Is there one way to write the permutation that takes $(1,2,...,n) \rightarrow (n,n-1,...,1)$ that is the same whether $n$ is even or odd?

Question

Is there one way to write the permutation that takes $(1,2,...,n) \rightarrow (n,n-1,...,1)$ that is the same whether $n$ is even or odd?

Are these formulas correct?

If $n$ is even then this permutation is given by $(1,n)(2,n-1)...(\frac{n}{2},\frac{n}{2}+1)=\prod_{k=1}^{\frac{n}{2}}(k,n-(k+1))$

If $n$ is odd then this permutation is given by $(1,n)(2,n-1)...(\frac{n-1}{2},\frac{n+2}{2})=\prod_{k=1}^{\frac{n-1}{2}}(k,n+1-k)$

Is there another way of expressing this permutation that is independent of whether $n$ is even or odd?? Perhaps without breaking it down to transpositions, using cycle notation or something? Thank you.

Answer 1

You could write it as $\sigma(k)=n-k+1.$