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Is there a nice bijective proof of the fact that the determinant of the $(n+1)$-by-$(n+1)$ Hankel matrix whose respective entries are the central binomial coefficients $0 \choose 0$, $2 \choose 1$, $\dots$, $4n \choose 2n$ is $2^n$ (e.g., via Lindstrom-Gessel-Viennot)?

My first thought was to apply LGV with lattice-paths joining sources $(0,0),(1,1),\dots,(n,n)$ to sinks $(n,n),(n+1,n+1),\dots,(2n,2n)$, but one of the hypotheses of LGV isn’t satisfied.

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    $\begingroup$ I haven't checked all the details but I think you can do this using the fact that $\binom{2n}{n}$ counts Dyck paths of length $2n$ in which up steps starting at height 0 are weighted by 2. This fact corresponds, via Flajolet's combinatorial approach to continued fractions, to the continued fraction $$ \frac{1}{\sqrt{1-4x}}= \cfrac{1}{1-\cfrac{2 x}{1-\cfrac{x}{1-\cfrac{x}{1-\cfrac{x}{1-\cfrac{x}{1-\dots}}}}}} $$ $\endgroup$
    – Ira Gessel
    Commented May 8 at 0:03
  • $\begingroup$ $2^n$ is the number of (shifted) sub-shapes of the shifted shape for the strict partition $(n,n-1,\ldots,1)$. Stembridge explained how to count sub-shapes (more generally, plane partitions) of shifted shapes using not determinants but Pfaffians; see e.g. Remark 4 of my question mathoverflow.net/questions/350445. I feel like it's possible this line of thinking could explain your Hankel determinant but I don't see all the details right now. $\endgroup$ Commented May 8 at 0:21
  • $\begingroup$ Again not an answer, but just a pointer to known similar results about determinants of binomial coefficients (for the general audience - I'm sure you know this Jim): in Federico Ardila's "Algebraic and geometric methods in enumerative combinatorics" (arxiv.org/abs/1409.2562) in Section 3.1.6 we have many examples of computing determinants via LGV to count e.g. plane partitions, fans of Dyck paths, tilings of Aztec diamonds, etc.; and in Section 3.2.6 we have an explanation of the connection between continued fractions and Hankel determinants (as in Ira Gessel's comment above). $\endgroup$ Commented May 8 at 15:08

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As Ira Gessel points out in his comment, the result follows easily via a weighted version of the standard LGV proof one uses for Hankel determinants of Catalan numbers. Continued fractions aren’t needed; at a purely combinatorial level, Gessel’s weighting scheme ensures that each excursion has multiplicative weight 2, and the product of these 2’s corresponds to the multiplicity of the map that turns lattice paths in a square into Dyck paths in a triangle by folding along the diagonal joining $(0,0)$ to $(2n,2n)$.

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