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I have tried to plot the following map:

$\qquad \begin{cases} y_{n+1}=r y_n^{x_{n+1}-x_{n+2}}\\ y_1=x_1-x_2 \quad \end{cases}$

using the code shown below. The error I have got is that

x[1] - x[2] can't be used as an iterator

However, it works if I try it analytically then how I can got its plot?

funcs = 
  RecurrenceTable[
    {y[n + 1] == r( y[n]^(x[n+1] - x[n+2])), 
     y[1] == x[1] - x[2]}, 
    y, {n, 1, 5}];
Plot[funcs /. r -> 1, {x[1] - x[2], 0, 1}, 
  Evaluated -> True,   PlotRange -> All]

Note This is just explantion about running of the below code , The below Code seems converge always to r for odd iteration and even iteration but the sequence I mean is diverge such that it take two distincts limit according the parity of iteration odd and even , As example of the above system we may take this example: $a_n=(1-\frac12)^{(\frac12-\frac13)^{...^{(\frac{1}{n}-\frac{1}{n+1})}}}$ here we have limit of a_n for n odd close to 0.56778606544394002098000796382530333102219963214866 and for even iteration we have 0.85885772008416606762434379473241623070938618180813 for more behavior of this sequence one can check the link of this question here in SE .Probably I'm wrong in the reformulation of the above general recursive sequence.

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    $\begingroup$ You need another equation to generate the values for x[n+1] $\endgroup$ – Bob Hanlon Sep 12 '20 at 3:40
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Somehow the sequence x[1], ..., x[6] must be defined. This can be done by giving each of the six values), giving a recursive formula or by defining function x over the domain [1, ..., 6]. I will use the last method to show you how a plot can be made.

x[n_] := Log2[1/(n + 1)]
Module[{r, yVals},
  r = 1;
  yVals =
    RecurrenceTable[
      {y[n + 1] == r (y[n]^(x[n + 1] - x[n + 2])), y[1] == x[1] - x[2]},
      y, {n, 1, 5}];
 ListPlot[yVals]]

plot

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