If $a_1 = 2$, $a_2 = 7$, and $a_{n+1} = \frac{a^2_n}{a_{n-1}}$ rounded to the nearest integer, then every $a_n$ for $n\ge2$ is odd

Question

A sequence ${a_i}$ is defined as follows: $a_1 = 2, a_2 = 7,$ $a_{n+1} =\left( \frac{a^2_n}{a_{n-1}} ~\text{rounded to the nearest integer}\right)$

For example, $a_3 ~\text{is}~ 49/2 = ~\text{round}(24.5), \text{or}~ 25$. Prove, for all $i\geq2$, that $a_i$ is always an odd number.

First we find the first few numbers of $a_i$

$a_1=2 , a_2=7$

$$ a_{2+1} = \frac{a_2^2}{a_{2-1}}=\frac{49}{2}=25$$

$$ a_{3+1}=\frac{25^2}{7}=89$$

$$ a_{4+1}=\frac{89^2}{25}=317$$

$$ a_{5+1}=\frac{317^2}{89}=1129$$

These are clearly all odd numbers however how do we prove that for all $i \geq 2$ they will be odd

Answer 1

Let's have a look at this website : Automated Proof (or Disproof) of Linear Recurrences Satisfied by Pisot Sequences

And in particular the second entry of their first output.txt file.

It shows that the recurrence relation is verified : $a_{n+1}=3a_n+2a_{n-1}$.

Thus since $a_{n+1}$ as same parity than $a_{n}$ for $n\ge2$, then it is the the same as $a_2=7$, thus odd.

---

Here is a transcript their proof

Let $(b_n)_n$ be the sequence defined by the recurrence relation $\begin{cases} b_1=2 \\ b_2=7 \\ b_{n+1}=3b_n+2b_{n-1} \end{cases}$

Whose characteric equation is $r^2-3r-2=0$ with $\Delta=17$ and roots $\lambda=\frac{3+\sqrt 17}{2}$ and $\mu=\frac{3-\sqrt 17}{2}$.

$\lambda\simeq3.561552813$ with $|\lambda|>1$.

$\mu\simeq-0.561552813$ with $|\mu|<1$ in particular $\mu^n\to 0$.

We will show that $(b_n)_n$ satisfies the relation verified by $(a_n)_n$ which is $\displaystyle{b_{n+2}=\lfloor\frac12+\frac{b_{n+1}^2}{b_n}\rfloor}$

If we set $\displaystyle{c_n=\frac{(b_{n+1})^2-b_{n+2}\,b_n}{b_n}}$ then we have to prove the equivalent $-\frac12\le c_n\lt\frac12$.

Yet $|c_n|\le A\ |\mu|^n$

for some fixed constant $A$ (independent of n), that can be easily determined, if desired. In particular the sequence $(c_n)_n$ tends to $0$.

The first calculated $c_n$ have values

-0.5000000000, 0.2857142857, -0.1600000000, 0.08988764045, 
-0.05047318612, 0.02834366696, -0.01591643870, 0.008937923329,     
-0.005019115773, 0.002818498599, -0.001582735814, ...

and as $n$ goes to infinity, $c_n$ tends to zero (and it is possible (but a waste of time) to find an $N_0$ such that $|c_n|$ is guaranteed to be less than $\frac12$ for $n\ge N_0$ (or any epsilon for that matter), and check that for $n\le N_0$ then $-\frac12\le c_n\lt\frac12$. QED.

---

Note: some explanations

$b_n=u\lambda^n+v\mu^n$ for constants $(u,v)$ that can be determined in fonction of $b_2$ and $b_1$ as for any linear recurrence equation of order 2 with distinct roots.

$\require{cancel}(b_{n+1})^2-b_nb_{n+2}=(u\lambda^{n+1}+v\mu^{n+1})^2-(u\lambda^{n+2}+v\mu^{n+2})(u\lambda^{n}+v\mu^{n})=(\cancel{u^2\lambda^{2n+2}}+\cancel{v^2\mu^{2n+2}}+2uv\lambda^{n+1}\mu^{n+1})-(\cancel{u^2\lambda^{n+2}}+uv\lambda^{n+2}\mu^{n}+uv\lambda^{n}\mu^{n+2}+\cancel{v^2\mu^{n+2}})=uv\lambda^{n}\mu^{n}(2\lambda\mu-\lambda^2-\mu^2)=-uv\lambda^{n}\mu^{n}(\lambda-\mu)^2$

$|b_{n+1}^2-b_nb_{n+2}|=|uv(\lambda-\mu)^2|\lambda|^{n}|\mu|^{n}$

$|b_n|=|u\lambda^{n}+v\mu^{n}|=|u||\lambda|^{n}|1+\frac vu(\frac{\mu}{\lambda})^{n}|\ge|u||\lambda|^{n}|1-|\frac vu||$ since $|\frac{\mu}{\lambda}|^n<1$.

rem: if $u=v$ we can modify this $|1-|v/u||$ by $|1-|v\mu/u\lambda||$

Thus $|c_n|\le \frac{|v(\lambda-\mu)^2|}{|1-|\frac vu||}|\mu|^{n}\le A|\mu|^n$ as stated in their proof.

Then we just have to manually check that for the corresponding $N_0$ the values of $(a_n)_{n\le N_0}$ agrees with the recurrence relation.