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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.

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up vote 0 down vote accepted

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

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