Some identities of Cauchy numbers associated with continued fractions (Q2422203)

The authors study the Cauchy numbers \(\{c_n\}_0^{\infty}\), defined by their generating function \[\frac{x}{\log{(1+x)}}=\sum_{n=0}^{\infty}\,c_n\frac{x^n}{n!}.\] Using the following continued fraction expansion of the generating function (mark that is not the regular C-fraction expansion) \[\frac{x}{\log{(1+x)}}=\frac{x|}{|1}+\frac{1^2x|}{|\ 2\ }+\frac{1^2x|}{|\ 3\ } +\frac{2^2x|}{|\ 4\ }+\frac{2^2x|}{|\ 5\ }+\frac{3^2x|}{|\ 6\ }+\] and explicit expressions for the numerators and denominators of the \(n\)th approximants, they prove the following result. Theorem 1. Let \(H_n=1+\frac{1}{2}+\cdot +\frac{1}{n}\) be the harmonic numbers, then we have \[ \sum_{\ell =0}^{\min{(k,n-1)}}\, \binom{2n-\ell}{2n-2\ell -1} \binom{2n-\ell -1}{n-\ell} \times \left((n+1)(H_n-H_{n-\ell})+\dfrac{2n-\ell -1}{2(n-\ell )}\right) \dfrac{c_{k-\ell}}{(k-\ell )!} = \] \[ \begin{cases} \dfrac{n+1}{2}\binom{2n-k+1}{n}\binom{n}{k}, & \quad 0\leq k\leq n-1 \\ \dfrac{n+1}{2}\binom{2n-k+1}{n}\binom{n}{k}- \dfrac{1}{2}\dfrac{c_{k-n}}{(k-n)!}, & \quad k=n \\ -\dfrac{1}{2}\dfrac{c_{k-n}}{(k-n)!}, & \quad n+1\leq k\leq 2n \end{cases}\] and \[ \sum_{\ell =0}^{\min{(k,n-2)}} \binom{2n-\ell -1}{2n-2\ell -2} \binom{2n-2\ell -3}{n-\ell -1} \left((H_n-H_{n-\ell -1}\right) \frac{c_{k-\ell}}{(k-\ell )!} = \] \[\begin{cases} \frac14 \binom{2n-k}{n} & \quad 0\leq k\leq n-2 \\ \frac14 \binom{2n-k}{n} \binom{n}{k} - \frac{H_n}{2} \frac{c_{k-n+1}}{(k-n+1)!} & \quad k=n-1,n \\ -\frac{H_n}{2}\frac{c_{k-n+1}}{(k-n+1)!} & \quad n+1\leq k\leq 2n. \end{cases} \] The layout of the paper is as follows: \S1. Introduction. \S2. Preliminaries. \S3. Main result: Contains the main result and its proof. \S4. Other continued fractions related to Cauchy numbers: Two other continued fractions for the generating function are mentioned. References (contains 13 items).