$n=2^m q,(2,q)=1$
结果好像是$(C_{2n}^1,C_{2n}^3,C_{2n}^5,\dots,C_{2n}^{2n-1})=2^{m+1}$
$C_{2n}^1=2n=2^{m+1}q$
$(C_{2n}^1,C_{2n}^3,C_{2n}^5,\dots,C_{2n}^{2n-1})|2^{m+1} q$
$C_{2n}^{2k+1}=\frac{2n}{2k+1}C_{2n-1}^{2k}=2^{m+1}\frac{qC_{2n-1}^{2k}}{2k+1}$
$2^{m+1}|C_{2n}^{2k+1}$
$2^{m+1}|(C_{2n}^1,C_{2n}^3,C_{2n}^5,\dots,C_{2n}^{2n-1})$
如果有一个$C_{2n}^{2k+1}$与q互质,那就会取等 |