Dear Aldo,

Last night I converted my ideas concerning your counting problem into
\TeX form. This is the contents of the \AmSTeX-file below. Since I
did everything pretty fast there might be misprints. But I think that
the arguments are correct (only possibly incomplete). I do not know
whether this solution is satisfying for you. In any case I can give
some expression.

I gave you
an incomplete information. I mailed two copies of my manuscript to
Prof\. Herzog. For your convenience I have also sent the \AmSTeX-file
of this manuscript to you.

  With best wishes,

            Christian

----------------------------------------------------------------------
\input format
\NoBlackBoxes
\TagsOnRight
\def\th{\vartheta}
\def\bvert{\big\vert}
\def\Bvert{\Big\vert}
\def\Expr{\operatorname{Expr}}
\def\less{\operatorname{less}}
\def\xmaj{\operatorname{xmaj}}
\def\ymaj{\operatorname{ymaj}}
\def\stat{\operatorname{stat}}
\def\A{{\Cal A}}
\def\B{{\Cal B}}
\def\E{{\Cal E}}
\def\F{{\Cal F}}
\def\S{{\Cal S}}
\def\x{{\bold x}}
\def\y{{\bold y}}
\def\RR{{\frak R}}
\def\v#1{{\vert#1\vert}}
\def\V#1{{\Vert#1\Vert}}

\def\Krat{1}

Your counting problem is the following: Count nonintersecting lattice
paths (with steps in East- and South-direction) by East-South-turns, 
but with one difference. If the end point
of a path is reached by an East step, then the end point also is a
turn. In my language (my paths always run in North-East direction!)
this means counting nonintersecting lattice
paths (with steps in North- and East-direction) by North-East-turns, 
where the end point is also a turn if it is reached by a North step.

Actually, your end points are more special, they all lie on a
diagonal (antidiagonal in my formulation of the problem), but my
combinatorial proof of my counting formula can also be used to attack
this modified problem.

Again a very general result can be proved for {\it non-crossing}
lattice paths. All the notation refers to the manuscript \cite{\Krat} that I have
already sent.

\proclaim{Theorem 1}Let $\Cal A_i=(A_1^{(i)},A_2^{(i)})$ and $\Cal E_i=(E_1^{(i)},E_2^{(i)})$ be lattice points satisfying
$$
A_1^{(1)}+1\le A_1^{(2)}+2\le\dotsb\le A_1^{(r)}+r,\quad
A_2^{(1)}\ge A_2^{(2)}\ge \dotsb\ge A_2^{(r)},\tag1$$
and
$$E_1^{(1)}\le E_1^{(2)}\le \dotsb\le E_1^{(r)},\quad
E_2^{(1)}\ge E_2^{(2)}\ge \dotsb\ge E_2^{(r)}.\tag2$$
The generating function $\sum _{\Cal P} w_{\bold x,\bold y}(\Cal P)$
(where the end points also contribute to the weight according to
\cite{\Krat, (11)} if they are
reached by a North step)
where the sum is over all families $\Cal P=(P_1,\dots, P_r)$ of non-crossing lattice paths $P_i:\Cal A_i\to\Cal E_i$,
is
$$\det_{1\le s,t\le
r}\bigg(\left\{\matrix \format\c\quad &\l\\
1&s=t\text { and }A_2^{(s)}=E_2^{(s)}\\
y_{E_2^{(t)}}f_{s-t+1}(\x,A_1^{(s)},E_1^{(t)},
\y,A_2^{(s)}+1,E_2^{(t)}-1)&\text {otherwise}\endmatrix\right\}\bigg),\tag3$$
where $f_m(\x,a,b,\y,c,d)=\sum _{k} ^{}
e_{k+m}(x_a,\dots,x_b)e_k(y_c,\dots,y_d)$ with $e_n(z_1,\dots,z_h)$
denoting the elementary symmetric function in the variables
$z_1,\dots,z_h$.
\endproclaim

\remark{Remark} You may observe that the only differences with
Theorem~4 of \cite{\Krat} are the slightly more restrictive condition
(2) and the slightly modified result (3). The exceptional definition
for the case $s=t$ and $A_2^{(s)}=E_2^{(s)}$ has the following
explanation: First, it is indeed different from the other expression
since the other expression for $s=t$ and $A_2^{(s)}=E_2^{(s)}$ would
have to be taken as 0. Secondly, if there is an $i$ with 
$A_2^{(i)}=E_2^{(i)}$ then the generating
function decomposes into the product of the generating function for
the first $i-1$ paths times the generating function for the last $r-i$
paths since because of the location of starting and end points ((1)
and (2)) no path $P_j$, $j\ne i$, can cross $P_i$. Indeed, if
$A_2^{(i)}=E_2^{(i)}$ the
entries of the determinant (3) are $0$ whenever $s\le i\le t$ and
$s\ne t$. So the determinant has block form and decomposes into the
product of its $(1,\dots,i-1\mid 1,\dots,i-1)$-minor times its
$(i+1,\dots,r\mid i+1,\dots,r)$-minor, which perfectly corresponds to
the above decomposition of generating functions.
\endremark


\demo{Proof} How do array representations \cite{\Krat, (5)} of paths look
like if we take this modified notion of North-East turn into account?
It is very simple. A path $P:(A_1,A_2)\to(E_1,E_2)$ can be represented 
(compare \cite{\Krat, (5)})
$$\matrix\format\r\quad &\c&\c&\c&\c&\c&\quad \l\\
 A_1\le&a_1&\ a_2&\ \dots&\ a_{k-1}&\ a_k&\le E_1\\
A_2+1\le& b_1&\ b_2&\ \dots&\ b_{k-1}&\ E_2&\endmatrix\ .\tag5$$
$i=1,\dots,r$. Observe the differences with \cite{\Krat, (5)}: The
upper bound on the first row is one more, and the last entry in the
second row must be the $y$-coordinate of the end point! This makes
sense since if the end point is a turn the $x$-coordinate of this
turn equals $E_1$ and always the last turn of a path has
$y$-coordinate $E_2$. Of course the array representation is incorrect
for the case that $A_2=E_2$ where there is only one path between
$(A_1,A_2)$ and $(E_1,E_2)$, namely the horizontal path which would
correspond to the empty array. However, I
already discussed the case $A_2^{(i)}=E_2^{(i)}$ in the Remark. So we
may ignore $A_2^{(i)}=E_2^{(i)}$ for the following and hence assume
$A_2^{(i)}\ne E_2^{(i)}$ for all $i=1,2,\dots,r$.

Now we can start with the proof. It is easy to see that (3) is the
generating function
$$\sum _{(\Cal P,\si)} ^{}\sgn\si\, w_{\x,\y}(\Cal P),\tag6$$
where the sum is over all pairs $(\Cal P, \si)$ of permutations $\si$
in $\frak S$, the symmetric group of order $r$, and families $\Cal
P=(P_1,\dots,P_r)$ of two-rowed arrays, $P_i$ being of type $\si(i)-i$
 and the bounds for the entries of $P_i$ being as follows,
$$\matrix \format\r&\quad \c&\ \c\quad &\l \\
A_1^{(\si(i))}\le&\dots\ a_{k_i-1}^{(i)}&a_{k_i}^{(i)}&\le E_1^{(i)}\\
A_2^{(\si(i))}+1\le&\dots\ b_{k_i-1}^{(i)}&E_2^{(i)}&
\endmatrix,\tag7$$
$i=1,\dots,r$. (Compare with \cite{\Krat, (21), (22)})

The nice thing is that we can use the same involution (24--27) here.
This is because in our situation (meaning: because of the slightly
more restrictive inequalities (2) for the coordinates of the end
points) this involution never moves the last entry in a second row.

Let $P_i$ and $P_{i+1}$ be given by
$$\matrix\format\r&\quad \c\ &\c\ &\c\ &\c\ &\c\ &\c\quad &\l\\
 A_1^{(\si(i))} \le&\dots&a_{I-1}&a_I&\dots&a_{k_i-1}&a_{k_i}&\le E_1^{(i)}\\
 A_2^{(\si(i))}+1 \le&\dots&b_{I-1}&b_I&\dots&b_{k_i-1}&E_2^{(i)}&
\endmatrix,\tag8$$
and $P_{i+1}$ be given by
$$\matrix\format\r&\quad \c&\ \c&\ \c&\ \c&\ \c&\ \c&\ \c\quad &\l\\
 A_1^{(\si(i+1))} \le&\innerhdotsfor2\after\quad 
&c_{J}&c_{J+1}&\dots&c_{k_{i+1}-1}&c_{k_{i+1}}&\le E_1^{(i+1)}\\
 A_2^{(\si(i+1))}+1 \le&\dots&d_{J-1}&d_J&\innerhdotsfor2\after\
&d_{k_{i+1}-1}&E_2^{(i+1)}&
\endmatrix,\tag9$$
with crossing point $S=(a_I,d_J)$, 
$$\gather  c_J<a_I\tag10a\\
b_{I-1}<d_J\tag10b
\endgather$$
and
$$1\le I\le k_i+1,\quad  0\le J\le k_{i+1}.\tag10c$$
(Compare \cite{\Krat, (24--26)}.) These two two-rowed arrays are
mapped to (compare \cite{\Krat, (27)})
$$\matrix \format\c\ &&\c\ \\
\dots&c_J&a_I&\dots&a_{k_i}\\
\dots&d_{J-1}&b_I&\dots&b_{k_i}\endmatrix,\tag11a$$
$P_{i+1}$ being given by
$$\matrix \format\c &&\ \c\\
\innerhdotsfor2\after\ &a_{I-1}&c_{J+1}&\dots&c_{k_{i+1}}\\
\dots&b_{I-1}&d_J&\innerhdotsfor2\after\
&d_{k_{i+1}}\endmatrix.\tag11b$$

The last entry in the second row of $P_{i+1}$,
$d_{k_{i+1}}=E_2^{(i+1)}$,
never can move since $J\le k_{i+1}$. When does $b_{k_{i}}=E_2^{(i)}$
move? Only in case that $I=k_i+1$. However, then, because of (10b) we have 
$E_2^{(i)}=b_{k_i}<d_J$, because of (9) we have $d_J\le E_2^{(i+1)}$,
and because of (2) we have $E_2^{(i+1)}\le E_2^{(i)}$. This is a
contradiction. Hence indeed the last entry in a second row never
moves, and so our involution maps pairs of two-rowed arrays (8), (9)
again to pairs of arrays under consideration. \quad \quad \qed

Now we may perform the usual ''unshifting``, 
i.e\. by replacing $A_1^{(i)}$ by $A_1^{(i)}-i+1$, $A_2^{(i)}$ by
$A_2^{(i)}+i-1$, etc\., and set $x_i=y_i=\sqrt z$ to get:
\proclaim{Corollary} Let 
$\Cal A_i=(A_1^{(i)},A_2^{(i)})$ and $\Cal E_i=(E_1^{(i)},E_2^{(i)})$ be lattice points satisfying
$$A_1^{(1)}\le A_1^{(2)}\le\dotsb\le A_1^{(r)},\quad
A_2^{(1)}>A_2^{(2)}>\dotsb>A_2^{(r)},$$
and
$$E_1^{(1)}< E_1^{(2)}<\dotsb< E_1^{(r)},\quad
E_2^{(1)}> E_2^{(2)}> \dotsb> E_2^{(r)}.$$
The NE-turn generating function $\sum _{\Cal P} ^{}z^{\text
{turns$(\Cal P)$}}$ (in our modified sense) 
for all families $\Cal P=(P_1,\dots,P_r)$ 
of nonintersecting lattice paths $P_i:\Cal A_i\to\Cal E_i$, is
$$\det_{1\le s,t\le r}
\bigg(\sum _{k} ^{}\binom{E_1^{(t)}-A_1^{(s)}+s-t+1}{k+s-t+1}
\binom{E_2^{(t)}-A_2^{(s)}-s+t-1}{k}z^{k+1}\bigg).\tag12$$
Again, the $s=t$-entry in case $A_2^{(s)}=E_2^{(s)}$ has to be taken
as 1.\quad \quad \qed
\endproclaim
Therefore a solution for your problem is something like
$$\sum _{E_1<\dots<E_r} ^{}
\det_{1\le s,t\le r}
\bigg(\sum _{k} ^{}\binom{E_t-A_1^{(s)}+s-t+1}{k+s-t+1}
\binom{-E_t-A_2^{(s)}-s+t-1}{k}z^{k+1}\bigg).\tag13$$
(Here the end points have the form $(E_i,-E_i)$.)

An alternative form would be a pfaffian which is obtained by
modifying Stembridge's "pfaffian theory``. The result would be
something like
$$\text {pf}(Q(A_1^{(i)},A_2^{(i)},A_1^{(j)},A_2^{(j)})),\tag14$$
where
$$\multline Q(A_1^{(i)},A_2^{(i)},A_1^{(j)},A_2^{(j)})\\
=\sum _{E_1<E_2} ^{}\det_{1\le s,t\le 2}\bigg(\sum _{k} ^{}
\binom{E_t-A_1^{(\nu(s))}+s-t+1}{k+s-t+1}
\binom{-E_t-A_2^{(\nu(s))}-s+t-1}{k}z^{k+1} \bigg),\\
\endmultline\tag15$$
where $\nu(1)=i$ and $\nu(2)=j$. Also here the exceptional definition
for entries has to be taken into account.
I have to check if any
simplification of the sums in (15) are possible.


\enddemo

\Refs

\ref\no \Krat\by C. Krattenthaler\paper Counting nonintersecting
lattice paths with turns \endref
\endRefs

\end
