Every central binomial coefficient, starting with the one in row 2, is divisible by two. There are a few proofs of this already on this site, like prove that $\frac{(2n)!}{(n!)^2}$ is even if $n$ is a positive integer

I looked at Pascal's triangle, and every other central binomial coefficient starting at the one in row 6, appears to be divisible by four. Is this true? Is there a nice way to prove this? (I attempted to use Pascal's identity, but didn't arrive to anything).

Side Note: It also looks like every 4th central binomial coefficient, starting at row 14, is divisible by 8. Is it a pattern that doubling the previous "starting row" and adding two gives the next "starting row", and doubling the previous "gap" gives the new gap for the next power of two? I checked a few things: central binomial coefficient in row 30 is divisible by 16, the central coefficient in row 62 is divisible by 32, and the one in row 126 is divisible by 64.


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