Do such unique representations of positive integers exist?

Question

It is well known that every positive integer $n>0$ can be represented uniquely in the form $$ n=2^k(2m+1), $$ for positive integers $k,m\geq0$. Does there exist one or more constants $c>1$ such that $$ 2^k(2m+c) $$ is a unique representation for positive integers greater than some lower bound?

Answer 1

For $c=3$ the expression is unique, though you can not get a lot of numbers this way. For example, no power of $2$ can be written this way.

To see that, for the numbers which can be expressed this way, the expression is unique: Just remark that $$2^{k_1}(2m_1+3)=2^{k_2}(2m_2+3)\implies k_1=k_2\implies 2m_1+3=2m_2+3\implies m_1=m_2$$

A similar argument goes through for any odd $c$. For even $c$ the argument fails since we can not conclude that $k_1=k_2$.

Answer 2

Assume $c$ is even. Then $$2^k(2m+c)=2^{k+1}\left(m+\frac{c}{2}\right)$$ Now assume $k>0$ and $m$ are fixed, so they represent a certain fixed number. $$2^{k+1}\left(m+\frac{c}{2}\right)=2^k\left(2m+c\right)=2^{k'+1}\left(m'+\frac{c}{2}\right)$$ where $k'=k-1$ and $m'=2m+\frac{c}{2}$. So the representation is not unique. It's also clear that it doesn't include odd numbers, since $2^{k+1}\mid n$. However, you can trivially represent any even number $n\geq c$ by setting $k=0$ and having $m=\frac{n-c}{2}$.

Now assume $c>1$ is odd. If $n=2^k(2m+c)$, then $2m+c$ is an odd number bigger than $1$ that divides $n$, which therefore can't be a power of $2$, hence your representation doesn't include these. However it's still certainly unique, by the same argument that lulu gave. The first implication only relies on the fact that $2m+c$ is odd.