# Find all positive integer numbers $a_1$ such that $a_n$ is a integer number for all $n \in \mathbb{Z^+}$

Let $$m$$ be a positive integer number and $$\left(a_n\right)$$ be a sequence such that $$a_1\in \mathbb{Z}^+$$ and $$a_{n+1}=\left\{\begin{matrix} a_n^2 +2^m \quad &\text{if} \quad a_n<2^m,\\ \frac{a_n}{2} &\text{otherwise}. \end{matrix}\right.$$ For each $$m\in \mathbb{Z}^+$$, find all positive integer numbers $$a_1$$ such that $$a_n$$ is a integer number for all $$n \in \mathbb{Z^+}$$.

First I assume that $$a_1 \geq 2^m$$ and assume $$a_1=2^kl$$, where $$l$$ is an odd number. Then $$l$$ must be less than $$2^m$$ since if not, then $$a_{k+1}=l$$ and $$a_{k+2}$$ is not an integer number. And since the assumption $$a_1 \geq 2^m$$, there must exist an $$h$$ such that $$a_{h+1}=2^{k-h}l<2^m$$ and $$a_h=2^{k-h+1}l >2^m$$. Then $$a_{h+2}=2^{k-h}l+2^m$$ and $$a_{h+3}=2^{k-h-1}l+2^{m-1}$$. Since $$2^{k-h}l<2^m$$, we see that $$a_{h+3}<2^{m-1}+2^{m-1}=2^m$$ and therefore $$a_{h+4}=2^{k-h-1}l+2^{m-1}+2^m$$. Then $$a_{h+5}=2^{k-h-2}l+2^{m-2}+2^{m-1}<2^{m-2}+2^{m-2}+2^{m-1}=2^m$$. Going on the process till $$k-h-1=0$$ yields the next term will not be an integer. This implies that $$l=0$$.
I will show that if $$a_1=2^m$$, then $$a_n$$ is a integer number. Indeed, $$a_2=2^{m-1}, a_3=2^{m-1}+2^m, a_4=2^{m-2}+2^{m-1}, a_5=2^{m-2}+2^{m-1}+2^m$$. We obtain by induction that $$a_{2k+1}=2^m+2^{m-1}+...+2^{m-k}$$ and $$a_{2k}=2^{m-1}+...+2^{m-k}$$.
The case $$a_1=2^k$$ with $$k>m$$ will done when $$a_{k-m+1}=2^m$$
The case $$a_1<2^m$$ will turn into the first case that $$a_2 \geq 2^m$$ and $$a_2=2^kl$$ with an odd number $$l$$.
This completes the proof.

Am I on the right track? It may be quite "hand-crafted". Help me if there is a mistake or a better way. Thank you.

• >We obtain by induction that $a_{2k+1}=2^m+2^m−1+...+2^{m−k}$ and $a_{2k}=2^{m−1}+...+2^{m−k}$.  And for sufficiently large k those aren't integers. – liaombro May 12 at 18:59
• I don't get it the formula is $a_{n+1}=a_n^2+2^m$ why are you using $a_{n+1}=a_n+2^m$ ? – Julian Mejia May 12 at 19:41
• It's a mistake. I forgot the formula. – RuaSun May 13 at 1:16

You should state what you are trying to prove. It appears the claim is that the only $$a_1$$ that works is $$2^m$$.
$$l$$ cannot be zero because then $$a_1=0$$, which is not allowed.
If you find any $$a_1$$ that is greater than or equal to $$2^{m-1}$$ you can multiply it by any power of $$2$$ and have a new $$a_1$$ that works. In particular, if $$2^m$$ works, so does any higher power of $$2$$.
$$2^m$$ does not work. $$a_2=2^{m-1}$$, then $$a_3=3\cdot 2^{m-1}$$ and every two iterations will remove a power of $$2$$ until you get to an odd number greater than $$2^m$$ and fail. For example, if $$m=4$$, the sequence is $$16,8,24,12,28,14,30,15,31,31/2$$