# Generalised Pascal Triangle with two-periodic formula

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?’

• – Lord Shark the Unknown Dec 20 '19 at 7:21
• See also the Wikipedia entry on Catalan's triangle. The table provided does include your triangle values, but with your rows appearing as rising diagonals. They moreover give a formula for the entries, suggesting a fairly definitive answer to your problem. – Semiclassical Dec 20 '19 at 7:25

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


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 $${6 \choose 2}-{6 \choose 1} = 15-6$$
As the comment by Semiclassical indicates, this is highly relevant to Catalan's triangle, defined by the recurrences $$C(n+1,k)=C(n+1,k-1)+C(n,k)$$ for $$1, $$C(n+1,n+1)=C(n+1,n)$$ for $$n\geq1$$, and the boundary conditions $$C(n,0)=1$$ for $$n\geq0$$ and $$C(n,1)=n$$ for $$n\geq1$$. There is an explicit formula for $$C(n,k)$$ given by $$C(n,k)=\frac{(n+k)!(n-k+1)}{k!(n+1)!}.$$ Now, if the $$k$$th entry in the $$n$$th row of your triangle is called $$D(n,k)$$, then it is easy to notice that $$D(n,k)=C(n-k+1,k-1).$$ (You can convince yourself by looking at how $$D(n,k)$$ are entries along the rising diagonal on the table of values provided in the Wikipedia page, or by verifying the defining recurrence is satisfied like so.) Consequently we have $$\boxed{D(n,k)=C(n-k+1,k-1)=\frac{n!(n-2k+3)}{(k-1)!(n-k+2)!}}.$$