Python one-liner using recursion
easyLine = lambda n : 1 if n==0 else int((2*(2*n-1)/n)*easyLine(n-1))
Because,
easyLine(n) = 2*(2*n-1)/n * easyLine(n-1) when n > 0 = 1 when n=0
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Python one-liner using recursion
Because,