Letting $a_1,a_2,\cdots,a_r$ be integers which are larger than or equal to $2$, let us define $$[a_1,a_2,\cdots,a_r]=\frac{1}{a_1-\frac{1}{a_2-\frac{1}{\ddots-\frac{1}{a_r}}}}$$

(Note that the negative signs are used)

Also, let $X, Y, Z$ be positive integers which satisfy $$Z\lt X+Y,\ Z\gt X,\ Z\gt Y$$ and let $$\frac XZ=[a_1,a_2,\cdots,a_r],\ \frac YZ=[b_1,b_2,\cdots,b_s].$$

Then, here is my question.

Question : Is the following true?

"There exist $r^{\prime}\le r, s^{\prime}\le s$ such that $$[a_1,a_2,\cdots,a_{r^{\prime}}]+[b_1,b_2,\cdots,b_{s^{\prime}}]=1$$ for any $(X,Y,Z)$."

Remark : Observing the initial numbers is not sufficient because the nearer to $1$ the value $\frac XZ+\frac YZ$ is, the harder it is to find the answer (see example 2).

Examples :

  1. $\frac XZ=\frac 37=[3,2,2]$ and $\frac YZ=\frac 57=[2,2,3]$ leads $[3]+[2,2]=\frac 13+\frac 23=1$ where $\frac 37+\frac 57=\frac 87\approx 1.143$

  2. $\frac XZ=\frac{901}{2067}=[3,2,2,4,2]$ and $\frac YZ=\frac{1170}{2067}=[2,5,2,2,3]$ leads $[3,2,2,4]+[2,5,2,2]=\frac{10}{23}+\frac{13}{23}=1$ where $\frac XZ+\frac YZ=\frac{2071}{2067}\approx 1.002.$

Motivation : I've got an algorithm to find $b_1,b_2,\cdots,b_s$ such that $$1-x=[b_1,b_2,\cdots,b_s]$$ for any given $x=[a_1,a_2,\cdots,a_r]$.

Algorithm : Supposing that $2^r$ represents $r$-consective $2$s, I'm going to write $$[a_1,a_2,\cdots,a_r]=[2^{q_1},p_1,2^{q_2},p_2,\cdots,2^{q_s},p_s,2^{q_{s+1}}]$$ where $p_i\ge 3\in \mathbb N, q_i\ge 0\in \mathbb Z$. For example, $[2,2,5,3,2,4]=[2^2,5,2^0,3,2^1,4,2^0]$.

Then, the algorithm is $$1-[2^{q_1},p_1,2^{q_2},p_2,\cdots,2^{q_s},p_s,2^{q_{s+1}}]$$ $$=[(q_1+2),2^{(p_1-3)},(q_2+3),2^{(p_2-3)},(q_3+3),2^{(p_3-3)},\cdots,(q_s+3),2^{(p_s-3)},(q_{s+1}+2)].$$

After getting this algorithm, I reached the above expectation. I can neither find any counterexample even by using computer nor prove that the expectation is true. Can anyone help?

Update : I crossposted to MO.


I'm posting an answer just to inform that the question has received an answer by Alexey Ustinov on MO.



Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.