Consider the following recurrence: $$ f(x) = \lfloor f(x-1)\rfloor \cdot(f(x-1) - \lfloor f(x-1)\rfloor) + 1 \space(x\geq 2, \space x\in \mathbb{N}), \space f(1) = c $$
How can one solve such a recurrence, i.e. finding a closed formula? I know some basic methods how to solve linear homogenous recurrence relations, e.g. with a spectral decomposition or by finding the characteristic equation, but I don't really know whether this can be applied to the above.
Edit: $c\in \mathbb{R}$.
For convenience I’ll write $a_n$ for $f(n)$ and $\{x\}$ for the fractional part of $x$. For real $x$ let $g(x)=\lfloor x\rfloor\{x\}+1$. Then the recurrence can be written
$$a_n=\lfloor a_{n-1}\rfloor\{a_{n-1}\}+1=g(a_{n-1})$$
for $n\ge 2$, where $a_1=c$. Clearly $g(x)=1$ if $x\in\Bbb Z\cup[0,1)$, and it’s not hard to prove that $g$ is continuous from the right (because the floor function is).
Suppose that $n\le x<n+1$, where $n\in\Bbb Z$, so that $\lfloor x\rfloor=n$, and let $\alpha=\{x\}=x-n$; then
$$x-g(x)=n+\alpha-n\alpha-1=(n-1)(1-\alpha)\,.\tag{1}$$
This is $0$ iff $n=1$, so $g(x)=x$ iff $x\in[1,2)$. In particular, if $a_n\in[1,2)$ for some $n\in\Bbb Z^+$, then $a_k=a_n$ for all $k\ge n$.
We have now shown that if $a_n\in\Bbb Z\cup[0,2)$ for some $n\in\Bbb Z^+$, then $\langle a_k:k\in\Bbb Z^+\rangle$ is eventually constant: at $1$ if $a_n\in\Bbb Z\cup[0,1)$, and at $a_n$ if $a_n\in[1,2)$. In particular, this is the case if $c=a_1\in\Bbb Z\cup[0,2)$.
Now suppose that $c>2$; it follows from $(1)$ that $\langle a_n:n\in\Bbb Z^+\rangle$ is strictly decreasing as long as $a_n\ge 2$. Clearly $a_n\ge 1$ for all $n\ge 2$, so either some $a_n<2$, or $\langle a_n:n\in\Bbb Z^+\rangle$ converges monotonically down to some $a\ge 2$. But then
$$g(a)=\lim\limits_{n\to\infty}g(a_n)=\lim\limits_{n\to\infty}a_{n+1}=a\ge 2\,,$$
since $g$ is continuous from the right, so $a\in[1,2)$. This contradiction shows that some $a_n<2$, and it follows that the sequence is eventually constant at some value in $[1,2)$.
Now suppose that some $a_n<0$, and let $m=\lfloor a_n\rfloor$. Then $m<m\{a_n\}$, so $m+1<a_{n+1}$, and an easy induction shows that $0=m+|m|<a_{n+|m|}$. In particular, if $c<0$, there is an $n\in\Bbb Z^+$ such that $a_n>0$, and it follows again that the sequence is eventually constant at some value in $[1,2)$.