The central binomial coefficients $\binom{2m}{m}$ have g.f. $\frac{1}{\sqrt{1-4z}}$ and lower bound $\frac{4^m}{\sqrt{4m}} \le \binom{2m}{m}$. I'm interested in a related integer series $$T(2m, m) = \sum_{r=1}^{m} \binom{2m-r-1}{m-1} F(r) = \sum_{j=0}^{2m-1} \binom{2m-j-1}{m+j}$$ which is the central coefficient of the pseudo Riordan array $(\frac{z}{1-z-z^2}, \frac{z}{1-z})$. $F(r)$ is the $r$th Fibonacci number, with $F(0)=0, F(1)=1$, and $T$ is an offset version of A105809.
Side note: This is clearly related to the binomial Riordan array because it shares the second g.f., which transforms between columns, and in consequence it also satifies the recurrence $T(n, k) = T(n-1, k-1) + T(n-1, k)$ for $k > 0$.
By a theorem of Paul Barry I can obtain a g.f. for the central coefficients of $T$ of $$\frac{z}{1-4z+z\sqrt{1-4z}}$$
Is it possible, either from one of the sums or from the g.f., to obtain a lower bound similar to the one I mentioned above for the central binomials? By "similar" I'm hoping for something of the form $\frac{4^m}{\sqrt{am+b}}$, and numerical experiments suggest that this is plausible.