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Hi, I'm new to sagemath.
Is there any way to so calculate/solve/find a short version of a recursive defined sequence?

E.g. I have a sequence like: (Fibonacci)

def f(n):                            
        if n == 0:  
                return 0  
        if n == 1:  
                return 1
        if n == 2:
                return 1
        else:
                return f(n-1)+f(n-2)

How can I compute a short form of $f_n$?

In this example case $f_n$ would be:
$f_n=\frac{1}{\sqrt{5}} (\frac{1+\sqrt{5}}{2})^n - \frac{1}{\sqrt{5}} (\frac{1-\sqrt{5}}{2})^n$

Hi, I'm new to sagemath.
Is there any way to so calculate/solve/find a short version of a recursive defined sequence?

E.g. I have a sequence like: (Fibonacci)

def f(n):                            
        if n == 0:  
                return 0  
        if n == 1:  
                return 1
        if n == 2:
                return 1
        else:
                return f(n-1)+f(n-2)

How can I compute a short form of $f_n$?

In this example case $f_n$ would be:
$f_n=\frac{1}{\sqrt{5}} (\frac{1+\sqrt{5}}{2})^n - \frac{1}{\sqrt{5}} (\frac{1-\sqrt{5}}{2})^n$

from sympy import Function,rsolve  
from sympy.abc import n  
u = Function('u')  
f = u(n-1)+u(n-2)-u(n)  
rsolve(f, u(n), {u(0):0,u(1):1})  

-sqrt(5)*(1/2 - sqrt(5)/2)**n/5 + sqrt(5)*(1/2 + sqrt(5)/2)**n/5