Laboratory

Iterate the map

Start at 1. The closed form would send it to itself. The boundary sends it to 2. After that, A(n) = sopfr(n) + 2^Ω(n) − Ω(n) takes over, and every orbit so far reaches 12.

Why the boundary at 1 is the whole argument: The Argument from One.

Iterate

unity

Factorisation

empty

Ω = 0

A(n)

boundary; closed form = 1

Motion

contracts

Closed form at 1

0 + 2⁰ − 0 = 1

Empty product. Frozen.

Boundary

A(1) = 2

The 1 argument. FN can begin.

Orbit of 1

The closed form at 1 is the empty product: 0 + 2⁰ − 0 = 1. The boundary refuses that reading and sends 1 to 2. The orbit meets a door at 11, then 12.

The three doors

A(11) = A(21) = A(25) = 12, and nothing else does. Try 21 or 25 to enter through a door that is not on the principal spine.

Principal spine

-1
0
1
2
3
4
6
7
8
11
12∞

The interval [−1, 12]

Fourteen positions. Filled: the spine. Open: 5, 9, 10 — recovered, not refused.

A(n) for n = 2 … 24

n	A(n)	motion
	3	+1
	4	+1
	6	+2
	6	+1
	7	+1
	8	+1
	11	+3
	8	-1
	9	-1
	12	+1
	12	fixed
	14	+1
	11	-3
	10	-5
	20	+4
	18	+1
	13	-5
	20	+1
	14	-6
	12	-9
	15	-7
	24	+1
	21	-3