Inequality on sequence of integers whose only prime factors are $2$ or $3$

Question

Let $M=\lbrace 2^i 3^j | i,j \geq 0\rbrace$ and denote by $m_k$ the $k$-th element of $M$ ; so $m_1=1,m_2=2,m_3=3,m_4=4,m_5=6\ldots$. Is it true that $3m_k\geq 2m_{k+1}$ for every $k>1$ ?

My thoughts : If $m_k$ is not a power of $3$, then $\frac{m_k}{2}$ is an integer and is in $M$ ; the same can be said about $\frac{3m_k}{2}$. By the definition of $m_{k+1}$ we then have $m_{k+1} \geq \frac{3m_k}{2}$ as wished. So we can assume that $m_k$ is a power of $3$. Similarly we can assume that $m_{k+1}$ is a power of $2$. Then I am stuck.

Answer 1

Hint: if $m_{k+1} = 2^p$ with $p \ge 2$, $\dfrac{3}{4} m_{k+1} \in M$.

Answer 2

Consider the following cases:

• $\small2|m_k \implies \frac32m_k\in{M} \implies m_k<m_{k+1}\leq\frac32m_k \implies \color\red{3m_k\geq2m_{k+1}}$
• $\small3|m_k \implies \frac43m_k\in{M} \implies m_k<m_{k+1}\leq\frac43m_k \implies 4m_k\geq3m_{k+1} \implies \color\red{3m_k\geq}\frac94m_{k+1}>\color\red{2m_{k+1}}$