Progressions and powers of 2, DeMO 2020

Here's a constructive solution: given $r,$ this shows how to construct infinitely many powers of $2$ in the sequence. In fact, I claim that (certain) powers of $2$ in the sequence correspond to zeroes in the binary decimal expansion of $1/r.$

The bounds on $r$ imply that if we write $1/r$ in binary, the decimal expansion starts with $0.1\ldots.$

We can always find a binary expansion of $1/r$ with infinitely many zeroes: if we couldn't, then the binary expansion would end with the string $01111\ldots,$ which is equivalent to the string $10000\ldots.$ Let us assume we have written the binary expansion of $1/r$ with infinitely many zeroes.

Suppose that the $(k+1)$ st digit after the decimal place in $1/r$ is $0.$ Then consider the binary expansion of $2^k/r,$ which equals an integer $n_k$ plus $.0\ldots.$ Then $\frac{2^k-1}r = \frac{2^k}r - \frac1{r}$ equals $\begin{aligned} n_k&.0\ldots \\ -0&.1\ldots \\ \hline \end{aligned}$ which is less than $n_k.$

That is, $\begin{aligned} \frac{2^k-1}r < n_k \le \frac{2^k}r \\ 2^k-1 < n_k r \le 2^k \end{aligned}$ which means that $\lceil n_k r \rceil = 2^k.$ This holds for infinitely many $k,$ so the answer is $\fbox{no}.$

First, notice that for all $n$ $a_{n+1}-a_{n}= \, 1 \quad \text{or} \quad 2 \text{} \qquad (1)$

Let us begin by assuming that there exists a real number $1<r<2$ such that the sequence $a_n$ contains only finitely many powers of $2$ . This will lead us to a contradiction. We will now consider such a $r$ .

There exists a natural number $p>3$ (we demand $>3$ since we don’t want to deal with the anomalies which might occur at very small numbers) such that for all integers $q \geq p$ , $\, a_q$ is not a power of $2$ . Now, let $k$ be the unique natural number such that $a_{k-1}<2^p<a_k$ . Because of fact $(1)$ and the definition of $a_n$ we find that $(k-1)r<2^p-1 \quad \text{and} \quad 2^p<kr<2^p+1.$ Define $0<d<1$ such that $kr=2^p+d$ . Subtracting this equation from the left inequality yields $-r<-d-1 \leftrightarrow r>d+1.$ We will proceed by showing that $r>2^{q+1}d$ for all natural numbers $q$ , which is our desired contradiction since $d>0$ is constant and $r<2$ . This will be done by induction on $q$ .

Base case $(q=0)$ : we have to show that $r>2^{0+1}d=2d$ . We have already shown that $r>1+d$ . Since $1>d$ , this immediately implies $r>1+d>d+d=2d$ .

Inductive step (showing that the statement holds for $q+1$ , assuming that it holds for $q$ ) : our definition of $d$ immediately implies $2^{q+1}kr=2^{p+(q+1)}+2^{q+1}d>2^{p+(q+1)}.$ Subtracting this from our inductive hypothesis ( $r>2^{q+1}d$ ) yields $2^{q+1}kr-r<2^{p+(q+1)}.$ Fact $(1)$ implies $2^{q+1}kr-r<2^{p+(q+1)}-1$ . Subtracting $2^{q+1}kr=2^{p+(q+1)}+2^{q+1}d$ from this implies $r>1+2^{q+1}d,$ which together with $r<2$ implies $r>2^{q+2}d$ and we are done.

First draw the graph for fn=ceil(2n) and Gn=ceil(n),we find that ,hn= 2^n, meets fn at 2 integral points n=1, n=2; which are consecutive elements of the series A2 , having no integral point in between and r<2. Hence there is no solution for r in the provided range (1,2).

This can be easily shown that for some $1<r<2$ we can either get infinitely many $2^m$ or nothing.

Suppose, we have some $1+\frac{p_0-1}{n}<r<1+\frac{p_0}{n_{0}} , p_0\in$ { ${1,2,3.....n_0}$ }, we have $\lceil {n_0}{r} \rceil =2^m$ . From, this we get $n_0+p_0=2^m$ ,

Now, if we multiply $n_0$ by $2$ ? Say, $n_1=2^k*{n_0}$ .

But, it's not mandatory that $n_1+p_1=2^{m+k}$ , for some $p_1$ . Obviously, we shall get here something important, $n_1+p_1=2^{m+k}-x, 2^{k-1}<x<2^{k}$

But, we can fine this out. There will exist infinitely many $k$ s(obviously they are large enough and shall form a sequence inductively $k<k_1<k_2<k_3<......$ ) such that we will have enough freedom to change $n_1$ to $m=n_1+l, l<2^{k}$ such that the $x$ will be vanished.

And in this way we can form infinitely many $(n,p)$ couple so that $\lceil{nr}\rceil$ becomes a power of two. So, their will be no case of finitely many...

We can see this problem logically.... The least integer function give the nearest next integer for any given number i.e $\displaystyle \lceil 8.6 \rceil = 9$

Hence $\displaystyle \lceil nr \rceil > n , \ \forall r \ \in \ (1,2)$ .

And as $\displaystyle n \ \in \ \N$ thus there cannot be confined set.

But $\displaystyle \ \forall r \ \in \ (0,1)$ we have a possiblity of having confined set giving rise to the possibility of having finite exponents of 2.

I did this way, please do ensure if it is correct.