What are some "natural" interpolations of the sequence $\small 0,1,1+2a,1+2a+3a^2,1+2a+3a^2+4a^3,\ldots $? (This is a spin-off of a recent question here)
In fiddling with the answer to that question I came to the set of sequences    
$\qquad \small \begin{array} {llll}
A(1)=1,A(2)=1+2a,A(3)=1+2a+3a^2,A(4)=1+2a+3a^2+4a^3, \ldots \\
B(1)=1,B(2)=1+3a,B(3)=1+3a+6a^2,B(4)=1+3a+6a^2+10a^3, \ldots \\
C(1)=1,C(2)=1+4a,C(3)=1+4a+10a^2,C(4)=1+4a+10a^2+20a^3, \ldots \\
\ldots \\
 \end{array} $
with some indeterminate a .         
We had the discussion often here in MSE, that interpolation to fractional indexes, say A(1.5)=?? is arbitrary, considering, that an initial solution composed with any 1 -periodic function satisfies the condition. But here the embedding in a set of sequences, which are constructed from binomial-coefficients might suggest some "natural" interpolation, such as for      
$\qquad \small K(1)=1, K(2)=1+a, K(3)=1+a+a^2, \ldots $     
the interpolation $\small K(r) = {a^r-1 \over a-1}$ seems the most "natural" which even can smoothly be defined for a=1. This observation made me to refer to "q-analogues" $\small [r]_a $ in my answer in the initiating MSE-question, but it's not obvious how to interpolate the shown sequences of higher orders A , B , C (I think they're not related to the "q-binomial-analogues" , for instance ).   
Q: So what would be some "natural" interpolation to fractional indexes for the sequences A, B, C, and possibly in general for sequences generated in the obvious generalized manner? 

Agreeing mostly with Henning's ansatz I got now the general form as
$$ A_m(n) = {1 \over (1-a)^m} - \sum_{k=0}^{m-1} \binom{n+m}{k}{a^{n+m-k} \over (1-a)^{m-k} } $$
I do not yet see, whether some examples of fractional indexes agree with the solutions of all three given answers so far, for instance: given a=2.0 what is A(1.5), B(4/3), C(7/5)? With my programmed version I get now 
$\qquad \small  A(1.5)\sim 9.48528137424  $
$\qquad \small  B(4/3) \sim 11.8791929545  $
$\qquad \small  C(7/5) \sim 18.4386165488  $
(No interpolation for fractional m yet)

[update 2] the derivative-versions of Sivaram and Michael arrive at the same values so I think, all versions can be translated into each other and mutually support each other to express a "natural" interpolation.
[update 3] I had an index-error in my computation call. Corrected the numerical results.

I'll build from Michael's work (thanks for doing the heavy lifting!) and start with
$$A_n(r)=\frac1{n!}\frac{\mathrm d^n}{\mathrm da^n} \frac{a^{r+n}-1}{a-1}$$
Let's switch back to the series representation, and swap summation and differentiation:
$$A_n(r)=\frac1{n!}\sum_{k=0}^{r+n-1} \frac{\mathrm d^na^k}{\mathrm d a^n}$$
and make a suitable replacement:
$$A_n(r)=\frac1{n!}\sum_{k=0}^{r+n-1} \frac{k!a^{k-n}}{(k-n)!}=\frac1{n!}\sum_{k=-n}^{r-1} \frac{(k+n)!a^k}{k!}=\frac1{n!}\sum_{k=0}^{r-1} \frac{(k+n)!a^k}{k!}$$
where a reindexing and removal of extraneous zero terms was done in the last two steps.
The result can then be expressed as:
$$A_n(r)=\frac{\mathrm{I}_{1-a}(n+1,r)}{(1-a)^{n+1}}=\frac{1-\mathrm{I}_a(r,n+1)}{(1-a)^{n+1}}$$
where $\mathrm{I}_z(r,n)$ is a regularized incomplete beta function.
In terms of the Gaussian hypergeometric function, we have
$$\begin{align*}
A_n(r)&=\frac{\mathrm{I}_{1-a}(n+1,r)}{(1-a)^{n+1}}\\
&=\frac{a^r}{(n+1)\mathrm{B}(n+1,r)}{}_2 F_1\left({{n+r+1, 1}\atop{n+2}}\mid1-a\right)\\
&=\binom{n+r}{n+1} {}_2 F_1\left({{1-r,n+1}\atop{n+2}}\mid1-a\right)
\end{align*}$$
where the third one is derived from the second one through the Pfaff transformation. In particular, the third one gives the representation
$$A_n(r)=r\binom{n+r}{r}\sum_{k=0}^{r-1}\binom{r-1}{k}\frac{(a-1)^k}{n+k+1}$$
which is valid for arbitrary $n$ and nonnegative integer $r$.
Other hypergeometric representations can be derived.
