Floor floor floor inside another floor

Question

Inspired from this question

$$ \aleph(x,n)=\lfloor x\lfloor x\lfloor x...\rfloor\rfloor\rfloor\ $$

where $\aleph$ is the inner floor function with $n$ times for $x$. For example;

$$ \aleph(x,3)=\lfloor x\lfloor x\lfloor x\rfloor\rfloor\rfloor\ $$

or

$$ \aleph(x,1)=\lfloor x\rfloor $$

so from the previous question we know that;

if $$ \aleph(x,3)=\frac{2020}{x} $$ then $$ x=-\frac{2020}{305} $$

so this time the question is

What is the maximum value of $n$ with the minimum value $x$ such that a solution exists if the same question is asked without the value of $n$ given?

such as;

$$ \aleph(x,n)=\frac{2020}{x} $$

---

For example; if $$ \aleph(x,9)=\frac{2020}{x} $$ then there is solution such as: $$ x=-\frac{2020}{979} $$

but this is not maximum value of $n$.

Answer 1

The maximum $n$, its smallest $x$ and the equality now satisfied are: $ \require{begingroup}\begingroup \def \a #1#2{ {\aleph} \!\!\: \left( {#1} , {#2} \right) } \def \b #1{ {#1}\d{#1} } \def \d #1{ {\large{{#1} \over 2019}} } \def \e { {\!\;\varepsilon} } \def \f #1{ \left\lfloor {#1} \right\rfloor } \def \l { \\[.3ex] } \def \x { {\-\b1} } \def \xd { {\-\!\:\d{2020}} } \def \xp { {\big( \x \big)} } \def \. #1{ {\,{#1}\,} } \def \- { {\scriptsize \raise.25ex -} } \def \+ { {\scriptsize \phantom +} } \def \={ \kern-.3em & \kern-.3em = \kern-.3em & \kern-.3em } $

\begin{array}{rcccccc} n \= \+4035 \\[1ex] x \= \xd \= \-\b{1} \\[2ex] \a{x}{n} \= \-2019 \= {\Large{ 2020 \over \-\,\LARGE{2020\over2019}~ }} \= \Large{2020 \over \Large \raise.3ex x} \end{array}

This solution uses the recurrence relation $\a{x}{i{+}1} = \f{x\,\a{x}{i}}$ observed in the definition $ \a{x}{i} = \underbrace{\f{x\,\f{x\,\f{...\f{x}}}}} _{\large \f{~i~\sf levels~}} $.  Here is how $x\,\a{x}{n} = 2020$ is reached:

\begin{matrix} \a{x}{1} \= \f{x } \= \f{\x } \= \-2 \l \a{x}{2} \= \f{x\,\a{x}{1}} \= \f{\+\b{2}} \= \+2 \l \a{x}{3} \= \f{x\,\a{x}{2}} \= \f{\-\b{2}} \= \-3 \l \a{x}{4} \= \f{x\,\a{x}{3}} \= \f{\+\b{3} } \= \+3 \\[-.3ex] &\vdots& &\vdots& &\vdots& \\ \a{x}{4033} \= \f{x\,\a{x}{4032}} \= \f{\-\b{2017}} \= \-2018 \l \a{x}{4034} \= \f{x\,\a{x}{4033}} \= \f{\+\b{2018}} \= \+2018 \l \a{x}{4035} \= \f{x\,\a{x}{4034}} \= \f{\-\b{2018}} \= \-2019 \\[2ex] \hline \raise1ex\strut \boldsymbol{x\,\a{x}{n}} \= x\,\a{x}{4035} \= \xp(\-2019) \= \boldsymbol{2020} \end{matrix}

For this solution the goal is taken to approach $~ \a{x}{i} = \large{2020 \over \large \raise.3ex x} ~$ as gradually as possible. This suggests examining values of $x$ that border between progressing and getting stuck along the recurrence relation $\a{x}{i{+}1} = \f{x\,\a{x}{i}}$.

It seems obvious that $\a{x}{i}$ should not overshoot 2020 for $i \.< n$ and that, for minimal progress, $x$ should be as close as possible to 0. The puzzle statement’s example of $\a{\-{2020\over979}}{9}$ opens the way for $x \.< 0$ but it is easier to get a feel for the puzzle with $x \.> 0$.

What is the smallest positive value of $x$ that does not get stuck? It is $x \.= 2$, as demonstrated in comparison to $1 \.\le x \.< 2$.

\begin{array}{rclcrcl} \a{2}{1} \= \f{2} && \a{x}{1} \= \f{x} \kern1em\textsf{for$~~1\le x<2$} \\ \= 2 && \= 1 \\[1.5ex] \a{2}{2} \= \f{2\f{2}} && \a{x}{2} \= \f{x\f{x}} \l \= 4 && \= \f{(x)(1)} \\ \small \textsf{(doubles from}~\rlap{\textsf{$\a{2}{1}$ to $\a{2}{2}$)}} && &\kern3em& \= 1 \\ && && & \small\llap {\textsf{(stuck at}}~\rlap{\textsf{$\a{x}{1}$)}} \end{array}

For this smallest positive candidate of $x \. = 2$, $~ \a{2}{i} $ grows exponentially to $\a{2}{11} = 2048$, which is too much, meaning that $n{=}10$ would be the largest possibility for $n$ if $x$ is some difficult-to-pin-down number near 2.

What, then, is the smallest (closest to zero) negative value of $x$ that does not get stuck? It is $x = \-1{-}\e$ with an infinitesimally positive $\e$, as demonstrated in comparison to $x \.= \-1$.

\begin{array}{rclcrcl} \a{\-1{-}\e}{1} \= \f{\-1{-}\e} && \a{\-1}{1} \= \f{\-1} \\ \= \-2 && \= \-1 \\[1.5ex] \a{\-1{-}\e}{3} \= \f{(\-1{-}\e)\f{(\-1{-}\e)\f{\-1{-}\e}}} && \a{\-1}{3} \= \f{(\-1)\f{(\-1)\f{\-1}}} \l \= \f{(\-1{-}\e)\f{(\-1{-}\e)(-2)}} && \= \f{(\-1)(1)} \l \= \f{(\-1{-}\e)\f{2{+}2\e}} &\kern1em& \= \-1 \l \= \f{(\-1{-}\e)(2)} && & \small\llap{\textsf{(stuck at}}~\rlap{\textsf{$\a{\-1}{1}$)}} \l \= \f{\-2{-}2\e} \\ \= \-3 \\ \small \textsf {($\,$increments by} ~\rlap{\textsf{$\-1$ from $\, \a{\-1{-}\e}{1} \,$ to $\, \a{\-1{-}\e}{3} \,$)}} \end{array}

Pursuing this candidate of $x = \-1{-}\e$ works as well as could be hoped! Not only is the progression of $\a{x}{i}$ linear rather than exponential but it also grows at only half the rate of $i$, as laid out for the solution’s $x$ near the top of this answer. All that remains is to choose an $x$ near −1 so that $x \, \a{x}{n} = 2020$.

Although $~ \a{\-1{-}\e}{4037} = \-2020 ~$ looks promising, it is too good to be true because $~ (\-1{-}\e)\,\a{\-1{-}\e}{4037} = 2020{+}2020\e > 2020 ~$ overshoots the target.

Thus, using $n \.= 4035$ and working from $~ \a{\-1{-}\e}{4035} = \-2019 ~$ means solving for $~ x = \-1{-}h ~$ in $~ (\-1{-}h)\,\a{\-1{-}h}{4035} = 2019{+}2019h = 2020 \,$. And there it is, $\, h \.= \d{1} \,$ so $\, x \.= \-\b{1} \,$.

$\endgroup$