Finding a general form of a recurrence We have that $T_n^2-T_n+1=T_{n+1}$ for n greater than $1$ ($n$ is an integer)
And we have to find the general form of the series.Also $T_2=2.$
Im new to this topic so i tried substitution but failed,any help would be appreciated.Thanks in advance.
 A: Ahh well that is a good question.
You just have to find the values of some of the first terms
As $T_2$ = 2
You can easily find the values of
$T_3$ = 3
$T_4$ = 7
$T_5$ = 43
As this here at first you would not see any pattern but I have been working on something like this so i recognise this pattern.
It is as follows that you have to multiply the previous term amd add 1 to it to get the next term.
For example
$T_4$ =($T_2$) ($T_3$) + 1
Or I guess if you notice it is a way of finding new primes as suggested by Euclid about the infinite primes his proof include this reccurence of multiplying the primes and adding one to it gives you a new prime.
At last to state there is no general formula for this reccurence as it is dealing with primes I don't think it should but if you got some please do let me know
A: I suggest $T_1=2$ instead of $T_2=2$. Then this is exactly the Sylvester's Sequence. It doesn't have a nice closed form, but one can claim that
$$
T_n=\left [c^{2^{n+1}} + 1/2\right]
$$
More information here.
