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How would one prove/disprove this? ...

Conjecture:
Suppose $p$, $q$ are distinct primes, and define $\ f(n) = n p, \ g(n) = \left \lfloor \frac{n}{q} \right \rfloor$ for all $n \in \mathbb{N_+}$; then
for all $x,y \in \mathbb{N_+}$, there exists a composition $F = g \circ g \circ \cdots \circ g \circ f \circ f \cdots \circ f$ such that $ \ y = F(x) $.

A weaker conjecture is implicit in the title question -- same as above, but allowing compositions with the functions in arbitrary order.

If this is something well-known or discussed elsewhere, my apologies (I did search); a reference would then be welcome.

(An analogous conjecture by Donald Knuth involving factorial and integer-squareroot functions suggested this question.)

r.e.s.
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1 Answers1

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The strong one is true. As an example, let us take $p=2, q=3, x=1$. Then the numbers we can reach are $\lfloor \frac {2^r}{3^s} \rfloor$ with $r,s \in \Bbb N_+$. Given $y$, we need to find $r,s$ such that $\log y \lt \log \frac {2^r}{3^s} \lt \log(y+1)$ or $\log y \lt r \log 2 - s \log 3 \lt \log (y+1)$ As we can approximate $\frac {\log 3}{\log 2}$ arbitrarily closely by a rational, we can do this. The same argument works for general $p,q,x$ as long as the logs are rationally independent.

Ross Millikan
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  • We need to find r,s such that log(y/x) <= r log p - s log q < log((y+1)/x). Sorry if this is obvious, but how does an arbitrarily-close rational approximation of log p / log q help? (We need r log p - s log q ≈ log(y/x), not ≈ 0.) – r.e.s. Oct 12 '13 at 01:11
  • The denominator $x$ is no issue, it is just a fixed offset. $\frac 12(\log (y) + \log (y+1))-\log x$ is some irrational number, and we can find an infinite number of pairs $r,s$ such that $\frac rs -\frac {\log p}{\log q}\lt \frac 1{s^2}$. Once $s$ is large enough, $\frac 1s \gt r\log q - s \log p$ will fit within the interval $(\log y-\log x,\log(y+1)-\log x)$ – Ross Millikan Oct 12 '13 at 03:44
  • The fact that I needed the $\frac 1s^2$ approximation shows it is not trivial. Thanks for asking. – Ross Millikan Oct 12 '13 at 14:02