Determinant of a matrix with generalized binomial coefficients Let
$$
A=
\begin{bmatrix}\binom{-1/2}{1}&\binom{-1/2}{0}&0&0&...&0\\ \binom{-1/2}{2}&\binom{-1/2}{1}&\binom{-1/2}{0}&0&&...\\...&&&\binom{-1/2}{0}\\ &&&...&&\binom{-1/2}{0}\\ \binom{-1/2}{n}&\binom{-1/2}{n-1}&\binom{-1/2}{n-2}&\binom{-1/2}{n-3}&...&\binom{-1/2}{1}\end{bmatrix}.
$$
How can I calculate $\det A$?
Thank you very much.
 A: I would replace $-\frac12$ by a positive integer $N$, prove the result $\binom {N+n-1} n$ by induction on $N$, and then since both sides are polynomials, plug in $N=-\frac12$.
(Call the determinant $f(n,N)$. Just add the $(n-1)$th row to the $n$th row, then the $(n-2)$th row to the $(n-1)$th row and so on, you will get $f(n,N)=f(n,N+1)-f(n-1,N+1)$ which enables us to give an induction proof for the result $\binom{N+n-1}n$.)
An alternative to the induction proof is to regard it as a special case of the second Jacobi-Trudi formula for Schur functions.
I want to prove that 
$$\det_{1\le i,j \le n} \binom{N}{i-j+1}=\binom{N+n-1}n.$$
For an elementary symmetric function $e_k(1,1,\dots,1)$ with $m$ variables equals $\binom{m}k$ whereas $h_k(1,1,\dots,1)=\binom{m+k-1}{k}$. The second Jacobi-Trudi formula gives
$$s_{\lambda}=\det_{i,j=1}^{\lambda_1} e_{\lambda^\ast_i+j-i}.$$
So, if we choose the partition $\lambda=(n)$ and evaluate at $N$ ones, we get:
$$\det_{i,j=1}^ne_{1+j-i}(1,1,\dots,1)=s_{(n)}(1,1,\dots,1)=h_n(1,1,\dots,1)=\binom{N+n-1} n$$ as desired.
Again, because it is a polynomial, we can plug in $N=\frac12$.

Edited to add:
Both arguments end the following way:
The determinant with $-\frac12$ replaced by $N$ is a polynomial in $N$ of a certain bounded degree (it is not necessary to actually determine it, the important thing is that the degree does not involve $N$, it is not hard to see that the degree is $n$, but it is also enough to bound it by a more obvious bound like $1+2+\dots+n$).
The resulting binomial coefficient is also a polynomial in $N$ (of degree $n$).
We have proved that the two polynomials are equal as long as we plug in positive integer values for $N$. But these are infinitely many values and two polynomials who agree on infinitely many values (or agree on degree plus one values) are identical.
Since the two polynomials are identical, we can plug in $-\frac 12$ (or $\pi$ or $2+i \dots$).
A: A quick computation in Mathematica suggests that $\det A_n = \binom{n - \frac{3}{2}}{n}$.  I would imagine that a proof would involve row operations to convert the matrix into lower triangular form, then taking the product of the main diagonal entries.  We would begin from the bottom row and work our way up.  Unfortunately I don't have the time for writing out a complete solution.
