TITLE: Generalization for Catalan number
QUESTION [6 upvotes]: I know that the number of ways to open and close $n$ parenthesis is the $n$-th Catalan number. What about the number of sequences of open and closed parenthesis of length $k$ that contains exactly $n$ well open and closed parenthesis?
Example: ))())() is a sequence of length 7 that contains exactly 2 well open and closed parenthesis.
Basic fact:
The number of sequences of open and closed parenthesis of length $k$ is equal to $2^k;$
The number of sequences of open and closed parenthesis of length $k$ that contains exactly $n$ well open and closed parenthesis is obviously greater that the the $n$-th Catalan number, but which is the precise value?

REPLY [9 votes]: We  start by  computing the  relevant generating  functions. Taking  a
binary string  or string  of parentheses  we extend  all well-balanced
subsequences to be  maximal. This leaves additional  filling which has
the property that it may not contain an open parentheses followed by a
closing  one.   We  parameterize  the  filling  by  the  variable  $q$
indicating its length, so that  there are $q+1$ admissible fillings of
length  $q$  with  $q+1$  places  for  insertion  of  a  well-balanced
non-empty  word. We  furthermore  have the  Catalan number  generating
function
$$C(z) = \frac{1-\sqrt{1-4z}}{2z}.$$
Inserting into the filling we find  the mixed OGF of binary words with
pairs of well-balanced  parentheses marked which is  (variable $z$ for
word length and $u$ for pairs of parentheses)
$$\sum_{q\ge 0} (q+1) z^q 
\sum_{p=0}^{q+1} {q+1\choose p} (C(uz^2)-1)^p
\\ = \sum_{q\ge 0} (q+1) z^q C(uz^2)^{q+1}
= \frac{C(uz^2)}{(1-zC(uz^2))^2}.$$
In this interpretation of the problem we have that an open parenthesis
paired with  a closed  one with  a sequence between  them that  is not
well-balanced do  not count.  The binomial  coefficient  indicates the
choice of slots in the filling where a non-empty well-balanced word is
placed. This step  may be skipped if we admit  the empty well-balanced
word in which case we proceed to the next line. 
Now  to  extract  coefficients  from this  we  recall  the  functional
equation of $C(z)$ which is
$$C(z) = 1 + z C(z)^2
\quad\text{or}\quad
z = \frac{C(z)-1}{C(z)^2}.$$
We do the coefficient on $[u^k]$  first ($k$ pairs of parentheses) and
view $C(uz^2)$ as a function of $u$ which yields
$$u = \frac{1}{z^2} \frac{C(uz^2)-1}{C(uz^2)^2}.$$
The coefficient extractor here is
$$[u^k] \frac{C(uz^2)}{(1-zC(uz^2))^2} =
\frac{1}{2\pi i} \int_{|u|=\epsilon}
\frac{1}{u^{k+1}} \frac{C(uz^2)}{(1-zC(uz^2))^2}
\; du.$$
We let $w=C(uz^2)$ so that $u=(w-1)/w^2/z^2$ and 
$$du = \frac{1}{z^2} 
\left(\frac{1}{w^2} - 2\frac{w-1}{w^3}\right) \; dw
= \frac{1}{z^2}  \frac{2-w}{w^3} \; dw$$
which yields for the integral
$$\frac{1}{2\pi i} \int_{|w-1|=\gamma}
z^{2k+2} \frac{w^{2k+2}}{(w-1)^{k+1}}
\frac{w}{(1-zw)^2} \frac{1}{z^2}  \frac{2-w}{w^3} \; dw
\\ =  \frac{z^{2k}}{2\pi i} \int_{|w-1|=\gamma}
\frac{w^{2k}}{(w-1)^{k+1}}
\frac{1}{(1-zw)^2} (2-w) \; dw.$$
We get two pieces here which are
$$\frac{z^{2k}}{2\pi i} \int_{|w-1|=\gamma}
\frac{w^{2k}}{(w-1)^{k+1}}
\frac{1}{(1-zw)^2} \; dw$$
and
$$-\frac{z^{2k}}{2\pi i} \int_{|w-1|=\gamma}
\frac{w^{2k}}{(w-1)^{k}}
\frac{1}{(1-zw)^2} \; dw.$$
We obviously require where $p\le k$
$$\frac{1}{p!} 
\left(w^{2k} \frac{1}{(1-zw)^2}\right)^{(p)}$$
This is by Leibniz
$$\frac{1}{p!} 
\sum_{q=0}^p {p\choose q} \frac{(2k)!}{(2k-q)!} w^{2k-q}
\frac{z^{p-q} (p-q+1)!}{(1-zw)^{p-q+2}}
\\ = \sum_{q=0}^p {2k\choose q} w^{2k-q}
\frac{z^{p-q} (p-q+1)}{(1-zw)^{p-q+2}}.$$
Evaluate at $w=1$ to get
$$\sum_{q=0}^p {2k\choose q}
\frac{z^{p-q} (p-q+1)}{(1-z)^{p-q+2}}.$$
The contribution from the two integrals now becomes
$${2k\choose k} \frac{z^{2k}}{(1-z)^2}
+ z^{2k} \sum_{q=0}^{k-1} {2k\choose q}
\left(\frac{z^{k-q} (k-q+1)}{(1-z)^{k-q+2}}
- \frac{z^{k-1-q} (k-q)}{(1-z)^{k-q+1}}\right)
\\ = {2k\choose k} \frac{z^{2k}}{(1-z)^2}
\\ + z^{2k} \sum_{q=0}^{k-1} {2k\choose q}
\left(\frac{z^{k-q} (k-q+1) - z^{k-1-q} (k-q)
+ z^{k-q} (k-q) }{(1-z)^{k-q+2}}\right)
\\ = {2k\choose k} \frac{z^{2k}}{(1-z)^2}
\\ + z^{2k} \sum_{q=0}^{k-1} {2k\choose q}
\left(\frac{z^{k-q} (2k-2q+1) - z^{k-1-q} (k-q) }
{(1-z)^{k-q+2}}\right)
\\ = z^{2k} \sum_{q=0}^{k} {2k\choose q}
\frac{z^{k-q} (2k-2q+1) - z^{k-1-q} (k-q) }
{(1-z)^{k-q+2}}.$$
Doing  the  coefficient  extraction  on  $[z^n]$  (word  contains  $n$
parentheses in total) we get two pieces, namely
$$\sum_{q=0}^{k} {2k\choose q}
[z^{n-3k+q}] \frac{2k-2q+1}
{(1-z)^{k-q+2}}
\quad\text{and}\quad
-\sum_{q=0}^{k} {2k\choose q}
[z^{n-3k+q+1}] \frac{k-q}
{(1-z)^{k-q+2}}$$
which produce 
$$\sum_{q=0}^{k} {2k\choose q}
(2k-2q+1)
{n-2k+1\choose k-q+1}
\quad\text{and}\quad
-\sum_{q=0}^{k} {2k\choose q}
(k-q)
{n-2k+2 \choose k-q+1}.$$
In order to join these we put
$${n-2k+1 \choose k-q+1} =
\frac{n-3k+q+1} {n-2k+2} {n-2k+2 \choose k-q+1}$$
to get
$$\sum_{q=0}^{k} {2k\choose q}
{n-2k+2\choose k-q+1}
\frac{(k-q+1)(n+2q+1-4k)}{n-2k+2}$$
and arrive at the closed form where clearly $n\ge 2k:$
$$\bbox[5px,border:2px solid #00A000]{
\sum_{q=0}^{k} {2k\choose q}
{n-2k+1\choose k-q}
(n+2q+1-4k).}$$
As a sanity check when $n=2k$ we should get ordinary Catalan
numbers. We obtain
$$\sum_{q=0}^{k} {2k\choose q}
{1\choose k-q}
(2q+1-2k)
= {2k\choose k} - {2k\choose k-1} 
= \frac{1}{k+1} {2k\choose k}$$
and the  check goes  through. We  verified all of  the above  in Maple
including an enumeration routine,  which is suprisingly effective even
on strings of, say, $16$  parentheses. Here is the generating function
for pairs of parenthesis among $8$ parentheses:
$$14\,{u}^{4}+84\,{u}^{3}+100\,{u}^{2}+49\,u+9$$
and here it is for $10$ parentheses:
$$42\,{u}^{5}+270\,{u}^{4}+375\,{u}^{3}+245\,{u}^{2}+81\,u+11.$$
For $11$ parentheses we get
$$264\,{u}^{5}+660\,{u}^{4}+660\,{u}^{3}+352\,{u}^{2}+100\,u+12.$$
The leading term is not a Catalan number when $n$ is odd. This was the
Maple code:

with(combinat);

CAT := n -> 1/(n+1)*binomial(2*n,n);

CAT_PARS :=
proc(n)
option remember;
local recurse, res;

    res := [];

    recurse :=
    proc(sofar, oc, cc)
        if oc = 0 and cc = 0 then
            res := [op(res), sofar];
            return;
        fi;

        if oc > 0 then
            recurse([op(sofar), `(`], oc-1, cc+1);
        fi;
        if cc > 0 then
            recurse([op(sofar), `)`], oc, cc-1);
        fi;
    end;

    recurse([], n, 0);
    convert(res, `set`);
end;

BALCOUNT :=
proc(parlist)
local posA, posB, intervA, intervB, len, subseq, count,
    res, n;

    subseq := [];

    n := nops(parlist);

    for posA to n do
        for posB from posA+1 to n do
            intervA := parlist[posA..posB];
            len := posB - posA + 1;

            if type(len, even) and
            intervA in CAT_PARS(len/2) then
                subseq := [op(subseq), [posA, posB]];
            fi;
        od;
    od;

    res := [];

    subseq :=
    sort(subseq,
         (a, b) -> a[2]-a[1] < b[2]-b[1]);
    count := nops(subseq);

    for posA to count do
        intervA := subseq[posA];
        for posB from posA+1 to count do
            intervB := subseq[posB];

            if intervB[1] <= intervA[1] and
            intervB[2] >= intervA[2] then
                break;
            fi;
        od;

        if posB = count + 1 then
            res := [op(res), intervA];
        fi;
    od;

    add((iv[2]-iv[1]+1)/2, iv in res);
end;

GFP :=
proc(n)
option remember;
local gf, idx, d, parlist;

    gf := 0;

    for idx from 2^n to 2*2^n-1 do
        d := convert(idx, base, 2);

        parlist :=
        subs([0=`(`, 1=`)`], d[1..n]);

        gf := gf + u^BALCOUNT(parlist);
    od;

    gf;
end;

GFA :=
proc(n)
local C;

    C := (1-sqrt(1-4*u*z^2))/(2*u*z^2);

    coeftayl(C/(1-z*C)^2, z=0, n);
end;

GFB :=
n -> add(u^k*add(binomial(2*k,q)*binomial(n-2*k+1, k-q)
         *(n+2*q+1-4*k), q=0..k), k=0..floor(n/2));

GFC :=
n -> add(u^k*(n+1-2*k)^2/(n+1)*binomial(n+1,k),
         k=0..floor(n/2));

Additional simplification. Splitting the sum into two
parts we get for the first part
$$(n+1-4k) \sum_{q=0}^{k} {2k\choose q}
{n-2k+1\choose k-q}
\\ = (n+1-4k) \sum_{q=0}^{k} {2k\choose q}
[z^{k-q}] (1+z)^{n-2k+1}
\\ = (n+1-4k) [z^k] (1+z)^{n-2k+1}
\sum_{q=0}^{k} {2k\choose q} z^q.$$
Now  we may  extend  $q$ beyond  $k$  because the  sum  term does  not
contribute to the coefficient extractor in that case, getting
$$(n+1-4k) [z^k] (1+z)^{n-2k+1}
\sum_{q\ge 0} {2k\choose q} z^q
\\ = (n+1-4k) [z^k] (1+z)^{n-2k+1} (1+z)^{2k}
= (n+1-4k) [z^k] (1+z)^{n+1}
= (n+1-4k) {n+1\choose k}.$$
The second part is
$$ 2\sum_{q=0}^{k} {2k\choose q} q
{n-2k+1\choose k-q} 
= 2\sum_{q=1}^{k} {2k\choose q} q
{n-2k+1\choose k-q} 
\\ = 4k\sum_{q=1}^{k} {2k-1\choose q-1}
{n-2k+1\choose k-q} 
= 4k\sum_{q=0}^{k-1} {2k-1\choose q}
{n-2k+1\choose k-q-1}
\\ = 4k\sum_{q=0}^{k-1} {2k-1\choose q}
[z^{k-1-q}] (1+z)^{n-2k+1}
\\ = 4k [z^{k-1}] (1+z)^{n-2k+1} 
\sum_{q=0}^{k-1} {2k-1\choose q} z^q
= 4k [z^{k-1}] (1+z)^{n-2k+1} 
\sum_{q\ge 0} {2k-1\choose q} z^q
\\ = 4k [z^{k-1}] (1+z)^{n-2k+1} (1+z)^{2k-1}
= 4k [z^{k-1}] (1+z)^{n}
= 4k {n\choose k-1}
\\ = 4k {n+1\choose k} \frac{k}{n+1}.$$ 
Adding these two we get the second closed form
$$\bbox[5px,border:2px solid #00A000]{
\frac{(n+1-2k)^2}{n+1} {n+1\choose k}.}$$
It is now immediate that we obtain a Catalan number when $n=2k$
since 
$$\frac{1}{2k+1} {2k+1\choose k}
=\frac{1}{2k+1} {2k\choose k} \frac{2k+1}{k+1}
= \frac{1}{k+1} {2k\choose k}.$$
We  keep in  mind here  that even  though the  closed form  is defined
including for $2k\gt n$ we must have that $2k\le n$ since $k$ balanced
parenthesis  require  at  least   $2k$  characters,  with  a  balanced
parenthesis participating in exactly one pair. Observe that the closed
form is symmetric with respect to $k\to n+1-k.$
 The algorithm to generate well-balanced words of parentheses seems
to have appeared  several times at stackexchange  sites, one reference
is           this           link           at           stackoverflow
com.