# Represent $1729$ using four fours only.

I keep trying Four fours puzzle for various numbers, i.e. express a number using four fours and only four fours along with any mathematical operation.

Today, I was thinking for Ramanujan number, i.e. $1729=10^3+9^3=12^3+1^3$. I think using $1729=12^3+1$ should be easier as $1^3=1$, and we need not worry about its power $3$.

Any help with this number is appreciated!

• Which operations are allowed? Just the basic 4? What makes you think this is possible? – gt6989b May 19 '16 at 10:26
• See this post for some of the basic operations, but you may use some other too as long as no other number is involved. It is most of the time possible. It is possible. Some people claim it can be done for every natural number, but there is no proof as such. – Departed May 19 '16 at 10:49
• $$\Gamma(\sqrt{4})+\sqrt{\sqrt{\sqrt{(4!/\sqrt{4})^{4!}}}}$$ Source: The Definitive Four Fours Answer Key, David A. Wheeler. First missing entry: 2179... – Did May 19 '16 at 11:50
• Basically, if logarithms are allowed, you can express any positive integers in the form of: $$n=-\sqrt{4}\cdot\frac{\ln(\ln \underbrace{\sqrt{\sqrt{\cdots{\sqrt{4}}}}}_{n}/\ln 4) }{\ln 4}$$ – Mc Cheng May 19 '16 at 12:37
• @BhaskarVashishth The paper mentions several other sources, mostly online, which you might want to check. But, explaining precisely the operations you allow and the ones you forbid seems to me to be a more urgent task... – Did May 19 '16 at 13:19

Let $n!!$ denote the double factorial of $n$ and $p_n$ the "$n$-th prime" function. (This might be stretching the rules a bit.) Then we can use the prime factorization of the Ramanujan number to obtain $$1729=p_4\cdot p_{4+\sqrt{4}}\cdot p_{4!!}$$

• I don't know whether the OP will consider it "within the rules", but this was very imaginative! +1 – MathematicianByMistake May 19 '16 at 11:46

Here is what I came up with. David Wheeler, mentioned above, does not allow ceiling or floor function, but if you do, this is also a way.

$1729=12^3+1=(\Gamma(4)\times \sqrt{4})^{\lceil(\sqrt{\Gamma(4)})\rceil}+ \Gamma(\sqrt{4})$

where $\Gamma(n)=(n-1)!$ and $\lceil \ \rceil$ is the ceiling function, i.e. for real number $x$, $\lceil x\rceil$ represents the smallest integer greater than or equal to $x$.

or

$1729=(\frac{4!}{\sqrt{4}})^{\lceil(\sqrt{\Gamma(4)})\rceil}+ \Gamma(\sqrt{4})$

If the "Terminal function" from this question is allowed ($n? = n + (n-1) + (n-2) + ... + 1$), here is one solution:

$$(4?^{√4})?-((√4)?^4)?$$