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?
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 $\displaystyle \geq 2$. I.e. if the fibonacci sequence is $\displaystyle F_1,F_2,...$ and your sequence here is $\displaystyle T_1,T_2,...$, set
$\displaystyle T_{n}=F_{n+2}=\frac{\varphi^{n+2}-\psi^{n+2}}{\sqrt5}$
for $\displaystyle n\geq 2$
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.