Generalizations of $\sum_{m=3n+2}^{\infty}\phi^m=\phi^{3n}$ and $\sum_{m=13n+1}^{\infty}(\sqrt2-1)^m=\dfrac{(\sqrt2-1)^{13n}}{\sqrt2}$ I noticed that the following identies hold with the help of wolfram alpha and oeis. I'm sure they're well-known, but I'd like to know how they generalize.
$$\sum_{m=3n+2}^{\infty}\phi^m=\phi^{3n}=(-1)^n(a_n-b_n\sqrt5),$$
where $\phi=\dfrac{\sqrt5-1}2,a_0=1,b_0=0,$ and $a_n$ and $b_n$ are the numerator and denominator respectively of the $n$th convergent of the continued fraction $\sqrt5=2+\dfrac1{4+\dfrac1{4+\dfrac1{4+\cdots}}}$
Similarly,
$$\sum_{m=13n+1}^{\infty}(\sqrt2-1)^m=\dfrac{(\sqrt2-1)^{13n}}{\sqrt2}=(-1)^n\dfrac{c_{13n}-d_{13n}\sqrt2}{\sqrt2},$$
where $c_0=1,d_0=0,$ and $c_n$ and $d_n$ are the numerator and denominator respectively of the $n$th convergent of the continued fraction $\sqrt2=1+\dfrac1{2+\dfrac1{2+\dfrac1{2+\cdots}}}$
Do similar identities exist for other values of $\dfrac{\sqrt{k^2+4}-k}2$ for positive integer $k$ (the previous examples are the cases $k=1,2$ respectively)?

Well, I came up with a much better solution than my previous one, but it's too substantial to add to my old answer.
We are looking for irrational numbers $x$ with convergents $\frac{p_n}{q_n}$ such that, if we define $f_x(n)=p_n-q_nx$, there exists some real number $c$ such that $f_x(n+1)=cf_x(n)$. Notice that, according to Wikipedia, if $x$ has continued fraction $[a_0;a_1,a_2,\ldots]$, then the following recurrences hold
$$p_n=a_np_{n-1}+p_{n-2}$$
$$q_n=a_nq_{n-1}+q_{n-2}$$
So, using these, we have
$$f_x(n+2)=a_{n+2}f_x(n+1)+f_x(n)$$
$$c^2f_x(n)=ca_{n+2}f_x(n)+f_x(n)$$
Or, since $f_x(n)$ can never be $0$ since $x$ is irrational, we divide out $f_x(n)$ and receive
$$c^2-ca_{n+2}-1=0$$
$$\frac{c^2-1}{c}=a_{n+2}$$
which must hold for any $n\geq 0$, so the continued fraction of $x$ must be of the form $[y;z,k,k,\ldots]$ for integers $x$, $y$, and $z$. In particular, since the continued fraction of $\phi_k=\frac{\sqrt{k^2+4}-k}2$ is $[0;k,k,\ldots]$ since is the unique value in $[0,1)$ satisfying $\phi_k=\frac{1}{k+\phi_k}$, this means that any real $x$ such that the "residue" of the convergents $f_x(n)$ is an exponential function must be of the form
$$x=y+\frac{1}{z+\phi_k}.$$
In fact, every such $x$ satisfies the condition; to show that, note that, here $f_x(0)=\frac{1}{z+\phi_k}$ and $f_x(1)=1-\frac{z}{z+\phi_k}$ and that the condition $cf_x(0)=f_x(1)$ holds whenever $c=-\phi_k$, which is obvious, given that $c$ satisfies $c^2-kc-1=0$ and $\phi_k$ satisfies $\phi_k^2+k\phi_k-1=0$ and both are the unique roots of their respective polynomials in $(-1,1)$. Thus, inductively, we have that $x=y+\frac{1}{z+\phi_k}$ is a necessary and sufficient condition for $f_x(n+1)=cf_x(n)$ for all $n$, and moreover, this implies
$$f_x(n)=(-\phi_k)^nf(0)=(-\phi_k)^n\frac{1}{z+\phi_k}.$$
Knowing this, and the fact that a geometric sum is also a exponential function, your identities may be proven with ease and further similar ones can be found.
I found an infinite family of solutions using Mathematica to look for solutions to the following equation $f(0)=\frac{1}{z+\phi_k}=\frac{\phi_k^a}{1-\phi_k}=\sum_{m=a}^{\infty}\phi_k^m$ for integer $a$ and $z$. There could be identities not found by solving this equation, but from the solution this has for $a=1$ and $z=k-1$ we get the following family: Let $X_k=\frac{1}{k-1+\phi_k}$ (or that, plus any integer; turns out the integer term doesn't matter) and $p_n/q_n$ be the convergents thereof in lowest form. Then
$$\sum_{m=n+1}^{\infty}\phi_k^m=(-1)^{n+1}(p_n-q_nX_k).$$
I also found ($a=2$,$z=3$) that for $X=\frac{1}{3+\phi_2}=1-\frac{1}{\sqrt{2}}$ with convergents $p_n/q_n$, we have
$$\sum_{m=n+2}\phi_2^m=(-1)^{n+1}(p_n-q_nX).$$
I've not found any other solutions, aside from the ones you list.
