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?

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
