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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.

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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 $$ C_n=4^n\frac{(2n-1)!!}{(2n)!!}\cdot \frac1{n+1}=\frac2{\pi}\int_0^{\pi/2} (2\sin x)^{2n}dx\int_0^{1} t^n dt. $$

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  • $\begingroup$ This is very nice. Other integral representations of $(C_n)$ as mixtures of geometric progressions are also known; see e.g. en.wikipedia.org/wiki/Catalan_number#Properties $\endgroup$
    – Iosif Pinelis
    Commented Dec 22, 2021 at 21:37
  • $\begingroup$ @IosifPinelis I guess they are distributionally the same:) $\endgroup$
    – Fedor Petrov
    Commented Dec 23, 2021 at 12:24
  • $\begingroup$ Indeed, they must be, because of the Weierstrass approximation theorem. :-) $\endgroup$
    – Iosif Pinelis
    Commented Dec 23, 2021 at 14:32

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