Problem: Given $n$ unique cards in a series from 1 to $n$ inclusive, arrange the cards such that drawing the first card, then placing the next card at the back of the deck, then drawing the next card and so on will result in an ordering of the set when all cards are drawn.
Question: How can we generalize the ordering of the cards?
Partial Solution: I've discovered the following relationships, though I don't believe this is quite enough to complete an ordering for an arbitrary number of cards.
Using an array to order the cards like so:
The $2^n$ row has value $n$ in the $n$th column (which is always the last column for that row)
For the $2n+1$ row, each $2n+1-2k$ column has value $2n+1-n-k=n+1-k$ for $k<n$
- Row $2^n+1$ has value $2^n+1$ in column 2
- For each interval of rows from $2^n+2$ through $2^{n+1}+1$, the $2k-2^{n+1}$ row has value $n$. This implies that $2^n+2<k \leq 2^{n+1}+1$.
- Each $2n$ row has value $n+1+k$ in column $4k+2$ if $k<n/2$
- Each $2n+1$ row has value $n+2+k$ in column $4k+4$
The explorations and rules are posted in this spreadsheet.
EDIT: The rules given seem to be enough to generalize indefinitely, but I have not proved this.