What is the exact value of $\sqrt{-3+2\sqrt{-5+3\sqrt{-7+4\sqrt{-9+\dots}}}}$

Question

What is the exact value of $$R=\sqrt{-3+2\sqrt{-5+3\sqrt{-7+4\sqrt{-9+\dots}}}}$$
I tried to solve it like $\sqrt{2+\sqrt{2+\sqrt{2+\dots}}}$, i.e. I tried to find the sequence function for this expression. I got that for $x=1$ expression became $$R=x\sqrt{-2x-1+(x+1)\sqrt{-2x-3+(x+2)\sqrt{-2x-5+\dots}}}$$ Now it is obvious that $$f(x)=x\sqrt{-2x-1+f(x+1)}$$ for some function $f$ such that $R=f(1)$. Problem is how to solve this recurrence equation. After squaring both sides it become more complicated. Is there an easy way to solve this recurrence equation. If you have better solution, please explain how to find the exact value of $R$.

Answer 1

Nested Radical

$$ c_0 = (n + a)^2 \quad {\small\textit{added for conciseness}}\\ x + n + a = \sqrt{c_0 + ax + x\sqrt{c_0 + a(x + n) + (x + n)\sqrt{c_0 + a(x + 2n) + (x + 2n)\sqrt{...}}}} $$

This equation was discovered by Ramanujan.

Your equation is a case of this type of nested radical with $x = 2,\ n = 1,\ a = -2$.

$$ x + n + a = 1 $$

Answer 2

Maple agrees that this converges to $1$ (or something very close to $1$).
The plot shows $z_1=\sqrt{-3}$, $z_2=\sqrt{-3+2\sqrt{-5}}$ and so on (joined in order by line segmants):

added

Note that $f(x)=x^2$ solves the difference equation that you should use: $$ f(x) = x\sqrt{-2x-1+f(x+1)} $$