Ok first some definitions:
Let a shifted diagram of some strict partition $\lambda$ be a Young tableau whose $i^{th}$ row is shifted $i-1$ spaces to the right, (I use french notation, and start counting rows at the bottom).
Let a standard shifted tableau be a filling of the shifted diagram using every digit from 1 to n, where $\lambda \vdash n$, such that entries are increasing from left to right along rows, and from bottom to top up columns.
The descent set of some filling $T\in SShT(\lambda)$ is all $i\in T$ such that $i+1$ is strictly above $i$, and the peak set is all $i\in Des(T)$ such that $i\neq 1$ and $i-1\notin Des(T)$.
What I want to do is count the number of ways that a specific peak set may occur within standard shifted tableaux of height 2, i.e. for $\lambda = (n-k,k), k<n/2$.
I have shown that the total number of fillings for such a diagram is given by ${n-2\choose k} - {n-2\choose k-2}$. I do this by showing such fillings are in bijection with Dyck paths from $(0,0)$ to $(n-2-k,k)$ only moving (1,0) or (0,1) and never touching the line $y=x+2$. The reason is that 1 and 2 must go in the first two boxes of the bottom row, and then we have to place $n-2$ numbers, $k$ of which go in the top row, and the top row numbers must be greater than the one directly below, which gives the not-touching-the-line condition.
What I want to do is try to enumerate the possible fillings which give a specific peak set. For example, among $SShT(7,5)$ the peak set $\{3,7,10\}$ may occur in 3 ways, as the following:
Obviously for some general peak set $\{p_1,...,p_k\}$ in this situation it must be the case that $p_i$ corresponds to a (1,0) path (unless it is 2, in which case we don't draw it) and $p_i+1$ must be a $(0,1)$ path, so essentially a peak forces an angled path of length 2 (1,0),(0,1) to occur. Also, the only time we can have any upwards paths is following a peak, since otherwise we would introduce a new peak.
There are some other obvious restrictions like if $p_1=2$ then $p_1+2$ cannot be $(0,1)$, or indeed if $p_1,p_2,p_3,...,p_i$ are all "minimal" in a sense (i.e. 2,4,6,...) then there is no freedom to choose anything until $p_i+2$.
Aside from writing down some rules and working out some examples I don't think I am very close to finding a nice closed form for this, does it even seem possible? I don't have much experience in this type of counting argument and so I am not sure if this is likely to work out nicely or not. Any hints or suggestions would be very much appreciated, or if you can see some reason why this won't work out nicely that too would be good to know!
Thanks very much.
