TITLE: Generalised Pascal Triangle with two-periodic formula
QUESTION [8 upvotes]: Consider a generalisation of pascal's triangle of the following form
$$1 \\ 1 \ 1 \\ 1\ 2\\ 1\ 3\ 2\\ 1\ 4\ 5\\ 1\ 5\ 9\ 5$$
That is to say, for even-numbered rows the $n$-th term is found by adding the $(n-1)$-th and $n$-th terms of the previous row, and for odd-numbered rows, the rule is similar, except that no new term is added on the end at all. (For example, there is no $1$ on the end of the third row above.)
Is it possible to find a explicit expression for the entries of this triangle? 
Motivation: This problem is related to the following problem of enumerative geometry: 'Let $n\geq 2$, given $(n-2)$-dimensional subspaces $H_1,\ldots, H_{2(n-2)}$ of $\mathbb{P}^n$ how many lines are there, in general, intersecting all spaces?’

REPLY [7 votes]: If you start with Pascal's triangle
1
1  1
1  2  1
1  3  3  1
1  4  6  4  1
1  5 10 10  5  1
1  6 15 20 15  6  1    

precede and follow with $0$s
0  1  0
0  1  1  0
0  1  2  1  0
0  1  3  3  1  0
0  1  4  6  4  1  0
0  1  5 10 10  5  1  0
0  1  6 15 20 15  6  1  0    

take differences between consecutive terms
1 -1
1  0 -1
1  1 -1 -1
1  2  0 -2 -1
1  3  2 -2 -3 -1 
1  4  5  0 -5 -4 -1
1  5  9  5 -5 -9 -5 -1

and finally drop the zero and negative values in the right-hand half
1 
1 
1  1 
1  2 
1  3  2  
1  4  5 
1  5  9  5 

you get your triangle.
So, one possibility for a formula is $${n \choose m}-{n \choose m-1}$$ assuming you start counting rows and columns that way starting from $0$.  For example the $9$ here is ${6 \choose 2}-{6 \choose 1} = 15-6$