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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.

Community
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Math enthusiast
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2 Answers2

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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

Ankit Kumar
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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.

FFjet
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  • What is c here?? A constant?? – Math enthusiast Dec 24 '20 at 07:21
  • Yeah, a constant. See here. It is also described in the link in my answer. – FFjet Dec 24 '20 at 07:22
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    Its a bit hard to read the site.....its format is bit strange....i would be greatly thankful if you could type the thing on that site here.....i know its inconvenient but it would be great if you could consider doing it. – Math enthusiast Dec 24 '20 at 07:38
  • I'll do that if I have time. – FFjet Dec 24 '20 at 08:01
  • the two quadratic recurrences that work out well are $x_{n+1} = x_n^2$ and $x_{n+1} = x_n^2 - 2,$ then shifts of these and so on. The bad news here is that completing the square gives $(x- \frac{1}{2})^2 + \frac{3}{4}$ – Will Jagy Dec 24 '20 at 20:33