeating set of values
Convergent set - values converge to either a single value or a repeating set of values

Convergent values take the form
f(a) = b
f(b) = c
f(c) = d
...
f(n) = a

With the above in mind, how would I apply your modulo suggestion?

Thanks

Ken

 

[Tide]

That's a little different than what I thought you were asking in your original post. I'm afraid you're stuck with iterating.