User blog:Edwin Shade/The Origami Function

There is a new function I have devised, which is to be notated F(n), where n is a whole number greater than or equal to 0. F(n) is the total amount of unique, well-defined final forms a square sheet of paper may be folded into with n-folds. By "unique" I mean that the final form of the paper cannot be rotated or flipped to another final form that uses the same amount of folds; and by "well-defined" I mean a fold that can be replicated precisely, and there can be no possible ambiguity as to the fold's folding.

Note that each crease counts as a fold, and so a petal-fold actually counts as five folds, even if it's name seems to imply it is one fold.

Because as the number of folds increases, so does the amount of reference points with which to define folds, it means F(n) grows slightly more than $$a^n$$, and grows at a rate approximately proportional to $$a^{b*n}$$, where a and b are constants.

F(0) is equal to 1, since there is only one way to fold a square with 0 folds, and that of course is not to fold at all.

F(1) is equal to 2, because a square may either be folded in half lengthwise or diagonally.

F(2) is about 10, I came to this result by trying all the possible 2-fold arrangements on a piece of paper nearby.

F(3) is about 100, this is a reasonable estimate. F(3) could probably be found by hand given a quarter-hour.

F(4) is about 1,000, this too is a reasonable estimate, and is the upper limit of what one could calculate brute-force using paper.

F(10) could be calculated by a computer, but it would most likely require millions of collective hours of computer time and represent the highest value of F(n) calculable by modern software.

An interesting consequence of the F(n) function is that for every well-defined origami creation, since you can enumerate all the different combinations of folds, there is a corresponding number that describes how to fold it exactly, even if that number is thousands of digits long.