Induct, If it is even add one to every $n_1$ in the representation of $\frac{n}{2}$, if it is odd remove $3^{\lfloor \log_3(n)\rfloor}$ and add one to every $n_1$ in $\frac{n-3^{\lfloor \log_3(n)\rfloor}}{2}$.

gghx wrote: Induct, If it is even add one to every $n_1$ in the representation of $\frac{n}{2}$, if it is odd remove $3^{\lfloor \log_3(n)\rfloor}$ and add one to every $n_1$ in $\frac{n-3^{\lfloor \log_3(n)\rfloor}}{2}$. What if there is a $3^{\lfloor \log_3(n)\rfloor}$ in the representation of $\frac{n-3^{\lfloor \log_3(n)\rfloor}}{2}$

This is a classic. I prove it by strong induction. Let $n=2k$ and assume all numbers $1,2,\dots,2k-1$ have such a representation. Letting $k=\textstyle \sum_{1\le i\le k}3^{m_i}2^{n_i}$, we get $2k= \textstyle \sum_{1\le i\le k}3^{m_i}2^{n_i+1}$. Assume now $n$ is odd. If $n$ is a power of $3$, we are done, so assume $a$ is such that $3^a<n<3^{a+1}$. Note that $n-3^a$ is even, less than $3^a$, and admits a form above. Adding $3^a$ to this, we complete the proof.

sman96 wrote: gghx wrote: Induct, If it is even add one to every $n_1$ in the representation of $\frac{n}{2}$, if it is odd remove $3^{\lfloor \log_3(n)\rfloor}$ and add one to every $n_1$ in $\frac{n-3^{\lfloor \log_3(n)\rfloor}}{2}$. What if there is a $3^{\lfloor \log_3(n)\rfloor}$ in the representation of $\frac{n-3^{\lfloor \log_3(n)\rfloor}}{2}$ $3^{\lfloor \log_3(n)\rfloor}$ is strictly more than $\frac{n-3^{\lfloor \log_3(n)\rfloor}}{2}$.

grupyorum wrote: This is a classic. I prove it by strong induction. Let $n=2k$ and assume all numbers $1,2,\dots,2k-1$ have such a representation. Letting $k=\textstyle \sum_{1\le i\le k}3^{m_i}2^{n_i}$, we get $2k= \textstyle \sum_{1\le i\le k}3^{m_i}2^{n_i+1}$. Assume now $n$ is odd. If $n$ is a power of $3$, we are done, so assume $a$ is such that $3^a<n<3^{a+1}$. Note that $n-3^a$ is even, less than $3^a$, and admits a form above. Adding $3^a$ to this, we complete the proof. How are you sure that theres no $m_i=a$ or $n_i=0$ in the representation of $n-3^a$ (I dont think that if you just take common when theres some $m_i$ or $n_i$, it will work, at least i wasnt able to take common factors and make it work. So if you would please elaborate this section)

Strong induction on n Base case is obvious If $n=6k= 2^1 * 3^1 * k $, then k can be represented as the given expression and we have to just add 1 to the power of 2's and 3's For $n= 6k \pm 2 = 2(3k \pm1)$, $3k \pm1$ can be expressed as the given expression, we just add 1 to the powers of 2 in the expression to get n Same goes for n=6k+3 Now for $n=6k \pm 1$ we chose $3^m$ s.t. $3^{m+1} > n$ then, $z= n-3^m $ is even, so we can express $\frac{z}{2}$ as above and multiply by 2 and add $3^m$ to get n I don't know if this is correct