Four Fours/Table of 4s
From charlesreid1
Back to Four Fours
One 4
Believe it or not, the rules allow you to do quite a bit with a single 4. The rules say you may combine 4s with any mathematical symbol except numbers. Thus, in addition to 4 alone, we also have:
$ 1 = i^{4} $
$ 2 = \sqrt{4} $
$ 24 = 4! $
The following fractions are also useful:
$ \dfrac{1}{4} $
$ \dfrac{1}{2} = \dfrac{1}{\sqrt{4}} $
$ \dfrac{1}{24} = \dfrac{1}{4!} $
Could possibly add constants (harmonic number), possibly special functions.
Once you allow variables like x into the mix, it's lights out.
One 4 with Variables
$ 5 = \dfrac{ \ln{ \left( \dfrac{ \ln{ \left( \sqrt{ \sqrt{ \sqrt{ \sqrt{ \sqrt{ x } } } } } \right) } }{ \ln{x} } \right) } }{ \ln{\sqrt{4}} } $
Two 4s
Note: we aren't using fourth roots or one-quarter powers very much, e.g., $ \sqrt{2} = \sqrt[4]{4} = 4^{\frac{1}{4}} $. Adding this would greatly expand the possibilities.
$ 1 = \dfrac{4}{4} $
$ 1 = \log_{4}(4) $
$ 2 = 4 - \sqrt{4} $
$ 2 = \dfrac{4}{\sqrt{4}} $
$ 3 = \sqrt{4} + i^{4} $
$ 3 = 4 - i^{4} $
$ 4 = \sqrt{4} \times \sqrt{4} $
$ 4 = \sqrt{4} + \sqrt{4} $
$ 4 = \dfrac{4}{i^4} $
$ 5 = 4 + i^{4} $
$ 5 = \sqrt{ 4! + i^4 } $
$ 6 = 4 + \sqrt{4} $
$ 6 = (4 - i^4)! $
$ 8 = 4+4 $
$ 8 = 4 \sqrt{4} $
$ 12 = \dfrac{4!}{\sqrt{4}} $
$ 16 = 4 \times 4 $
$ 16 = 4^{\sqrt{4}} = (\sqrt{4})^4 $
$ 16 = \sqrt{4^4} $
$ 20 = 4! - 4 $
$ 22 = 4! - \sqrt{4} $
$ 23 = 4! - i^4 $
$ 24 = \sqrt{ (4!)^{\sqrt{4}} } $
$ 25 = 4! + i^4 $
$ 26 = 4! + \sqrt{4} $
$ 28 = 4! + 4 $
$ 44 $
$ 48 = 4! \times \sqrt{4} $
$ 96 = 4! \times 4 $
$ 120 = (4 + i^4)! $
$ 256 = 4^4 $
$ 576 = (4!)^{\sqrt{4}} $
$ 720 = (4 + \sqrt{4})! $
$ 331,776 = (4!)^4 $
$ 16,777,216 = \sqrt{4}^{4!} $
$ 281,474,976,710,656 = 4^{4!} $
$ 1,333,735,776,850,284,124,449,081,472,843,776 = {4!}^{4!} $
(For those keeping score at home, that's 1 decillion 333 nonillion 735 octillion 776 septillion 850 sextillion 284 quintillion 124 quadrillion 449 trillion 81 billion 472 million 843 thousand 776)
Three 4s
These lists blow up pretty fast... as you can see, focusing on using a smaller number of 4s can force you to be creative. This makes it possible to combine 4 4's beyond the integers from 1 to 20, and keep on going...
$ 2^n = (\sqrt{4})^n $
where $ n $ is any number expressible with two 4's.
$ 2 = \dfrac{4+4}{4} $
$ 2 = \sqrt{4} \times \left( \dfrac{4}{4} \right) $
$ 3 = \dfrac{ \ln{(4+4)} }{ \ln{\sqrt{4}} } $
$ 4 = 4 \times \left( \dfrac{4}{4} \right) $
$ 4 = 4 + 4 - 4 $
$ 5 = 4 + \dfrac{4}{4} $
$ 7 = 4 + 4 - i^{4} $
$ 7 = 4 + \sqrt{4} + i^{4} $
$ 8 = 4 \times \left( \dfrac{4}{\sqrt{4}} \right) $
$ 9 = 4 + 4 + i^{4} $
$ 10 = 4 + 4 + \sqrt{4} $
$ 11 = \dfrac{ 4! - \sqrt{4} }{ \sqrt{4} } $
$ 12 = 4+4+4 $
$ 18 = 4 \times 4 + \sqrt{4} $
$ 20 = 4 \times 4 + 4 $
$ 26 = 4! + 4 - \sqrt{4} $
$ 28 = 4! + \sqrt{4} + \sqrt{4} $
$ 30 = 4! + 4 + \sqrt{4} $
$ 30 = \dfrac{(4+\sqrt{4})!}{4!} $
$ 30 = \dfrac{(4 + i^4)!}{4} $
$ 32 = 4(4+4) $
$ 32 = 4^{\sqrt{4}} \sqrt{4} $
$ 36 = (4+\sqrt{4})^{\sqrt{4}} $
$ 46 = 4! + 4! - \sqrt{4} $
$ 47 = (4!)(\sqrt{4}) - i^4 $
$ 49 = (4!)(\sqrt{4}) + i^4 $
$ 50 = 4! + 4! + \sqrt{4} $
$ 60 = \dfrac{(4+i^4)!}{\sqrt{4}} $
$ 64 = \dfrac{4^4}{4} $
$ 64 = (\sqrt{4})^{4 + \sqrt{4}} $
$ 74 = (4A)_{4 \cdot 4} $
(That is, 74 is 4A in hexidecimal, or base 16 = 4 * 4)
$ 75 = (4B)_{4 \cdot 4} $
$ 76 = (4C)_{4 \cdot 4} $
$ 77 = (4D)_{4 \cdot 4} $
$ 78 = (4E)_{4 \cdot 4} $
$ 79 = (4F)_{4 \cdot 4} $
$ 80 = 4(4! - 4) $
$ 116 = ( (4 + i^4)! - 4 $
$ 118 = ( (4 + i^4)! - \sqrt{4} $
$ 119 = ( (4 + i^4)! - i^4) $
$ 120 = \dfrac{(4+i^4)!}{i^4} $
$ 121 = ( (4 + i^4)! + i^4) $
$ 122 = ( (4 + i^4)! + \sqrt{4}) $
$ 124 = ( (4 + i^4)! + 4 $
$ 180 = \dfrac{(4+\sqrt{4})!}{4} $
$ 252 = 4^4 - 4 $
$ 254 = 4^4 - \sqrt{4} $
$ 258 = 4^4 + \sqrt{4} $
$ 260 = 4^4 + 4 $
$ 1,024 = 4^4 \times 4 $
$ 4,096 = (4+4)^{4} $
$ 65,536 = (4 \times 4)^{4} $