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.

  • $\begingroup$ Have you tried to mimic a proof of the bound for the central binomial coefficient? $\endgroup$ – Gerry Myerson Jul 30 '14 at 9:08
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    $\begingroup$ Since the dominant singularity is at $-2 + \sqrt{5}$ you have $T(2m,m) \sim (2+\sqrt{5})^n \cdot \dots$, so a(ny) lower bound of order $4^m$ from some moderately sized $m$ on seems likely, yes. $\endgroup$ – Raphael Jul 30 '14 at 15:30
  • $\begingroup$ @GerryMyerson, the central binomial coefficient has a rather neat product representation which allows an easy ad hoc proof of the bound. The only way I can see to get a similar product representation is to factor out a central binomial coefficient from each of the binomial coefficients in the second sum. Then bounding with $\frac{1}{\sqrt{m-2j-1}}\ge\frac{1}{\sqrt{m}}$ I get a non-Gosper-summable hypergeometric, and crudely bounding the terms to get a summable geometric sequence gives a worse bound than simply taking the first term of the original sum. $\endgroup$ – Peter Taylor Jul 30 '14 at 17:52
  • $\begingroup$ @Raphael, can you explain why there's a singularity at $-2+\sqrt{5}$? I see singularities at $-2-\sqrt{5}$ and $\frac{1}{4}$. $\endgroup$ – Peter Taylor Jul 31 '14 at 7:14
  • $\begingroup$ Huh. In the partial fraction expansion the denominator $1-4z-z^2$ occurs which has that zero; apparently its contribution cancels out. My bad. Anyway, I don't see how to get to strict lower bounds from an asymptotic so asymptotic singularity analysis (at least as far as I know it) it out of the question. $\endgroup$ – Raphael Jul 31 '14 at 8:03

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