Is there a smart precise way to determine LCM of $C(n,1) , C(n,2) , C(n,3) ,\ldots, C(n,k)$ :
Here $C(n,k)$ denotes no. of combinations of n objects taken k at a time.
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so $C\left(n,k\right)={n \choose k}$? – Max Oct 07 '15 at 11:50
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1I'm voting to close this question as off-topic because it is highly likely to be from an ongoing competition. – zhoraster Oct 07 '15 at 15:48
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OEIS says that $\operatorname{lcm}(C(n,1) , C(n,2) , C(n,3) ,\ldots, C(n,n)) = \operatorname{lcm}(1,2,\dots,n+1)/(n+1)$. This question contains a proof. However, no simpler expression is known for $\operatorname{lcm}(1,2,\dots,n)$.
It seems unlikely that there is a simple expression for the general case $\operatorname{lcm}(C(n,1) , C(n,2) , C(n,3) ,\ldots, C(n,k))$.
