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I know about the following identity: $$\displaystyle \text{lcm} \left( {n \choose 0}, {n \choose 1}, ... {n \choose n} \right) = \frac{\text{lcm}(1, 2, ... n+1)}{n+1}$$ 1) Is there any method to find $$\displaystyle \text{lcm} \left( {n \choose 0}, {n \choose 1}, ... {n \choose k} \right)$$ for any $k \le n$ without evaluating all the binomial coefficients?

2)Let me define $$g(n,k)=\displaystyle \text k{n \choose k}$$ $$\displaystyle \text{lcm} \left( g(n,0), g(n,1), ... g(n,n) \right)=\text{lcm}(1, 2, ... n)$$ Can we find $$\displaystyle \text{lcm} \left( g(n,0), g(n,1), ... g(n,k) \right)$$ for any $k \le n$ ?

EDIT: $$\displaystyle \text{lcm} \left( {n \choose 0}, {n \choose 1}, ... {n \choose k} \right)=\frac{\text{lcm}(n-k+1,n-k+2, ... n+1)}{n+1}$$ $$\displaystyle \text{lcm} \left( g(n,0), g(n,1), ... g(n,k) \right)=\text{lcm}(n-k,n-k+1, ... n)$$

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    For 1)Refer to this question: http://math.stackexchange.com/questions/1442/is-there-a-direct-proof-of-this-lcm-identity?rq=1 – Juan Jellyfish Oct 02 '15 at 20:26
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    @JuanJellyfish , thanks for the reference. But what about $g(n,k)$? –  Oct 03 '15 at 02:08

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