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Part of 2012 Silk Road

Problems(1)

gcd of binomial (2n, 2i+1) from 0<=i<i+1

Source: SRMC 2012

9/3/2018
Let n>1n > 1 be an integer. Determine the greatest common divisor of the set of numbers {(2n2i+1):0≤i≤n−1}\left\{ \left( \begin{matrix} 2n \\ 2i+1 \\ \end{matrix} \right):0 \le i \le n-1 \right\} i.e. the largest positive integer, dividing (2n2i+1)\left( \begin{matrix} 2n \\ 2i+1 \\ \end{matrix} \right) without remainder for every i=0,1,...,n–1i = 0, 1, ..., n–1 . (Here (ml)=Cml=m!l!(m−l)!\left( \begin{matrix} m \\ l \\ \end{matrix} \right)=\text{C}_{m}^{l}=\frac{m\text{!}}{l\text{!}\left( m-l \right)\text{!}} is binomial coefficient.)
number theorygreatest common divisorbinomial coefficients