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?

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- Apr 1st 2013, 04:27 PMmikewienermFinding 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? - Apr 1st 2013, 07:01 PMGusbobRe: Finding Formula for given sequence
- Apr 2nd 2013, 05:38 AMmikewienermRe: 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.

- Apr 2nd 2013, 05:46 AMGusbobRe: Fibonacci Sequence Formula
After the first term, this sequence is

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

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

for $\displaystyle n\geq 2$ - Apr 2nd 2013, 06:08 AMHallsofIvyRe: 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 $\displaystyle F_n= r^n$, the equation becomes $\displaystyle r^n= r^{n-1}+ r^{n-2}$. Dividing through by $\displaystyle r^{n-2}$ gives you $\displaystyle r^2= r- 1$ or $\displaystyle r^2- r+ 1= 0$. Solving that $\displaystyle r= \frac{1\pm\sqrt{5}}{2}$. That is, $\displaystyle F_n= A\left(\frac{1+\sqrt{6}}{2}\right)^n+ B\left(\frac{1- \sqrt{6}}{2}\right)^n$ for some constants A and B. That is exactly the solution you would get for the "usual" Fibonnacci formula. Putting $\displaystyle F_1= A\frac{1+ \sqrt{5}}{2}+ B\frac{1- \sqrt{6}}{2}= 0$ and $\displaystyle A\left(\frac{1+ \sqrt{5}}{2}\right)^2+ B\left(\frac{1- \sqrt{5}}{2}\right)^2= 2$ rather than setting them equal to the "usual" 1, 1, solve for A and b. - Apr 6th 2013, 04:10 AMisparksRe: 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