Even Stevens

Every day for the past $n$ days Even Stevens brings home his groceries in a plastic bag. He stores these plastic bags in a cupboard. He either puts the plastic bag into the cupboard with the rest, or else he takes an even number of the existing bags (which may either be empty or previously filled with other bags themselves) and places these into the current bag.
After 4 days there are 5 possible packings and if the bags are numbered 1 (oldest), 2, 3, 4, they are:
Four empty bags,
1 and 2 inside 3, 4 empty,
1 and 3 inside 4, 2 empty,
1 and 2 inside 4, 3 empty,
2 and 3 inside 4, 1 empty.
Note that 1, 2, 3 inside 4 is invalid because every bag must contain an even number of bags.
Define $f(n)$ to be the number of possible packings of $n$ bags. Hence $f(4)=5$. You are also given $f(8)=1\,385$.
Find $f(24\,680)$ giving your answer modulo $1\,020\,202\,009$.

This problem concerns a sequence known as the Even Stevens sequence, requiring a combination of combinatorial techniques and modular arithmetic to generate and compute the nth term.

The sequence is reminiscent of the “ballot numbers”, “Catalan numbers” and “Wedderburn-Etherington numbers” (sequence A001190 in the OEIS).

To clarify the behavior for the problem on hand:
on day 1 we have one bag, only one way to pack it (leave it empty).
On day 2, we have two bags; we can leave them both empty or put bag 1 in bag 2, so there are 2 ways.

For each new day, the process is to look at all the arrangements for the previous day, and count the ways to add a new bag. We can leave the new bag empty, or we can choose an even number of already-packed sets of bags to go into the new bag.

For efficiency, we can generate a recurrence relation based on previous terms to compute $f(n)$.

Firstly, let’s define an auxiliary function $g(n)$, where $g(n)$ counts the number of ways to pack $n$ bags such that we use all bags and the outermost bag is not empty. It’s easy to see that $g(n) = \sum_{k=0}^{n-2} f(k)*f(n-2-k)$ for all $n \geq 2$.

By counting the cases where the outermost bag is empty and where it’s not separately, we can derive a formula: $f(n) = f(n-1) + g(n)$ for all $n\geq2$.

Now we can use the above recurrence relation to compute the sequence terms. For efficiency, you should compute all terms $g(n)$ and $f(n)$ simultaneously. Another thing to note is – since we are interested in $f(24680) \mod{1020202009}$, take modulo $1020202009$ at each step to prevent overflow and reduce computation times.

Begin from $f(1)=1, f(2)=2, f(3)=2, f(4)=5, f(8)=1385$, iteratively compute till $f(24680) \mod{1020202009}$.

Lastly, remember that this problem may demand significant computation power depending upon your hardware, as it involves generating a large sequence in order to compute modular result for $f(24680)$.

Further, this problem could also be simplified by noticing patterns or employing some clever algorithmic optimizations, which are out of the scope of this general solution.

More Answers: