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When I was in lower school I remember playing the Four $4$'s game and we got to around $10$. For those of you who don't know the Four $4$'s game is one where you have to use four $4$'s and you have to construct an equation using multiplication, division, addition and subtraction.

This new game, however, is a bit more interesting. Firstly, you can only use the previously mentioned operations, as well as raising to a power (provided it uses the number $4$ such as $4^4$), you can use the factorial operation, and you can take the square root (ignore the $\frac12$). There is no concatenation or decimals, as well as no logarithms, functions (such as the Gamma Function), primordial numbers, or double, triple etc factorials. And the question is: What is the smallest Natural Number you can't create?

Just wanted to clarify by giving an example: $3 = \sqrt{4} + \sqrt{4} - \frac{4}{4}$

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  • $\begingroup$ What numbers have you created? $\endgroup$ – saulspatz Jul 13 '18 at 14:31
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    $\begingroup$ It seems like one should be able to brute-force it. Write a formal language for the acceptable strings, and evaluate them all; if I'm not mistaken, there are only finitely many. Wrong - I forgot the unary operators. You can take factorials forever. $\endgroup$ – saulspatz Jul 13 '18 at 14:44
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    $\begingroup$ I cannot get $\Large{39,\; 41, \;43}$. Who can get any of them? $\endgroup$ – Oleg567 Jul 13 '18 at 15:12
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    $\begingroup$ $${\large{33}} = \dfrac{64+2}{2} = \dfrac{\left(\sqrt{\sqrt{2}}\right)^{24}+\sqrt{4}}{\sqrt{4}} = \dfrac{ \left(\sqrt{ \sqrt{ \sqrt{4} } }\right) ^{4!} + \sqrt{4}}{\sqrt{4}}.$$ $\endgroup$ – Oleg567 Jul 13 '18 at 15:36
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    $\begingroup$ Update: $\large{41}$ is possible: $${\large 41} = \sqrt{1681} = \sqrt{\dfrac{24+40320}{24}} = \sqrt{\dfrac{4!+(4+4)!}{4!}}.$$ $\endgroup$ – Oleg567 Jul 13 '18 at 16:04
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Let's record (at least one of possible) solutions in the table.

In additional column we will mark which extension was used: $\hat{}$ or $\sqrt{\phantom{w}}$ or $!$ (in addition to mentioned classic $+\;-\;\times \;/$ version).

\begin{array}{|l|l|l|} \hline number & formula & extension \\ \hline 1 & 4-4+4/4 & \\ 2 & (4\cdot 4) /(4+4) & \\ 3 & (4+4+4) / 4 & \\ 4 & 4 + (4-4)*4 & \\ 5 & (4 + 4\cdot 4)/4 & \\ 6 & 4 + (4+4)/4 & \\ 7 & 4+4 - 4/4 & \\ 8 & 4+4+4-4 & \\ 9 & 4+4 + 4/4 & \\ 10 & 4 + 4+ 4/\sqrt{4} & \sqrt{\phantom{w}} \\ 11 & (4!+4!-4) /4 & ! \\ 12 & 4\cdot(4 - 4/4) & \\ 13 & (4!+4!+4) / 4 & ! \\ 14 & 4\cdot 4 - 4/\sqrt{4} & \sqrt{\phantom{w}} \\ 15 & 4\cdot 4 - 4/4 & \\ 16 & 4+4+4+4 & \\ 17 & 4\cdot 4 + 4/4 & \\ 18 & 4\cdot 4 + 4/\sqrt{4} & \sqrt{\phantom{w}} \\ 19 & 4! - 4 - 4/4 & ! \\ 20 & 4\cdot(4 + 4/4) & \\ 21 & 4! - 4 + 4/4 & ! \\ 22 & 4\cdot 4 + 4 + \sqrt{4} & \sqrt{\phantom{w}} \\ 23 & (4\cdot 4! - 4)/4 & ! \\ 24 & 4+4 + 4\cdot 4 & \\ 25 & 4! + 4\cdot(4-4)! & ! \\ 26 & 4! - 4 + 4!/4 & ! \\ 27 & 4! + 4!/(4+4) & ! \\ 28 & 4\cdot(4+4) - 4 & \\ 29 & 4! + 4 + 4/4 & ! \\ 30 & 4(4+4) - \sqrt{4} & \sqrt{\phantom{w}} \\ 31 & ((4!/4)! +4!)/4! & ! \\ 32 & 4\cdot 4 + 4\cdot 4 & \\ 33 & \left(\sqrt{\sqrt{\sqrt{4^{4!}}}}+\sqrt{4}\right) / \sqrt{4} & \hat{},!,\sqrt{\phantom{w}} \\ 34 & 4(4+4) + \sqrt{2} & \sqrt{\phantom{w}} \\ 35 & 4! + (4!-\sqrt{4})/\sqrt{4} & !,\sqrt{\phantom{w}} \\ 36 & 4\cdot(4+4) + 4 & \\ 37 & 4! + (4!+\sqrt{4})/\sqrt{4} & !,\sqrt{\phantom{w}} \\ 38 & 4! - \sqrt{4} + 4\cdot 4 & !, \sqrt{\phantom{w}} \\ 39 & ????????????????????? & \\ 40 & 4 (4+4+\sqrt{4}) & \sqrt{\phantom{w}} \\ 41 & \sqrt{(4!+(4+4)!)/4!} & !,\sqrt{\phantom{w}} \\ 42 & 4! + 4! - 4!/4 & ! \\ 43 & ????????????????????? & \\ 44 & 4! + 4 + 4\cdot 4 & ! \\ 45 & (4!/4)!/ (4\cdot 4) & ! \\ 46 & 4!+4!-4/\sqrt{4} & !,\sqrt{\phantom{w}} \\ 47 & 4!+4! - 4/4 & ! \\ 48 & 4 \cdot (4+4+4) & \\ 49 & 4!+4! + 4/4 & ! \\ 50 & 4!+4!+4/\sqrt{4} & !,\sqrt{\phantom{w}} \end{array}


To know which (real) numbers admit $4$ $4$'s formula, one can use this simple algorithm:

build set $A_1$ of numbers, which can be obtained from one $4$. Since $\sqrt{\phantom{w}}$ and $!$ are allowed (unary "operations"), the set $A_1$ isn't finite. We will stop at some reasonable step. Say, let the set $A_1$ is $$ A_1 = \{ 4, 4!, \sqrt{4}=2, \sqrt{4!}, \sqrt{\sqrt{4}}, \sqrt{\sqrt{4!}}, \sqrt{\sqrt{\sqrt{4}}}, \sqrt{\sqrt{\sqrt{\sqrt{4}}}} \}. $$

Then we build set $A_2$: set of numbers which can be obtained from two $4$'s: for any $a\in A_1$, for any $b\in A_1$ add to $A_2$ following numbers:

  • $a+b$,
  • $|a-b|$,
  • $a\cdot b$,
  • $a/b$ (if $b\ne 0$),
  • $b/a$ (if $a\ne 0$),
  • $a^b$ (if $a\ne 0$),
  • $b^a$ (if $b\ne 0$).

Then update the set $A_2$ with factorials (of its old small integer elements) and square roots of its old elements. We can repeat this update few times.

And clear the set $A_2$ of duplicates.

The set $A_3$ is derived this way from $A_1$ and $A_2$.

The set $A_4$ (final set!) has $2$ parts:
a) part $1$ - derived from $A_2$ and $A_2$;
b) part $2$ - derived from $A_1$ and $A_3$.


There is small chance that when extend $A_1$ with more roots/factorials ($A_2,A_3,A_4$ too), we can describe more numbers from the range $1,\ldots, 100$; but chances are really small since repeated roots tend to $1$, repeated factorials tend to $\infty$.

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  • $\begingroup$ Wow, this great. I wonder if there's a pattern between say using 3's instead of 4's. Or what is the first number you can't make from 5 4's etc? I think it's an interesting way to see the relation between the maximum number possible, and the binary and unary operations. $\endgroup$ – John Miller Jul 15 '18 at 21:48
  • $\begingroup$ If use four $3$'s, then it's hard to get $23$, $31$, $32$ for me. About five $4$'s: I'll able to answer a bit later. $\endgroup$ – Oleg567 Jul 17 '18 at 13:28
  • $\begingroup$ @John Miller: update: on four 3's: $23 = \dfrac{(3!)!}{3!\cdot 3!}+3$, $\;31 = \sqrt{(3!)! + \dfrac{(3!)!+3}{3}}$, so numbers $32,44,47$ are first hard examples there. On five 4's: such first hard examples are: $135, 159, 161, 189$ (according to my calculations). $\endgroup$ – Oleg567 Jul 18 '18 at 21:27

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