cp's OEIS Frontend

Take a square sheet of paper and fold it first vertically and then horizontally so that the bottom right corner stays in place. After each fold, unfold the paper and draw a line along each crease that is indented inwards (along which water would flow); upward creases (ridges) are not marked.

After two folds, we again have a (smaller and thicker) square, and we repeat the process.

After n individual folds, when the paper is unfolded the lines form a planar graph G(n). The numbers of regions, vertices, edges, and connected components in G(n) are given in the present sequence, A146528 (still to be confirmed), A342761, and A342762.

The number of vertices of degree 1 after n+1 folds appears to be A274230(n).

We ignore the folk theorem that says no sheet of paper can be folded more than seven times.

Column 1 of A183986

Based on the conjectured recursion formula, it is also the number of notches in a sheet of paper when you fold it n times and cut off the four corners (see A274230). - Philippe Gibone, Jul 06 2016

From Gus Wiseman, Jun 26 2022: (Start)

Also the number of integer compositions of n + 1 with an even part other than the first or last. For example, the a(3) = 1 through a(5) = 12 compositions are:

(121) (122) (123)

(221) (141)

(1121) (222)

(1211) (321)

(1122)

(1212)

(1221)

(2121)

(2211)

(11121)

(11211)

(12111)

The odd version is A274230.

(End)

A number's prime signature (row n of A124010) is the sequence of positive exponents in its prime factorization.

Also Heinz numbers of non-weakly alternating non-strict integer partitions, where we define a sequence to be weakly alternating if it is alternately weakly increasing and weakly decreasing, starting with either. These partitions are counted by A349796. This sequence involves the somewhat degenerate case where no strict increases are allowed.

From Tom Karzes, Jul 05 2016: (Start)

This is a three-dimensional analog of the holes-in-sheet-of-paper sequence A274230.

In d dimensions, assuming the axes for folding are selected in a round-robin fashion, the number of times a given dimension is folded is:

floor((n+i)/d)

where i runs from 0 (for the last dimension to be folded) through d-1 (for the first dimension to be folded).

The corresponding number of internal dividing lines/planes/etc. is (2^floor((n+i)/d)-1). The number of internal d-way intersections, which corresponds to the number of holes, is:

Product_{i=0..d-1}(2^floor((n+i)/d)-1)

where d is the number of dimensions and n is the total number of folds.

Note that the first several nonzero entries in these sequences are the powers of 3. Specifically, in d dimensions, the first d entries are 0, followed by the first (d+1) powers of 3.

It's not hard to see why this is so. The first nonzero entry occurs at d folds, and the value is 1. This is when you've folded once along each dimension.

After that, the next d folds each divide 2 old partitions into 4 new ones, i.e., they change the internal folds from 1 to 3. So for the next d entries you just multiply the previous entry by 3 (or more generally, by 3/1).

After that you multiply by 7/3 for the next d entries, then 15/7, then 31/15, etc. Each time you're just replacing one of the old factors with a new one, where each factor is one less than a power of two.

Here's an alternative formulation that avoids the iterated product.

For a given number of folds, there are only two factors, each raised to some exponent (with the sum of the exponents being the dimension d):

v1 = 2^(n/d)-1

v2 = 2^(n/d+1)-1

p1 = d-mod(n,d)

p2 = mod(n,d)

holes = (v1^p1)*(v2^p2)

This flattens to:

((2^(n/d)-1)^(d-mod(n,d))) * ((2^(n/d+1)-1)^(mod(n,d)))

(End)

This is a four-dimensional analog of the holes-in-sheet-of-paper sequence A274230. See A274230 and A274626 for further information.