if you keep iterating you will go blind

Ok, we start by identifing the set we're working in as X which is to say, the music space. This is the numerical space created by using musical pieces as coresponding to fininte line segments or sequences. Let the euclidian metric be defined in this numerical set. This is a metric space, by the definition and will be mapped into itself. K sub zero is defined as the original piece of music. Next we compress K sub zero to one third is original langth, this is K0c. To get K sub 1 we played K0c three times then overlay a reversed K0c, this is K0cr, onto the middle iteration of K0c. K2 is defined as three iterations of K1c with K1cr overlayed the middle iteration of K1c. Naturally K3 is done the same way. Three iterations of K2c and K2cr overlayed the second iteration. The basic formula for Kn is:

K(n-1)c+{K(n-1)c*K(n-1)cr}+K(n-1)c

where * signifies overlay rather than multiplication.
The aformetioned being a finite set of strictly contractive transformations

Kn : X -> X, 0 _< n _< 4

with respective contractivity factors Sn. Sn is defined as one third to the nth power. Then we define the transformation K :H(X)->H(X), where H(X) is the collection of nonempty, compact subsets of X, K(M) is the union of all the contraction mappings from 0 to 4 created by Kn of M, for any M in H(X).

Providing that K is a strict contraction, it follows from the contraction maping system that if the metric space is compleat, then K has a unique fixed point A in H(X) satisfying the selfcovering condition (listed as def. 2.2 in Discrete Iterated Function Systems by Mario Peruggia p.17). This is used as the outline supporting this explination. This leads there to be a hyperbolic iterated function system with all of its defined properties. By def. 2.4, the collection {X, K-1, K0, K1, K2, K3, K4} where K1,...,K4 are strict contractions on X is called a hyperbolic iterated function system with condensation, if K-1 maps every set that is an element to H(X) into the same fixed nonempty, compact set C. This implies a unique set A as previously mentioned which is the attractor of the IFS with condensation. A satisfies the property the attractor is given by the union of the N images of it self under (K0, ..., K4, condensation set C) so that K-1 maps every nonempty, compact subset of the plain into a segment of the Koch snowflake.
And Given that the Koch Snowflake is selfcovering at all iterations
K0 has 1 or 4 or 16
K1 has 7 or 28
K2 has 34 or more
K3 has 166 or more
K4 has 752

Knumber of coverings = number of external sides in the previous iteration + number of internal divisions
each iteration the internal division is equal to 1/4 the size of the previous iteration's triangles

Kinternal = ((Kn-1)*4)
Kexternal = Kn -1
Knoc = Kinternal + Kexternal

Whereby verifying the self covering property of the attractor, previously mentioned. And in four iterations we have an image with affined copies of itself, that are selfcovering. K-1 is unique affined transformation of X into itself and is uniquely determined by any three points and their images under transformation, provided such points are not colinear.(YEAH!)

Right so the angle on each side is the nearest completely interior triangle translated to the nearest segment. The only angles involed are 60, 120 and 240.
60*2 = 120
120*2 = 240
so 60*4 = 240
This means that you know exactly what angle you're looking at by how many triangles touch it. All the triangles are equalateral so it doesn't matter what size the triangles are because they all have 60 degree angles. This implies that an algorhythm of the Koch curve would contain various multiples of 60. Divide the triangle into four congruent equalateral triangles. Take the top one turn it 60 degrees and translate it to the middle third of the nearest segment on which it point outward. repeat process for 120 and 240 degrees. If you start with the first rotated triangle the trip is only 60 degrees and the sixty again to reach 240. Inorder to create the iteration 2 you rotate 60 degres clockwise and paste it counterclockwise. So the algorhythm would contain a alternating copy and paste function. Now, beginning with the division stage, repeat the entire process. This will produce the next two iterations. ad infinitum, ad nauseum. Or until you're blind...whichever comes first.(w00t!)