1. ## Finding Formula for given sequence

I have the following sequence: 0 2 3 5 8 13 21 34..........
This is very close to the fibonacci sequence, but not exactly. Any ideas?

2. ## Re: Finding Formula for given sequence

Originally Posted by mikewienerm
I have the following sequence: 0 2 3 5 8 13 21 34..........
This is very close to the fibonacci sequence, but not exactly. Any ideas?
$F_0 = 0, F_1=2, F_3=3, F_n=F_{n-1}+F_{n-2}\quad\forall n\geq 3$

3. ## Re: Finding Formula for given sequence

This is a recursive formula. I was hoping someone could help me find a forumla where I can find F(n) without having to find the previous values. For example, finding F(50) would be very tedious.

4. ## Re: Fibonacci Sequence Formula

After the first term, this sequence is exactly the same as the Fibonacci sequence (with the first three terms truncated). Surely you can use the same formula to calculate terms $\geq 2$. I.e. if the fibonacci sequence is $F_1,F_2,...$ and your sequence here is $T_1,T_2,...$, set

$T_{n}=F_{n+2}=\frac{\varphi^{n+2}-\psi^{n+2}}{\sqrt5}$

for $n\geq 2$

5. ## Re: Fibonacci Sequence Formula

What Gusbob said, of course!

But also you can apply the same ideas that lead to the "usual" Fibonacci formula. If you "look for" a solution of the form $F_n= r^n$, the equation becomes $r^n= r^{n-1}+ r^{n-2}$. Dividing through by $r^{n-2}$ gives you $r^2= r- 1$ or $r^2- r+ 1= 0$. Solving that $r= \frac{1\pm\sqrt{5}}{2}$. That is, $F_n= A\left(\frac{1+\sqrt{6}}{2}\right)^n+ B\left(\frac{1- \sqrt{6}}{2}\right)^n$ for some constants A and B. That is exactly the solution you would get for the "usual" Fibonnacci formula. Putting $F_1= A\frac{1+ \sqrt{5}}{2}+ B\frac{1- \sqrt{6}}{2}= 0$ and $A\left(\frac{1+ \sqrt{5}}{2}\right)^2+ B\left(\frac{1- \sqrt{5}}{2}\right)^2= 2$ rather than setting them equal to the "usual" 1, 1, solve for A and b.

6. ## Re: Fibonacci Sequence Formula

Let G represent the Golden Ratio
Then Fn=(G^n+(G-1)^n))/Sqrt 5
Where ^ represents is to the power of
Another method is to use the Binomial Expansion by observing Pascal Triangle