Show that for any $n$, $sum_{k=0}^infty frac{1}{3^k}binom{n+k log(3)/log(2)}{k}=c 2^n$ for some constant $c$

Let $0 < b < 2$, $0 < a < 2^{-b}$, and $t > 0$. We will evaluate the sum $$sum_{k=0}^infty a^k binom{bk + t}{k}$$ and will use the result to prove @Bartek's conjecture.

Recall that for any $alpha in mathbb{C}$, the function $z mapsto z^alpha := e^{alpha log z}$, where $log$ denotes the principal branch of the logarithm, is defined and holomorphic on $mathbb{C} setminus (-infty, 0]$, and satisfies $z^{alpha + beta} = z^alpha z^beta$ and $z^{kalpha} = (z^alpha)^k$ when $k$ is a positive integer. For $alpha > 0$, we can define $z^alpha = 0$ at $z = 0$ by continuity, and we have $$(1 + z)^alpha = sum_{n=0}^infty binom{alpha}{n} z^n$$ on the closed unit disk, and the series converges absolutely there (see e.g. here). Using this expression, one can write $$binom{alpha}{n} = frac{1}{2pi i} int_C frac{(1 + z)^alpha}{z^{n+1}} ,dz$$ where $C$ is the unit circle, with a small indent to avoid the point $z = -1$. This means $$binom{bk + t}{k} = frac{1}{2pi i} int_C frac{(1 + z)^{bk+t}}{z^{k+1}} ,dz.$$ Now, assuming the indent of $C$ is small enough so that $r leq |z| leq 1$ on $C$ for some $r > 2^b a$, we have $|frac{a(1+z)^b}{z}| leq r^{-1}2^b a < 1$ on $C$, hence the series $$sum_{k=0}^infty frac{a^k (1 + z)^{bk+t}}{z^{k+1}} = sum_{k=0}^infty frac{(1 + z)^t}{z}left(frac{a(1 + z)^b}{z}right)^k = frac{frac{(1+z)^t}{z}}{1 - frac{a(1+z)^b}{z}} = frac{(1+z)^t}{z - a(1 + z)^b}$$ converges absolutely on $C$, uniformly, and thus we have $$frac{1}{2pi i} int_C frac{(1+z)^t}{z - a(1 + z)^b} ,dz = sum_{k=0}^infty frac{1}{2pi i} int_C frac{a^k(1 + z)^{bk+t}}{z^{k+1}} ,dz = sum_{k=0}^infty a^k binom{bk+t}{k}.$$ It remains to evaluate the integral on the left. Note that $(1 + z)^t$ and $z - a(1 + z)^b$ are holomorphic on an open set containing $C$ and its interior.

Lemma: $z - a(1 + z)^b$ has a unique zero $r$ in the closed unit disk, and this satisfies $0 < r < 1$.

Proof: Suppose $z_0$ is a zero in the closed unit disk. Clearly $z_0$ cannot be negative or zero. Let $arg : mathbb{C} setminus (-infty, 0] to (-pi, pi)$ denote the argument function. Note that for $c > 0$ we have $arg(cz) = arg(z)$, $|arg(c + z)| leq |arg (z)|$, and if $c arg (z) in (-pi, pi)$, then $arg (z^c) = c arg (z)$. Also note that when $|z| = 1$, $arg(1 + z) = frac{1}{2} arg(z)$. Using these facts, we have begin{align*} |arg(1 + z_0)| &= |arg(1 - |z_0| + (|z_0| + z_0))| \ &leq |arg(|z_0| + z_0)| \ &= |arg(1 + z_0/|z_0|)| \ &= (1/2)|arg(z_0)| end{align*} which is less than $pi/2$, hence since $b < 2$, we have $$|arg(a(1 + z_0)^b)| = b|arg(1 + z_0)| leq frac{b}{2} |arg(z_0)|$$ which is strictly less than $|arg(z_0)|$ when $arg(z_0) neq 0$, so in that case we cannot have $z_0 = a(1 + z_0)^b$. Thus any zero must be positive real. But $x - a(1+x)^b$ has positive derivative on $[0, 1]$, and is negative at $x = 0$ and positive at $x = 1$, so there is a unique $r$ in $(0, 1)$ with $r - a(1+r)^b = 0$. [end proof.]

One can check that the residue of $frac{(1+z)^t}{z - a(1+z)^b}$ at $z = r$ is $frac{(1 + r)^t}{1 - ab(1 + r)^{b-1}} = frac{(1+r)^t}{1 - frac{br}{1+r}} = frac{(1+r)^{t+1}}{1 + r - br}$, and since $z = r$ is the unique pole of this function in $C$, we have

$$sum_{k=0}^infty a^k binom{bk+t}{k} = frac{1}{2pi i} int_C frac{(1+z)^t}{z - a(1 + z)^b} ,dz = frac{(1+r)^{t+1}}{1 + r - br}.$$

We can also run this process backwards: for $0 < r < 1$, and a given $0 < b < 2$, $frac{r}{(r+1)^b}$ is a strictly increasing function of $r$, so $frac{r}{(r+1)^b} < frac{1}{(1+1)^b} = 2^{-b}$, and thus setting $a = frac{r}{(r+1)^b}$, so $r = a(1+r)^b$, by the above we have $$sum_{k=0}^infty left(frac{r}{(r+1)^b}right)^k binom{bk+t}{k} = frac{(1+r)^{t+1}}{1 + r - br}.$$ Because $frac{r}{(r+1)^b}$ is increasing on $[0, 1]$, we can take the limit as $r to 1$ of both sides, giving $$sum_{k=0}^infty 2^{-bk} binom{bk+t}{k} = frac{2^{t+1}}{2-b}$$ as desired.

This is a bit too long for a comment but if we consider a function: $$f(x)=sum_{k=0}^{infty}a^k{bk+x choose k}$$ Then because of the identity: $${n+1 choose k}={n choose k} + {n choose k-1}$$ We have the following: $$f(x+1)=sum_{k=0}^{infty}a^k{bk+x+1 choose k}=sum_{k=0}^{infty}a^kleft[{bk+x choose k}+{bk+x choose k-1}right]=f(x)+sum_{k=0}^{infty}a^{k+1}{bk+x+b choose k}=f(x)+af(x+b)$$ Which is a functional equation that would be perfectly satisfied by the family: $$f(x)=c cdot d^x$$ Of exponential functions where the exponent $d$ would have to satifsy: $$d=1+ad^b$$ Which in fact is satisfied by $d=2$ when $a=frac{1}{3}$ and $b=log_23$. This equation seems however to not be enough to determine the function uniquely (even if we have that $f$ is increasing and continuous which should be easy to prove).
EDIT
After playing with the expression for a while I've noticed that although in general $f$ behaves exponentially only asymptotically when $d=2$ is the solution (equivalently when $a=2^{-b}$) $f$ becomes actually exponential. Moreover, the constant seems to be equal to $frac{2}{2-b}$. Thus, I conjecture: $$sum_{k=0}^{infty}2^{-bk}{bk+x choose k} = frac{2^{x+1}}{2-b}$$ For all real $x$ and all $0 le b le 2$ (outside this interval the sum does not converge).