Permutation Its Own Inverse?
Max A.
maxale at gmail.com
Thu Jul 13 01:44:13 CEST 2006
On 7/12/06, franktaw at netscape.net <franktaw at netscape.net> wrote:
> Yes, it is (its own inverse - it isn't all that easily answered).
>
> It is easiest to subtract out the triangular numbers, making each row
> its own permutation:
>
> 1
> 2,1
> 3,2,1
> 4,2,3,1
> 5,3,2,4,1
> 6,3,2,4,5,1
> 7,4,3,2,5,6,1
> ...
>
> We can now see the rule: if ceiling(n/ceiling(n/k)) = k, then T(n,k) =
> ceiling(n/k); otherwise T(n,k) = k. Equivalently, for each pair i, j
> with i*j >= n, but (i-1)*j and i*(j-1) < n, we have T(n,i) = j, T(n,j)
> = i. For any other k, T(n,k) = k.
Could you please give a proof for this statement?
Thanks,
Max
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