I have a sequence $(u_n)$ so that $u_1=a$ and $u_{n+1}=4u_n (1-u_n)$, for all $n=1,2,\ldots$. How many solutions are there in this equation $u_{2018}=0$?

I tried

u[1] := a
u[y_] := 4*u[y - 1] (1 - u[y - 1])
NSolve[u[11] == 0, a, Reals] // Length

With $n=1$ to $n=11$, I guess the number of solutions are $2^{n-1}$. With $u[2018]$, I don't know how to find. How can I get the answer?

  • 1
    $\begingroup$ To start you off: RSolve[{u[y] == 4*u[y - 1] (1 - u[y - 1]), u[1] == a}, u, y] $\endgroup$ – J. M. will be back soon Apr 10 '18 at 9:03
  • $\begingroup$ Consider what adding 1 to n does to the degree u[n+1] vs that of u[n]. $\endgroup$ – Daniel Lichtblau Apr 10 '18 at 14:25

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