I was talking with friends about silly questions involving what numbers you can get using only a single digit "3" and unary operations. We eventually conjectured that using only factorials and square roots you can get arbitrarily close to any number greater than or equal to $1$. But we are having trouble proving or disproving the conjecture.

Precisely, start with the number $3$. Then take its factorial $m$ times, and then take the square root of its result $n$ times. Ie, $(3!!\ldots !)^{\frac{1}{2^n}}$ where there are $m$ factorials. Let the set of numbers achievable in this way be $X$. Is $X$ dense in $[1,\infty)$?

The only progress we have made is to show that any interval $[x,x^2]$ for $x>1$ there is a limit point $a \in [x,x^2)$ of $X$. This is true because for any $z > x^2$ we can square root it an appropriate number of times to get it in $[x,x^2)$. We can do this for infinitely many points of the sequence $3,3!,3!!,\ldots$. And all points we get through this process are distinct. Let $F(m)$ be $3$ with $m$ factorials. If $F(m)^{\frac{1}{2^n}} = F(a)^{\frac{1}{2^b}}$ then we can raise each side to a power and get an expression of the form $F(m) = F(a)^{\frac{1}{2^c}}$. But factorials are not squares. (To see this for $q!$, note that there is a prime in $[q/2,q]$ by bertrand's postulate. This prime appears only once in the factorization of $q!$). So all numbers we get are distinct. We have infinitely many distinct points in $X \cap [x,x^2)$ and therefore there is a limit point.

Other than that, we can't figure anything out. It feels like $X$ should be dense. Consider some interval $[x,x^2]$. Take lots of factorials of $3$. Then take square roots of that until it falls in $[x,x^2]$. It feels like the points we get will be somewhat uniformly distributed around $[x,x^2]$ and therefore dense.

  • $\begingroup$ I don't see any reason why it should be dense nor do I see any reason why it should avoid a number. There is also no way to check anything because $\log\log(3!!!!) \mod \log 2$ already is pretty impossible to compute. $\endgroup$ – mercio Mar 4 '11 at 11:43
  • $\begingroup$ I think you may be right. If you look at trying to hit numbers in the range $[2,4)$ this way, then taking $\log_2(\log_2(3!! \ldots !))$ will give you an integer and a fraction, say $n+x$ with $0 \le x < 1$. The integer $n$ will be the number of square roots needed while $2^{2^x}$ will tell you where you end up in $[2,4)$. So starting with 3 you get to 3 after zero square roots, starting with 3! you get to 2.4494... after one square root, starting with (3!)! you get to 2.2759... after three, starting with ((3!)!)! you get to 2.6691... after twelve. I don't see a pattern in $x$. $\endgroup$ – Henry Mar 7 '11 at 8:36

A related conjecture is posted at this link-that with the floor function you can get all naturals. I remember seeing a proof to that version with $\pi$ instead of $3$ a while ago, but can't find it now.

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  • $\begingroup$ Nice, in fact, at the bottom of the page, there is a conjecture by Michael Hartley which is actually close to what Moor Xu is conjecturing. Proving Moor Xu's conjecture would prove Michael Hartley's conjecture. $\endgroup$ – Raskolnikov Mar 4 '11 at 16:43

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