Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

How to convert $p_n$ to an expression in terms of $n$ if $3p_{n-1}^2 - p_{n-2}=p_{n}$ and $p_0=5, p_1=7$?

This is a problem I haven't been able to finish for two days, please help. This question haven't answered for a day, please help!

share|improve this question
6  
I don't know if there's any way to solve this, to be honest. –  anon Oct 23 '11 at 4:04
3  
Why do you think it can be solved with generating functions? And didn't someone else post this problem today or yesterday? –  Thomas Andrews Oct 23 '11 at 4:37
    
Have you tried calculating the first few terms and seeing if there's any pattern? or seeing if those terms are in the Online Encyclopedia of Integer Sequences? –  Gerry Myerson Oct 23 '11 at 5:27
    
@Thomas: it definitely wasn't "someone else". ;) I tried doing something, but I think I'll let a mod deal with this now... –  J. M. Oct 23 '11 at 5:40
1  
If I start your series with $1,3$ it is not far off $3^{(2^n-1)}$. It is only a factor $100$ smaller at $n=9$ out of $10^{243}$. I got that by ignoring the $-p_{n-2}$ part. –  Ross Millikan Oct 24 '11 at 21:17

1 Answer 1

up vote 2 down vote accepted

As said by others, there is no closed form expression of $p_n$ for general $n$. Nevertheless, one can show that, for any $1\leqslant k\leqslant n$, $$ (3p_k-1)^{2^{n-k}}\leqslant3p_n\leqslant(3p_k)^{2^{n-k}}. $$ This already indicates roughly the growth rate of $(p_n)_n$, for example in the sense that the sequence $(q_n)_n$ defined by $q_n=2^{-n}\log p_n$ is positive and bounded. But one can have more.

To wit, $2^n(q_n-q_{n-1})\to\log3$ hence $(q_n)_n$ is ultimately increasing and converges to a finite positive limit $q$. The value of $q$ might depend on the first values of $(p_n)$ but this, and other similar elementary computations, yields the existence, for every large enough positive $p_0$ and $p_1$ ($p_0\geqslant2$ and $p_1\geqslant2$ is enough), of a finite $\color{red}{\alpha(p_0,p_1)>1}$ whose value might depend on $p_0$ and $p_1$, such that, when $n\to\infty$, $$ \color{red}{p_n=3\cdot\alpha(p_0,p_1)^{2^n}\cdot(1+o(1))}. $$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.