BUMP....

One of the nicer problems I've ever seen. Note that every natural number has a Zeckendorf Representation. Then, $k$ is the result when we truncate the last digit of this representation (and clearly shift everything down) - call this $\rho(n)$. For clarity, take the number $12=8+3+2=F_6+F_4+F_2=\overline{101010}_\phi$. Then, we claim that $k=10101_\phi=F_5+F_3+F_1=8$, and indeed we can get to it by \[(1,1)\mapsto(4,3)\mapsto(12,8)\] First, note that clearly this is true for $n=1$. We now show that both these operations preserve this. Thus, assume the pair is $(x,\rho(x))$. If we apply the first operation, and $x=\sum F_{y_i}$, then \[x+\rho(x)=\sum F_{y_i}+\sum F_{y_i-1}=\sum \left(F_{y_i}+F_{y_i-1}\right)=\sum F_{y_i+1}\]so the operation is clearly preserved. Note that this indeed is the maximum representation as, by assumption, also is a Zeckendorf Representation. Now, if we apply the second operation, note \[x+\rho(x)+1=\sum F_{y_i}+\sum F_{y_i-1}+1=\sum \left(F_{y_i}+F_{y_i-1}\right)+1=F_1+\sum F_{y_i+1}\]In addition, note \[2x+\rho(x)+1=2\sum F_{y_i}+\sum F_{y_i-1}=\sum F_{y_i}+\sum F_{y_i+1}=\sum F_{y_i}+\sum F_{y_i+1}+1=F_2+\sum F_{y_i+2}\]Thus, indeed $\rho(2x+\rho(x)+1)=x+\rho(x)+1$, as the second is clearly a Zeckendorf representation. To finish, we just need to show that every number can be generated. We induct on this: Base Case. $n\leq 1$. This is obvious. Induction Hypothesis. Assume for some $k$, and for all $n<F_k$, we can reach $(n,\rho(n))$. We shall show for all $F_k\leq m<F_{k+1}$, we can reach $(m,\rho(m))$. Induction Step. Suppose that $m=\overline{a_na_{n-1}\ldots a_2}$ is the Zeckendorf representation. If $a_2=0$, then taking the first operation on the pair $(\rho(m),\rho^2(m))$ suffices. Otherwise, we know $a_3=0$, so the second operation on the pair $(\rho^2(m),\rho^3(m))$ should work. This completes the induction step.

Wow. Very nice solution indeed. It seems more intiutive when you think the first move adds a $0$ of the number’s Zeckendorf Representation. Latter move adds $01$ of the numbers Zeckendorf Representation.