# Floor floor floor inside another floor

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 $$n$$.

• Not sure if I understand the question. Are you asking for the maximum value of $n$ such that a solution exists? May 6 '20 at 14:45
• Nice extension but it would be handier to define ℵ(x) as x⌊x⌊x⌊x...⌋⌋⌋. May 7 '20 at 6:13
• I did misread $\aleph()$ as $x\lfloor\ldots\rfloor$ at first but now appreciate how it is always an integer (especially as $\aleph$ visually resembles N)
– humn
May 7 '20 at 6:59
• @Xi'an my intention was to give a small hint in the question by seperating last $x$ multiplication from the function.
– Oray
May 7 '20 at 7:29
• Such a nice and neat original variation on the 4-floor puzzle. Daunting at first glance, this turns out to need no calculator or trial-and-error, just a willingness to start at the extremes of possibility.
– humn
May 7 '20 at 18:17

## 1 Answer

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 $$\, x \.= \-\b{1} \,$$.

$$\endgroup$$