Let $h(n,t) = \sum\limits\_{j = 0}^n {\binom
 {\lfloor {\frac{n}{2}} \rfloor }{j}\binom
 {\lfloor {\frac{n+1}{2}}\rfloor }{j}t^j \\ }.$


I am interested in the Hankel determinants $${D\_k}(n,t) = \det \left( {h(k + i + j,t)} \right)\_{i,j = 0}^{n - 1}.$$
These can easily be computed for $0 \leq k \leq 3.$


It seems that $${D\_4}(n,t) = {t^{\lfloor {\frac{{{n^2}}}{4}} \rfloor }}b(n,t)$$ with $b(n,t) = \sum\limits\_{j = 0}^{2n} \min({\binom{3+j}{3},\binom{2n+3-j}{3}})t^j.$


In order to prove this, I need the identity
$$a{(n,t)^2} = b(n - 1,t)\sum\limits\_{j = 0}^{n - 1} {{t^j}} - tb(n - 2,t)\sum\limits\_{j = 0}^n {{t^j}} $$
with $a(n,t) = \sum\limits\_{j = 0}^{n - 1} {(j + 1){t^j}} .$


Any idea how to prove this identity?



An interesting Hankel determinant
Denote $a\_n=a(n,t)$ and $b\_n=b(n,t)$. To help avoiding the **min** function, write
$$b\_n=\binom{n+3}3t^n+\sum\_{j=0}^{n-1}\binom{3+j}3\left[t^{2n-j}+t^j\right].$$
Notice that $a\_n=\frac{nt^{n+1}-(n+1)t^n+1}{(1-t)^2}$ and $\sum\_{j=0}^nt^j=\frac{1-t^{n+1}}{1-t}$. Your identity takes the form
$$(nt^{n+1}-(n+1)t^n+1)^2=(1-t)^3[(1-t^n)b\_{n-1}-t(1-t^{n+1})b\_{n-2}].$$
Now, as Mark Widon mentioned, try to read-off the coefficients of $t^k$.


**UPDATE.** Resorting back the original formulation of the claim
$$a\_n^2=b\_{n-1}\sum\_{j=0}^{n-1}t^j-t\,b\_{n-2}\sum\_{j=0}^nt^j,$$
I was able (after lots of routine algebraic simplification and reorganization) to rewrite the right-hand side as
\begin{align\*}
\left(\sum\_{j=0}^{n-1}(j+1)t^j\right)^2
&=\sum\_{j=0}^{n-1}\binom{3+j}3t^j+\sum\_{j=0}^{n-2}\left[\beta\_n-\beta\_{j+1}-\binom{n-j}3\right]\,t^{n+j} \\
&=\sum\_{j=0}^{n-1}\binom{3+j}3t^j+\sum\_{j=0}^{n-2}
\frac{(n - j - 1)(j^2 + 4jn + n^2 + 5j + 7n + 6)}6\, t^{n+j}
\end{align\*}
where $\beta\_k=\frac{k(k+1)(2k+1)}6$ (the sum of squares function).


Once we got this far, the next step is to compare the coefficients of $t^k$.


