Count NE lattice paths (0,0)->(n,n) with precisely K+1 points on main diagonal Equivalent to the number of correct bracket sequences consisting of $K$ concatenated correct sequences.
Tried something like this
$$
\sum_{c_1+\dots+c_K=n}h(c_1)h(c_2)\dots h(c_K), h(t)=2C^{t}_{2t-2}
$$
but no luck with folding that sum.
Any ideas?
 A: Assuming your $C_i$ are Catalan numbers, you can use Generating functions for this. Consider first the variables $d_i=c_i-1$
$$\sum _{c_1+\cdots c_k=n}\prod _{i=1}^kC_{c_i-1}=\sum _{d_1+\cdots d_k=n-k}\prod _{i=1}^kC_{d_i},$$ this looks like the Cauchy product so recall that
$$\mathcal{C}(x)=\frac{2}{1+\sqrt{1-4x}}=\sum _{k=0}^{\infty}C_kx^k,$$ and by doing the product $k$ times we get  $$\left (\frac{2}{1+\sqrt{1-4x}}\right )^k=\sum _{n =0}^{\infty}\left(\sum _{d_1+\cdots +d_k=n}\prod _{i=1}^kC_i\right )x^n,$$
so you want the coefficient of $x^n$ in
$$\left (\frac{2x}{1+\sqrt{1-4x}}\right )^k=x^k\mathcal{C}(x)^k.$$
Recall that $x\mathcal{C}(x)^2=\mathcal{C}(x)-1,$ and so using it over and over again you will expect to see just one Catalan numbers. In fact this result is in lemma 27 of https://www.sciencedirect.com/science/article/pii/S0196885814000219?via%3Dihub
A: The number of paths from $(0,0)$ to $(n,n)$ with $k+1$ points on the main diagonal is 
$$
\bbox[3pt, border:1px solid black]{2^{k}\frac{k}n\binom{2n-k-1}{n-1}.}
$$
Proof: The generating function for lattice paths which start at $(0,0)$ and end at $(n,n)$, and do not touch the main diagonal in between, is
$$
\sum_{n\ge 0}2C_{n-1}x^{n}=2x\sum_{n\ge 1}C_{n-1}x^{n-1}=2x\cdot \frac{1-\sqrt{1-4x}}{2x}=1-\sqrt{1-4x}:= f(x)
$$
Therefore, the generating function for the concatenation of $k$ such paths is
$$
f(x)^k=\big(1-\sqrt{1-4x}\big)^k
$$
so the answer to your question is the coefficient of $x^n$ in $f(x)^k$, which I write as $[x^n]f(x)^k$.
Notice that $f(x)$ satisfies the equation 
$$
-\tfrac14\big(f(x)^2-2f(x)\big)=x
$$
This means that $f$ is the compositional inverse of $g(x):=-\frac14 x(x-2)$, so we can use the Lagrange inversion formula (as stated in the Wikipedia article on formal power series) to find the coefficient of $x^n$ in $f(x)^k$:
\begin{align}
[x^n]f(x)^k
  &=\frac{k}n[x^{-k}]g(x)^{-n}
\\&=\frac{k}n[x^{-k}](-\tfrac14)^{-n}x^{-n}(x-2)^{-n}
\\&=(-4)^n\frac{k}n[x^{n-k}](x-2)^{-n}
\\&=(-4)^n(-2)^{-n}\frac{k}n[x^{n-k}](1-x/2)^{-n}
\\&=2^{n}\frac{k}n\binom{-n}{n-k}(-\tfrac12)^{n-k}
\\&=2^{k}\frac{k}n\binom{2n-k-1}{n-k}
\end{align}
