A general closed form for a family of nonlinear radical recurrences

Set

$$y_n=x_n-c.$$

Then

$$y_{n+1}=\sqrt[r]{y_n^r+\lambda}.$$

Since $r$ is odd,

$$y_{n+1}^r=y_n^r+\lambda.$$

Let

$$u_n=y_n^r.$$

Then

$$u_{n+1}=u_n+\lambda.$$

Thus $(u_n)$ is an arithmetic progression, and therefore

$$u_n=u_1+(n-1)\lambda.$$

Since

$$u_1=(x_1-c)^r,$$

we get

$$(x_n-c)^r=(x_1-c)^r+(n-1)\lambda.$$

Hence

$$x_n=c+\sqrt[r]{(x_1-c)^r+(n-1)\lambda}.$$

Now suppose that $\lambda>0$. Then

$$x_n\leq M$$

is equivalent to

$$(x_1-c)^r+(n-1)\lambda\leq(M-c)^r.$$

Therefore

$$n-1\leq\frac{(M-c)^r-(x_1-c)^r}{\lambda}.$$

Hence the largest possible value of $n$ is

$$n_{\max} = 1+ \lfloor ((M-c)^r-(x_1-c)^r)/\lambda \rfloor.$$

For the general form, suppose

$$x_{n+1}=\Phi^{-1}(\Phi(x_n)+\lambda).$$

Applying $\Phi$ gives

$$\Phi(x_{n+1})=\Phi(x_n)+\lambda.$$

Thus

$$\Phi(x_n)=\Phi(x_1)+(n-1)\lambda.$$

Therefore

$$x_n=\Phi^{-1}(\Phi(x_1)+(n-1)\lambda).$$

Consider the sequence defined by

$$a_1=1$$

and

$$a_{n+1} = \sqrt[3]{a_n^3-3a_n^2+3a_n+3}+1.$$

We want to determine the largest positive integer $n$ such that

$$a_n\leq 11.$$

First observe that

$$a_n^3-3a_n^2+3a_n+3 = (a_n-1)^3+4.$$

Hence the recurrence becomes

$$a_{n+1} = 1+\sqrt[3]{(a_n-1)^3+4}.$$

This is exactly the general recurrence

$$x_{n+1} = c+\sqrt[r]{(x_n-c)^r+\lambda}$$

with

$$c=1,\qquad r=3,\qquad \lambda=4,\qquad x_1=1.$$

Therefore

$$a_n = 1+\sqrt[3]{4(n-1)}.$$

Now impose the condition

$$a_n\leq 11.$$

Then

$$1+\sqrt[3]{4(n-1)}\leq 11,$$

so

$$\sqrt[3]{4(n-1)}\leq 10.$$

Cubing both sides,

$$4(n-1)\leq 1000.$$

Thus

$$n-1\leq 250,$$

and hence

$$n\leq 251.$$

Therefore the largest possible value is

$$\boxed{251}.$$

Indeed,

$$a_{251} = 1+\sqrt[3]{1000} = 11.$$