I am looking at basic recurrence relations with initial conditions and I would like to know if it is possible to prove some pattern that I am seeing.
Given the conditions
$$\bigg\lbrace_{x_{n+1}=(r a_n)^{1/k}}^{x_0=1}\;:r\in\Bbb R, k\in\Bbb N \wedge r,k \gt 1$$
Starting at $x_1$, I get the five first terms as
$$\left\{r^{\frac{1}{k}},\left(r^{\frac{1}{k}+1}\right)^{\frac{1}{k}},\left(r \left(r^{\frac{1}{k}+1}\right)^{\frac{1}{k}}\right)^{\frac{1}{k}},\left(r \left(r \left(r^{\frac{1}{k}+1}\right)^{\frac{1}{k}}\right)^{\frac{1}{k}}\right)^{\frac{1}{k}},\left(r \left(r \left(r \left(r^{\frac{1}{k}+1}\right)^{\frac{1}{k}}\right)^{\frac{1}{k}}\right)^{\frac{1}{k}}\right)^{\frac{1}{k}},\;\ldots\right\}$$
I see that the base $r$ is constant and that the exponents behave as the sequence
$$\left\{a_n \;:\;a_{n+1}=\frac{1}{k}a_n+\frac{1}{k},\;a_1=\frac{1}{k}\right\}$$
for which the first five terms are
$$\left\{\frac{1}{k},\frac{1}{k^2}+\frac{1}{k},\frac{\frac{1}{k^2}+\frac{1}{k}}{k}+\frac{1}{k},\frac{\frac{\frac{1}{k^2}+\frac{1}{k}}{k}+\frac{1}{k}}{k}+\frac{1}{k},\frac{\frac{\frac{\frac{1}{k^2}+\frac{1}{k}}{k}+\frac{1}{k}}{k}+\frac{1}{k}}{k}+\frac{1}{k},\,\ldots\right\}$$
What I see is that all terms are rational and greater than zero. Also, the numerator of $a_{n+1}$ seems to be the sum of the numerator and denominator of the $a_n$ term and the denominator of the $a_{n+1}$ term is the denominator of $a_n$ multiplied by $k$ such that
$$\left\{\frac{1}{k},\frac{k+1}{k^2},\frac{k^2+k+1}{k^3},\frac{k^3+k^2+k+1}{k^4},\frac{k^4+k^3+k^2+k+1}{k^5},\,\ldots\right\}$$
which I think is the same as the terms of the convergent geometric series
$$\frac 1k \sum_{n=0}^{\infty}\left(\frac 1k\right)^n \;=\; \frac 1{k-1}$$
Since the terms of the sequence are rational numbers and I am restricting $k\in\Bbb N:k\gt 1$, can I use induction? I checked that the first three cases work.
$k=2$ works $\left\{\frac{1}{2},\frac{3}{4},\frac{7}{8},\frac{15}{16},\frac{31}{32},\,\ldots\right\}$, $k=3$ works $\left\{\frac{1}{3},\frac{4}{9},\frac{13}{27},\frac{40}{81},\frac{121}{243},\,\ldots\right\}$ and $k=4$ also works $\left\{\frac{1}{4},\frac{5}{16},\frac{21}{64},\frac{85}{256},\frac{341}{1024},\,\ldots\right\}$.
Can I prove that the numerator of $a_{n+1}$ is the sum of the numerator and denominator $a_n$ and the denominator of $a_{n+1}$ is the product of the denominator of $a_n$ and $k$, by induction? If so, I don't see how to formulate a statement in such a way that I could proceed with a proof.
Thank you for your help and your time, I appreciate it very much.