Timeline for Dealing cards numbered $1$ to $n$ into piles
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 6, 2021 at 2:45 | comment | added | Sam Hopkins | Similarly, $A(2m,m)=2m(m!)^2$. | |
| Feb 6, 2021 at 2:20 | comment | added | Sam Hopkins | Likewise, $A(2m-1,m)=m!(m-1)!$ is easily seen: choose any permutation of $1,3,5,\ldots,2m-1$ and concatenate it with any permutation of $2,4,\ldots,2m-2$. | |
| Feb 6, 2021 at 2:16 | comment | added | Sam Hopkins | Nevermind, I see that for any sequence of length $n-1$ of max's or min's, there's a unique valid choice of $\sigma(1)$, so $A(n,1)=2^{n-1}$ is indeed clear. | |
| Feb 6, 2021 at 1:58 | comment | added | Sam Hopkins | I don't see an obvious argument why $A(n,1)=2^{n-1}$. Of course, you can construct a permutation counted by $A(n,1)$ by choosing, for each $i=2,\ldots,n$, $\sigma(i)$ to either be one more than the max or one less than the min of $\sigma(1),\ldots,\sigma(i-1)$, so $2^{n-1}$ makes sense as a rough guess; but this ignores two competing complications: the choice of $\sigma(1)$ and the fact that we might hit $n$ or $1$ in $\sigma(1),\ldots,\sigma(i-1)$ before $i=n$. | |
| Feb 6, 2021 at 1:41 | history | answered | Sam Hopkins | CC BY-SA 4.0 |