Represent $1729$ using four fours only.

Question

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!

Answer 1

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 $$1729=p_4\cdot p_{4+\sqrt{4}}\cdot p_{4!!}$$

Answer 2

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

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

Answer 3

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

$$(4?^{√4})?-((√4)?^4)?$$