Background

The footpaths across this little wood were legitimised as definitive rights-of-way in July 2020 by an order of HCC under the Wildlife and Countryside Act 1981 (see here - worth a look). So the second anniversary of that month seemed like an appropriate occasion to place a cache.

The container is in a camouflaged film pot about 3-4m up an oak tree, but it's a very easy reach and doesn't require any tree climbing or special tools. The route to the tree may be slightly prickly depending on the season but I managed it in shorts without losing more than a thimbleful of blood.

The puzzle

No tricks or red herrings this time - what you see is what you get.

What better item to construct a puzzle than the log strip with which we are all so familiar. Surely one can't construct a D4 puzzle from such a humble thing.

Try this...

• Take a log strip and lay it crossways on the table in front of you.
• Pick up the righthand end and fold it across to the left so that it lies on top of the left hand end.
• Flatten the fold down to make a sharp crease.
• Now pick up the fold that you have made and fold that across to the left in the same way.

Repeat this process a few times making a sharp fold each time and then unfold the strip.

You will see that this process has created quite a few creases in the log strip. Can you work out how many creases there will be after N folds?

Now look a little closer. Some of the folds are ins and some of them are outs. That's interesting. What will the pattern be after N folds?

The task

I'm not going to be too cruel with the number of folds because I'd like this puzzle to be available to those without programming experience. But I will put a credit below for the most elegant code (preferably in Python) messaged to me that calculates the pattern of creases for an arbitrary number of folds.

So consider a certain number of folds... the number that gives a decent number of creases: more than 50 but less than 100.

Label the creases in the obvious way and convert the resulting binary into a 19 digit decimal number. Label the first digit 'A' and the cache is at N51 44.RHN W00 22.OMJ.

Congratulations to Amberel for the most elegant coding solution so far