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Conjecture: $\prod\limits_{k=0}^{n}\binom{2n}{k}$ is divisible by $\prod\limits_{k=0}^n\binom{2k}{k}$ only if $n=1,2,5$.
The diagram shows Pascal's triangle down to row $10$.
I noticed that the product of the blue numbers is divisible by the product of the orange numbers. That is (including the bottom centre number $...
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Factoring the factorials
Just for the fun of it, I've started factoring $n!$ into its prime divisors, and this is what I got for $2\leq n\leq20$:
$$\begin{align}
2! &= 2^\color{red}{1} &S_e=1\\
3! &= 2^\color{red}...
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