On divisibility concerning binomial coefficients

Abstract

Let k and n be positive integers. We mainly show that $$(ln+1)|k(\genfrac{}{}{0ex}{}{kn+ln}{kn}),$$ $$2\left(\genfrac{}{}{0ex}{}{kn}{n}\right)|(\genfrac{}{}{0ex}{}{2n}{n})C_{2n}^{(k-1)}$$, $$\left(\genfrac{}{}{0ex}{}{kn}{n}\right)|(2k-1)C_n(\genfrac{}{}{0ex}{}{2kn}{2n}),$$ $$\left(\genfrac{}{}{0ex}{}{2n}{n}\right)|(k+1)C_n^{(k-1)}(\genfrac{}{}{0ex}{}{2kn}{kn}),$$ $$2^{k-1}\left(\genfrac{}{}{0ex}{}{2n}{n}\right)|(\genfrac{}{}{0ex}{}{2(2^k-1)n}{(2^k-1)n})C_n^{(2^k-2)},$$ $$(6n+1)\left(\genfrac{}{}{0ex}{}{5n}{n}\right)|(\genfrac{}{}{0ex}{}{3n-1}{n-1})C_{3n}^{(4)},$$ and $$\left(\genfrac{}{}{0ex}{}{3n}{n}\right)|(\genfrac{}{}{0ex}{}{5n-1}{n-1})C_{5n}^{(2)},$$ where C_n denotes the Catalan number $\left(\genfrac{}{}{0ex}{}{2n}{n}\right)/(n+1)$, and C_m^{(h)} refers to the Catalan number $\left(\genfrac{}{}{0ex}{}{(h+1)m}{m}\right)/(hm+1)$ of order h.

18 pages. The current (1.12) is new

References in corpus (1)

• Products and sums divisible by central binomial coefficients