Number of Taxicab routes in a triangular city I am assuming a triangle that is "almost" half a rectangular city with taxicab geometry. I am trying to find the number of paths in this triangular city.
Assuming that the ride starts from the corner of the city. If we move p steps in one direction, and q steps in the perpendicular direction, the number of paths in case of a rectangular city is known and is given by:
$$\binom{p+q}{p} ~ or ~ \binom{p+q}{q}$$
For example, assume the 45-degree rotated city in the following figure (left).
Complete and half cities

If we start from the point where the arrow is pointing, the numbers at the cross-points refer to the number of possible paths from the start point to each cross-point.
Now, assume the figure on the right side. Again, the numbers at the cross-points refer to the number of possible paths from the start point to each cross-point. I obtained these numbers using a combination of counting and observation.
The main observation is that the number of paths in the triangular city is a ratio of the rectangular city. Take for example the 7'th row, we find the following (Can you explain why?):
924/(7/1)=132
462/(7/2)=132
210/(7/3)=90
84/(7/4)=48
28/(7/5)=20
7/(7/6)=6
1/(7/7)=1
This applies to all rows.
Now to my question. Assume the following shape of a city, where the inlets are at the left edge, and the outlets are at the bottom edge.
My problem

What I want to do is to find the number of paths from any of the inlets to any of the outlets. Hopefully a formula, and a proof.
In Figure 2 is what I got so far. If the outlet is less than or equal to the input, this can be directly obtained using the formula for the rectangular case.
Assuming that this is correct, note that up to the diagonal, numbers are following the rules of Pascal's triangle. After the diagonal, which represents the boundary of the city, it does not follow the same rules, but there is a pattern.
1st diagonal after the half (subtract 1)
6= (6+1)-1
20=(6+15)-1
34=(15+20)-1
2nd (subtract 6)
20=(20+6)-6
48=(20+34)-6
62=(34+34)-6
3rd (subtract 20)
4th (subtract 48)
5th (subtract 90)
6th (subtract 132)
Which are the numbers in the row corresponding to inlet 7 (Again, can you explain why?).
 A: Imagine connecting your "inlets" together to the left of your city and the "outlets" at the bottom, like this (shown for 4 of each instead of 7 of each):
*
|
+-->:
|   ::.
+-->::::
|   :::::. <- dotted area corresponds to your drawing
+-->:::::::
|   ::::::::.
+-->::::::::::
    v  v  v  v
    |  |  |  |
    +--+--+--+--*

Then each route from some inlet to some outlet corresponds to a route from * to * in the expanded graph. But the expanded graph is just a "half-city" of size 2 more than the one you started with, except that the bottom left intersection is removed.
So the number of solutions is one less than the number from top to bottom in a "half-city" of size $n+2$.
The number of paths through a "half-city" of size $n$ is the $n$th Catalan number -- what you're counting is the number of strings of $n$ "left"s and $n$ "right"s such that no prefix contains more rights than lefts.
A: The sequence $1,6,20,48,90,132,132$    can be generated as $\displaystyle {5+i \choose i-1}\dfrac{8-i}{7}$ with $i$ running from $1$ to $7$, though it would be slightly more conventional to write $\displaystyle {n+k \choose k}\dfrac{n-k+1}{n+1}$ with $n=6$ and $k$ running from $0$ to $6$.
The right hand part of the first diagram is known as Catalan's triangle, with the sequence $1,1,2,5,14,42,132,\ldots$ being known as Catalan numbers: they appear frequently in mathematics.
