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The four fours puzzle is a recreational math puzzle whose aim is to express whole numbers using four occurrences of the digit 4 and a specified set of operators. A common variety permits the following:

  • Constants: decimals (e.g. 44, which uses two 4s), leading decimal point (e.g. $.4 = \frac{2}{5}$) and recurring decimals (e.g. $.\overline{4} = \frac{4}{9}$).
  • Unary operators: negation $-x$, square root $\sqrt{x}$ and integer factorial $x!$.
  • Binary operators: +, -, ×, ÷, exponentiation $x^y$ and arbitrary roots $\sqrt[y]{x}$.

If we extend the puzzle to arbitrary reals then it is clear by countability that some numbers can't be expressed precisely (though by Gelfond–Schneider some transcendental numbers can). However, for some potentially unexpressable numbers it is possible to express arbitrarily close estimates. For example $$e_n = \sqrt{\left(\frac{4!^{(n)} + \sqrt{4}}{4!^{(n)}} \right) ^{4!^{(n)}}}$$ where $$4!^{(n)} = 4\overbrace{!...!}^{n\textrm{ times}}$$ is a sequence of four fours expressions whose limit is $e$.

Is there a similar expressible limit expression for $\pi$? There are a number of relevant limits at Pi: Limit representations but I can't think of how to express any of them using just four 4s.

Note that proving the hypothesis in Repeated Factorials and Repeated Square Rooting would show that arbitrarily close estimates are possible for every number using just one 4, but without necessarily providing them explicitly.

Note also that if we permit usage of the $\ln$ operator, then we can approximate any number using just three 4s. For example, the following is a sequence of four fours expressions whose limit is $a \geq 1$:

$$a_n = \overbrace{\sqrt{\sqrt{\cdots \sqrt{- \frac{\ln ( \ln \overbrace{\sqrt{\sqrt{\cdots \sqrt{4}}}}^{\lfloor a^{2^n}\rfloor \textrm{ times}} / \ln 4 )}{\ln \sqrt{4}}}}}}^{n \textrm{ times}}$$

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Uri Granta
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    The formula above for $e$ contains a typo. Instead of $\sqrt{4}$ it should be $\sqrt{4!^{(n)}}$ as in the exponent. – Giovanni Resta Feb 18 '16 at 14:50
  • The iterated factorial seems unnecessarily complicated. I think you could instead just say $$e=\lim_{n\to\infty}\sqrt{\left(4^n+\sqrt4\over4^n\right)^{4^n}}$$ – Barry Cipra Feb 18 '16 at 14:53
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    @BarryCipra n itself can't appear in the expression as it would need to be produced using 4s somehow. Repeated factorials allow arbitrarily big numbers using just one 4. – Uri Granta Feb 18 '16 at 14:57
  • @GiovanniResta Thanks (though actually it was a different typo): I was going for (1 + 2/n)^(n/2). I'll fix. – Uri Granta Feb 18 '16 at 14:59
  • @UriZarfaty, I'm afraid I don't understand what you mean. Your displayed expression for $e$ has $n$'s in it. I'm just offering an alternative that doesn't use the factorial symbol. – Barry Cipra Feb 18 '16 at 14:59
  • @BarryCipra Sorry if I'm being unclear. I was providing a sequence of expressions that gradually get closer to e: sqrt((4+2)/4)^4, sqrt((4!+2)/4!)^4!, sqrt((4!!+2)/4!!)^4!!, etc. None of these use n by themselves. – Uri Granta Feb 18 '16 at 15:02
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    @UriZarfaty So, if I understand correctly, you want a sequence of numbers $a_n$ such that $a_n\rightarrow\pi$ and every term $a_n$ can be expressed using just four fours? – Wojowu Feb 18 '16 at 15:04
  • @Wojowu Yes, precisely. I'll see if I can clarify that in the question. – Uri Granta Feb 18 '16 at 15:16
  • If that's the case, I don't see how come countability of anything is an obstacle to representing all reals. If we have shown that (say) every dyadic rational can be expressed using four fours, then every real number would be a limit of a sequence like in the question. – Wojowu Feb 18 '16 at 15:18
  • @Wojowu Countability is certainly not a barrier for limit representations, only for exact representations. In fact if the link at the end of the question is correct then every number can be represented as a limit of expressions that consist of one 4 and a number of factorials and square roots. – Uri Granta Feb 18 '16 at 15:21
  • Ah, I see now what you meant. – Wojowu Feb 18 '16 at 15:23
  • @UriZarfaty, ah, I see what you mean now. – Barry Cipra Feb 18 '16 at 15:23

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If you extend the domain of the factorial operator to the reals, then:

$$4\cdot \left(\frac{\sqrt{\sqrt{\sqrt{\ldots\sqrt{\sqrt{4}}}}}}{\sqrt{4}}!\right)^{\sqrt{4}}$$ approaches $\pi$ as the number of square roots increases.

RavenclawPrefect
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    This is an interesting solution. I wonder if it's possible to abuse the formula $$\pi=\lim_{n\rightarrow \infty}n\sqrt{-1}(1-\sqrt[n]{-1})$$ to get more arbitrarily close four-fours approximations of $\pi$? – Christian E. Ramirez Oct 28 '22 at 08:56