Generalizing Binomial Coefficients / Pascal's Triangle

[Leroy Quet]

An ascii-image for visualization:
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How about this generalization of binomial coefficients?:

Standard BCs:
If we had, as is arranged in the ascii-image above, a 'row' 
of one box with a porous bottom, below which is a row of 2 equal-sized 
porous-bottom boxes, below which is a row of 3 boxes, etc...
And we pour 2^(n-1) units of water into the top box, and the water drips 
down until all of it is in the boxes at the bottom of our triange at a 
row of n boxes (which contain the water),
then how many units (assuming the water dripped uniformly) are in each of 
the boxes of the bottom row?

Box j (where the farthest left box is box-0,
the farthest right box is box-{n-1})
contains C(n-1,j) units of water, in theory at least.
(C(n-1,j) = (n-1)!/(j!(n-1-j)!).)

But what if the k_th row (the top row, traditionally with 1 box, is row-0;
the row with traditionally (n+1) boxes is row-n), instead of (k+1) boxes,
contained
b(k) boxes (where {b(k)} is any sequence of positive integers)??

But the boxes in any row k are all 1/b(k) in width and are next to each 
other directly below the boxes of row-{k-1}.

I get that, recursively, the generalized C is such that:

C(n,j) =

sum{k = R to S}  C(n-1,k),

where R = floor(B(n-1)*j/B(n));

and S = floor((B(n-1)*j +B(n-1) -1)/B(n)).

What about a closed form? (at least for some interesting {B(k)}'s)

Specifically, what if B(k) = F(k), the k_th Fibonacci number?


What has been written before about such a generalization of binomial 
coefficients?

thanks,
Leroy Quet