Rather than come up with a cumbersome explanation of what I mean, here is a picture of the beginning of the triangle:

$$\begin{matrix} 1 & & & \\ 0 & 1 \\ 1 & 0 & 1 \\ 0 & 2 & 0& 1\\ 2 & 0 & 3 & 0 & 1 \end{matrix} $$

So, as I've suggested in the title, it is similar to Pascal's triangle, but we ignore any entries to the left of the center, and don't let them contribute to subsequent entries. I'm looking to find a closed formula for the entries. Let's call the entry in the $n^{th}$ row and $k^{th}$ column $a_{n,k}$. Then it's clear that if $2\nmid n+k$, we have that $a_{n,k}=0$.

I was trying to think of a combinatorial interpretation of this, based on the fact that we can think of the $n,k$ entry of Pascal's triangle as the coefficient of $x^ky^{n-k}$ in $(x+y)^n$. Our $a_{n,k}$ correspond to the number of ways of writing $x^k$ from $(x+1/x)^n$ where if we think of choosing what we multiply out one by one, at each step we have chosen more $x$'s than $1/x$'s. This didn't lead me anywhere however.

Any suggestions are appreciated. Admittedly it's been quite a while since I've done any combinatorics.


2 Answers 2


This triangle, read by rows, forms the OEIS sequence A053121. This fantastic site gives (among other things) explicit expressions for elements. Here is one of them:

$a(n, m) := 0$ if $n < m$ or $n-m$ odd, else $a(n, m) = (m+1)\cdot\text{binomial}(n+1, (n-m)/2)/(n+1)$


Seems to be a version of the Catalan triangle. Or maybe this is closer.

  • $\begingroup$ I've actually never used that site before, thanks! $\endgroup$
    – user16544
    Nov 7, 2011 at 3:19

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