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This question arose on sci.math, but since almost all the competent mathematicians there have migrated here, I thought I'd give the question a wider audience. Starting with an integer $t>2$, perform the following operations

  1. Compute $t!, (t!)!, ((t!)!)!, \ldots$ repeated $ n$ times (allowing $n=0$ to result in $t$)
  2. To that result, $r$, compute $\sqrt{r}, \sqrt{\sqrt{r}}, \dots$ repeated $ m$ times (with the same convention as above).
  3. Return the floor of the result of step 2.

Call this function $S(t, n, m)$, meaning "start with $t$, do $n$ repeated factorials, then do $m$ repeated square roots, and finally take the floor". The original question was, given a positive integer, $x$, is there always a pair $(n,m)$ such that $S(4, n, m) = x$? One followup is obvious: does the answer depend on choice of starting value $t$?

[I had a hard time figuring what tags I should use, so feel free to retag this to your heart's content.]

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Restated, let $f(s)=s!$. Then for any $x$, does there exist $m,n$ such that $$x^{2^m}\leq f^n(t) <(x+1)^{2^m}$$ –  Thomas Andrews Sep 7 '12 at 20:21
    
Im pretty sure the answer is yes : there is always such a pair (n,m). Im also pretty sure about no : the answer does not depend on the choice of starting value t. Intuition clearly dictates this imnsho. Seems a lot like a collatz type problem. But i like collatz way better. Thomas Andrews shows that one can create similar problems without needing superfunctions of factorials and sqrt , but still different from collatz. I feel the OP question is therefore trying to run before you can walk if you know what i mean. –  mick Sep 7 '12 at 20:48
    
Consider the function $$\phi(x) = e^{e^{x\ln 2}}.$$ Then $\sqrt{\phi(x)}=e^{\frac12{e^{x\ln2}}}=e^{e^{x\ln2-\ln 2}}=\phi(x-1)$. Also consider $f(x)=\phi(\Gamma(\phi^{-1}(x)+1))$. Then the question is whether a sequence defined recursivly by $x_{n+1}=f(x_n)$ is dense modulo 1 (that is if $x_n-\lfloor x_n\rfloor$ is dense in $[0,1]$) –  Hagen von Eitzen Mar 8 '13 at 17:50
    
This is very close to a famous conjecture attributed to Knuth. In the original form, one is allowed to interleave factorials and sqrt-floor operations, so this variant would be strictly harder to affirm. –  Erick Wong Mar 8 '13 at 18:04
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