A sequence $a\_n$ is called *log-convex* if $\mathcal{L}(a\_n):=a\_{n+1}a\_{n-1}-a\_n^2\geq0$ for all $n$; it is *infinitely log-convex* provided that all the iterates $\mathcal{L}^k(a\_n)$ are still log-convex, $k\geq1$. Here $\mathcal{L}^2(a\_n)=\mathcal{L}(\mathcal{L}(a\_n))$, etc.


Consider in particular the well-known sequence of Catalan numbers $C\_n=\frac1{n+1}\binom{2n}n$. It has been established that both $C\_n$ and some of its "quantum" versions are log-convex.


I would like to ask:



> 
> **QUESTION.** Is it true that $C\_n$ is infintely log-convex? It appears to be so.
> 
> 
> 



Is $C_n$ infinitely log-convex?
I think so. If $a\_n=\int\_X f^n(x)d\mu(x)$ for a certain positive function $f$ on a measure space $(X,\mu)$, then $$a\_{n-1}a\_{n+1}-a\_n^2=\int\_{X\times X} f^{n-1}(x)f^{n+1}(y)d\mu(x)d\mu(y)-\int\_{X\times X} f^{n}(x)f^{n}(y)d\mu(x)d\mu(y)\\=\frac12\int\_{X\times X} f^{n}(x)f^{n}(y)\left(\frac{f(x)}{f(y)}+\frac{f(y)}{f(x)}-2\right)d\mu(x)d\mu(y)$$
has also a similar representation.
But
$$
C\_n=4^n\frac{(2n-1)!!}{(2n)!!}\cdot \frac1{n+1}=\frac2{\pi}\int\_0^{\pi/2} (2\sin x)^{2n}dx\int\_0^{1} t^n dt.
$$


