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 $c$ I stumbled upon this from playing around and it seems to be true but I don't know how to approach it.
 A: 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 $z^{k\alpha} = (z^\alpha)^k$ 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
$$\binom{\alpha}{n} = \frac{1}{2\pi i} \int_C \frac{(1 + z)^\alpha}{z^{n+1}} \,dz$$
where $C$ is the unit circle, with a small indent to avoid the point $z = -1$. This means
$$\binom{bk + t}{k} = \frac{1}{2\pi i} \int_C \frac{(1 + z)^{bk+t}}{z^{k+1}} \,dz.$$
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
$$\frac{1}{2\pi i} \int_C \frac{(1+z)^t}{z - a(1 + z)^b} \,dz = \sum_{k=0}^\infty \frac{1}{2\pi i} \int_C \frac{a^k(1 + z)^{bk+t}}{z^{k+1}} \,dz = \sum_{k=0}^\infty a^k \binom{bk+t}{k}.$$
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
$$\sum_{k=0}^\infty a^k \binom{bk+t}{k} = \frac{1}{2\pi i} \int_C \frac{(1+z)^t}{z - a(1 + z)^b} \,dz = \frac{(1+r)^{t+1}}{1 + r - br}.$$
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.
A: 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:
$$f(x+1)=\sum_{k=0}^{\infty}a^k{bk+x+1 \choose k}=\sum_{k=0}^{\infty}a^k\left[{bk+x \choose k}+{bk+x \choose k-1}\right]=f(x)+\sum_{k=0}^{\infty}a^{k+1}{bk+x+b \choose k}=f(x)+af(x+b)$$
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).
