Every bud on the main cardiod is pinched off by a pair of rays. The table below provides the ray angles for some of the buds on the main cardiod. In this table, p and q are the bud label, as per Devaney, and a_{1} and a_{2} are the angles of the two landing rays that pinch off that bud from each side.
p | q | a_{1} | a_{2} |
1 | 2 | 1/3 | 2/3 |
1 | 3 | 1/7 | 2/7 |
1 | 4 | 1/15 | 2/15 |
1 | 5 | 1/31 | 2/31 |
1 | ... | ... | ... |
1 | n | 1/(2^{n}-1) | 2/(2^{n}-1) |
... | ... | ... | ... |
2 | 5 | 9/31 | 10/31 |
2 | 7 | 17/127 | 18/127 |
2 | 9 | 33/511 | 34/511 |
2 | 11 | 65/2047 | 66/2047 |
2 | ... | ... | ... |
2 | 2n+1 | (2^{n+1}+1)/(2^{2n+1}-1) | (2^{n+1}+2)/(2^{2n+1}-1) |
... | ... | ... | ... |
3 | 7 | 41/127 | 42/127 |
3 | 8 | 73/255 | 74/255 |
3 | 10 | 145/1023 | 145/1023 |
3 | 11 | 273/2047 | 274/2047 |
3 | 13 | 545/8191 | 546/8191 |
3 | 14 | 1057/16383 | 1058/16383 |
3 | ... | ... | ... |
3 | 3n+1 | (2^{2n+1}+2^{n+1}+1)/(2^{3n+1}-1) | (2^{2n+1}+2^{n+1}+2)/(2^{3n+1}-1) |
3 | 3n+2 | (2^{2n+2}+2^{n+1}+1)/(2^{3n+2}-1) | (2^{2n+2}+2^{n+1}+2)/(2^{3n+2}-1) |
... | ... | ... | ... |
4 | 4n+1 | (2^{3n+1}+2^{2n+1}+2^{n+1}+1)/(2^{4n+1}-1) | ... |
4 | 4n+3 | (2^{3n+3}+2^{2n+2}+2^{n+1}+1)/(2^{4n+3}-1) | ... |
The 'main sequence' buds on the main cardiod have periodicity n and have Farey labels 1/n. We define the codeword for these buds to be n-1 zeros followed by 1. Its length is of course n binary digits or bits. If we put a decimal point in front of the codeword, it would equal 1/2^{n}. If we made the codeword repeat itself an infinite number of times, then its value would be 1/(2^{n}-1). The Bernoulli map of this codeword has period n, the same as the period of the bud. And, of course, the smaller of the two ray angles for this bud is just the value of the repeated codeword, viz. 1/(2^{n}-1).
The codewords for other buds are given by concatenating the codewords of their Farey parents, with the left parent contributing the codeword on the right, and the right parent the left codeword. Thus, the codeword for 2/(2n+1) = 1/n {+} 1/(n+1) is (n-1) zeros, 1, n zeros,1, or expressed as a fraction, (2^{n+1}+1)/2^{2n+1}. The length of this codeword is of course, (2n+1), the same as the Farey fraction denominator (which is the bud periodicity). The ray angle is given by repeating the codeword over and over, viz. (2^{n+1}+1)/(2^{2n+1}-1). It is not hard to verify that by concatenating codewords, we can get every other value in the table above.
Of course, this begs the question, what features are other codewords associated with? The answer is, of course, the tails.
Thus, for example, the tail of the bud at (re,im)=(-1.75,0) is rayed by 3/7 and 4/7. Now pick a number, any number, that points at an interesting feature on the main bulb. Say its binary expansion is 0.b_{1}b_{2}b_{3}b_{4}... Create a new binary expansion by concatenating the two strings (011) and (100). Whenever b_{k} is zero, use (011), and if its one, use (100). Thus, for example, on the main bulb, the ray 1/3 = 0.0101010101... pinches off the west bud. Substituting, we get 0.011100011100011100 .. = 4/9. Indeed, as the picture shows, that's where the 4/9th's ray goes.
Generally, the two rays that point at the tail-end are given by p/(2^{n}-1) and q/(2^{n}-1) for some p, q and n. One the main sequence, we have q=p+1. The two codewords for the above substitution are then p/2^{n} for 0 and q/2^{n} for 1.
More generally, we have 1/(2^{n}-1) whose continued fraction is [0;n,1,(n-1),1,(n-1),1,(-1),1,...]. The value of this continued fraction is 2/(n+1+sqrt(n^{2}+2n-3)). The table below gives some other relationships.
Farey Number | Expansion | Value | Notes |
1/(2^{n}-1) | [0;n,1,(n-1),1,(n-1),1,(n-1),1,...] | 2/(n+1+sqrt(n^{2}+2n-3)) | |
2/(2^{n}-1) | [0;(n-1),1,(n-1),1,(n-1),1,(n-1),1,...] | 2/(n-1+sqrt(n^{2}+2n-3)) | |
2^{m}/(2^{n}-1) | [0;(n-m),1,(n-1),1,(n-1),1,(n-1),1,...] | 2/(n-2m+1+sqrt(n^{2}+2n-3)) | m < n |
1/n | [0;n] | 1/2^{n} |