Math::PlanePath::RationalsTree -- rationals by tree
use Math::PlanePath::RationalsTree; my $path = Math::PlanePath::RationalsTree->new (tree_type => 'SB'); my ($x, $y) = $path->n_to_xy (123);
This path enumerates rational fractions X/Y in reduced form, ie. X and Y having no common factor.
The rationals are traversed by rows of a binary tree which effectively represents a coprime pair X,Y by steps of a subtraction-only greatest common divisor algorithm proving them coprime. Or equivalently by bit runs with lengths which are the quotients in the Euclidean GCD algorithm, which are also the terms in the continued fraction representation of X/Y.
The SB, CW, Bird, Drib and AYT trees all have the same set of X/Y fractions in a row, but in a different order due to different encodings of the N value, high to low or low to high and possible bit flips. The L tree has a shift which visits zero as 0/1 too.
The bit runs mean that N values are quite large for relatively modest sized rationals. For example 167/3 is N=288230376151711741, a 58-bit number. The tendency is for the tree to travel out to large rationals while yet to fill in small ones. The worst is the integer X/1 has N with X many bits, and similarly 1/Y has Y bits.
See examples/rationals-tree.pl in the PlanePath sources for a printout of all the trees.
The default tree_type=>"SB" is the tree of Moritz Stern and Achille Brocot. The rows are fractions of increasing value.
tree_type=>"SB"
N=1 1/1 ------ ------ N=2 to N=3 1/2 2/1 / \ / \ N=4 to N=7 1/3 2/3 3/2 3/1 | | | | | | | | N=8 to N=15 1/4 2/5 3/5 3/4 4/3 5/3 5/2 4/1
Writing the parents between the children as an "in-order" tree traversal to a given depth has all values in increasing order too,
1/1 1/2 | 2/1 1/3 | 2/3 | 3/2 | 3/1 | | | | | | | 1/3 1/2 2/3 1/1 3/2 2/1 3/1 ^ | 4/3 next level = (1+3)/(1+2)
New values are a "mediant" (x1+x2)/(y1+y2) formed from the left and right parent in this flattening. So the next level 4/3 is left parent 1/1 and right parent 3/2 forming mediant (1+3)/(1+2)=4/3. At the left end is imagined a preceding 0/1 and at the right a following 1/0, so as to add 1/level and level/1 at the ends for a total 2^level many new values.
Plotting the N values by X,Y is as follows. The unused X,Y positions are where X and Y have a common factor. For example X=6,Y=2 has common factor 2 so is never reached.
10 | 512 35 44 767 9 | 256 33 39 40 46 383 768 8 | 128 18 21 191 384 7 | 64 17 19 20 22 95 192 49 51 6 | 32 47 96 5 | 16 9 10 23 48 25 26 55 4 | 8 11 24 27 56 3 | 4 5 12 13 28 29 60 2 | 2 6 14 30 62 1 | 1 3 7 15 31 63 127 255 511 1023 Y=0 | ---------------------------------------------------- X=0 1 2 3 4 5 6 7 8 9 10
The X=1 vertical is the fractions 1/Y which is at the left of each tree row, at N value
Nstart = 2^level
The Y=1 horizontal is the X/1 integers at the end each row which is
Nend = 2^(level+1)-1
tree_type=>"CW" selects the tree of Neil Calkin and Herbert Wilf,
tree_type=>"CW"
"Recounting the Rationals", http://www.math.upenn.edu/~wilf/website/recounting.pdf
As noted above, the values within each row are the same as the Stern-Brocot, but in a different order.
N=1 1/1 ------ ------ N=2 to N=3 1/2 2/1 / \ / \ N=4 to N=7 1/3 3/2 2/3 3/1 | | | | | | | | N=8 to N=15 1/4 4/3 3/5 5/2 2/5 5/3 3/4 4/1
Going across by rows the denominator of one value becomes the numerator of the next. So at 4/3 the denominator 3 becomes the numerator of the 3/5 to the right. These values are Stern's diatomic sequence.
Each row is symmetric in reciprocals, ie. reading from right to left is the reciprocals of reading left to right. The numerators read left to right are the denominators read right to left.
A node descends as
X/Y / \ X/(X+Y) (X+Y)/Y
Taking these formulas in reverse up the tree shows how it relates to a subtraction-only greatest common divisor. At a given node the smaller of P or Q is subtracted from the bigger,
P/(Q-P) (P-Q)/P / or \ P/Q P/Q
Plotting the N values by X,Y is as follows. The X=1 vertical and Y=1 horizontal are the same as the SB above, but the values in between are re-ordered.
tree_type => "CW" 10 | 512 56 38 1022 9 | 256 48 60 34 46 510 513 8 | 128 20 26 254 257 7 | 64 24 28 18 22 126 129 49 57 6 | 32 62 65 5 | 16 12 10 30 33 25 21 61 4 | 8 14 17 29 35 3 | 4 6 9 13 19 27 39 2 | 2 5 11 23 47 1 | 1 3 7 15 31 63 127 255 511 1023 Y=0 | ------------------------------------------------------------- X=0 1 2 3 4 5 6 7 8 9 10
At each node the left leg is X/(X+Y) < 1 and the right leg is (X+Y)/Y > 1, which means even N is above the X=Y diagonal and odd N is below. In general each right leg increments the integer part of the fraction,
X/Y right leg each time (X+Y)/Y = 1 + X/Y (X+2Y)/Y = 2 + X/Y (X+3Y)/Y = 3 + X/Y etc
This means the integer part floor(X/Y) is given by the number of trailing 1-bits of N. For example 7/2 is at N=23 binary "10111" which has 3 trailing 1-bits for floor(7/2)=3.
N values for the SB and CW trees are converted by reversing bits. So at a given X,Y position
SB N = 1abcde SB <-> CW by reversing bits CW N = 1edcba except the high 1-bit
For example at X=3,Y=4 the SB tree has N=11 = "1011" binary and the CW has N=14 binary "1110", a reversal of the bits below the high 1.
N to X/Y in the CW tree can be calculated keeping track of just an X,Y pair and descending to X/(X+Y) or (X+Y)/Y using the bits of N from high to low. The relationship between the SB and CW N's means the same can be used to calculate the SB tree by taking the bits of N from low to high instead.
tree_type=>"AYT" selects the tree described (independently is it?) by D. N. Andreev and Shen Yu-Ting.
tree_type=>"AYT"
http://files.school-collection.edu.ru/dlrstore/d62f7b96-a780-11dc-945c-d34917fee0be/i2126134.pdf
Shen Yu-Ting, "A Natural Enumeration of Non-Negative Rational Numbers -- An Informal Discussion", American Mathematical Monthly, 87, 1980, pages 25-29. http://www.jstor.org/stable/2320374
Their constructions are a one-to-one mapping between integer N and rational X/Y as a way of enumerating the rationals. It's not designed to be a tree as such, but the result is the same 2^level rows as the above trees. The X/Y values within each row are again the same, but in a further different order.
N=1 1/1 ------ ------ N=2 to N=3 2/1 1/2 / \ / \ N=4 to N=7 3/1 1/3 3/2 2/3 | | | | | | | | N=8 to N=15 4/1 1/4 4/3 3/4 5/2 2/5 5/3 3/5
Each fraction descends as follows. The left is an increment and the right is reciprocal of the increment.
X/Y / \ X/Y + 1 1/(X/Y + 1)
which means
X/Y / \ (X+Y)/Y Y/(X+Y)
The left leg (X+Y)/Y is the same as in the CW has on the right. But Y/(X+Y) is not the same as the CW (the other there being X/(X+Y)).
As per the CW tree right leg, the AYT left leg increments the integer part, so
floor(X/Y) = count trailing 0-bits of N plus one extra if N=2^k
N=2^k is one extra because its trailing 0-bits started from N=1 where floor(1/1)=1 whereas any other odd N has floor(X/Y)=0.
The Y/(X+Y) right leg forms the Fibonacci numbers F(k)/F(k+1) at the end of each row, ie. at Nend=2^(level+1)-1. And as noted by Andreev, successive right leg fractions N=4k+1 and N=4k+3 add up to 1, ie.
X/Y at N=4k+1 + X/Y at N=4k+3 = 1 Eg. 2/5 at N=13 and 3/5 at N=15 add up to 1
Plotting the N values by X,Y gives
tree_type => "AYT" 10 | 513 41 43 515 9 | 257 49 37 39 51 259 514 8 | 129 29 31 131 258 7 | 65 25 21 23 27 67 130 50 42 6 | 33 35 66 5 | 17 13 15 19 34 26 30 38 4 | 9 11 18 22 36 3 | 5 7 10 14 20 28 40 2 | 3 6 12 24 48 1 | 1 2 4 8 16 32 64 128 256 512 Y=0 | ---------------------------------------------------- X=0 1 2 3 4 5 6 7 8 9 10
N=1,2,4,8,etc on the Y=1 horizontal is the X/1 integers at Nstart=2^level=2^X. N=1,3,5,9,etc in the X=1 vertical is the 1/Y fractions. Those fractions always immediately follow the corresponding integer, so N=Nstart+1=2^(Y-1)+1 in that column.
In each node the left leg (X+Y)/Y > 1 and the right leg Y/(X+Y) < 1, which means odd N is above the X=Y diagonal and even N is below.
The tree structure corresponds to Johannes Kepler's tree of fractions (see Math::PlanePath::FractionsTree). That tree starts from 1/2 and makes fractions A/B with A<B by descending to A/(A+B) and B/(A+B). This is the same as the AYT tree with
A = Y AYT denominator is Kepler numerator B = X+Y AYT sum num+den is the Kepler denominator X = B-A inverse Y = A
tree_type=>"HCS" selects continued fraction terms coded as bit runs 1000...00 from high to low, as per Paul D. Hanna and independently Jerzy Czyz and Will Self.
tree_type=>"HCS"
http://oeis.org/A071766 http://www.cut-the-knot.org/do_you_know/countRatsCF.shtml http://www.dm.unito.it/~cerruti/doc-html/tremattine/tre_mattine.pdf
Jerzy Czyz and William Self, "The Rationals Are Countable: Euclid's Proof", The College Mathematics Journal, volume 34, number 5, November 2003, page 367.
This arises also in a radix=1 variation of Jeffrey Shallit's digit-based continued fraction encoding (see "Radix 1" in Math::PlanePath::CfracDigits).
If the continued fraction of X/Y is
1 X/Y = a + ------------ 1 b + ----------- 1 c + ------- ... + 1 --- z
then the N value is bit runs of lengths a,b,c etc.
N = 1000 1000 1000 ... 1000 \--/ \--/ \--/ \--/ a+1 b c z-1
Each group is 1 or more bits. Using a+1 for the first group makes it 1 or more, since a=0 occurs for any X/Y<=1. Using z-1 in the last group ensures it's 1 or more since z>=2.
N=1 1/1 ------ ------ N=2 to N=3 2/1 1/2 / \ / \ N=4 to N=7 3/1 3/2 1/3 2/3 | | | | | | | | N=8 to N=15 4/1 5/2 4/3 5/3 1/4 2/5 3/4 3/5
The result is a bit reversal of the AYT tree N values.
AYT N = binary "1abcde" AYT <-> HCS by bit reversal HCS N = binary "1edcba"
For example at X=4,Y=7 the AYT tree has N=11 binary "10111" and the HCS has N=30 binary "11110", a reversal of the bits below the high 1.
Plotting by X,Y gives
tree_type => "HCS" 10 | 768 50 58 896 9 | 384 49 52 60 57 448 640 8 | 192 27 31 224 320 7 | 96 25 26 30 29 112 160 41 42 6 | 48 56 80 5 | 24 13 15 28 40 21 23 44 4 | 12 14 20 22 36 3 | 6 7 10 11 18 19 34 2 | 3 5 9 17 33 1 | 1 2 4 8 16 32 64 128 256 512 Y=0 | +----------------------------------------------------- X=0 1 2 3 4 5 6 7 8 9 10
N=1,2,4,etc in the row Y=1 are powers-of-2 for integers X/1 having just a single group of bits N=1000..000.
N=1,3,6,12,etc in the column X=1 are 3*2^(Y-1) corresponding to continued fraction 1/Y = 0 + 1/Y so the high group is a single bit for the 0 term and then final group length Y-1, so bits N=11000...00.
tree_type=>"Bird" selects the Bird tree by Ralf Hinze
tree_type=>"Bird"
"Functional Pearls: The Bird tree", http://www.cs.ox.ac.uk/ralf.hinze/publications/Bird.pdf
It's expressed recursively, illustrating Haskell programming features. The subtrees are tree plus one reciprocal on the left, and tree reciprocal plus one on the right,
1/(tree + 1) and (1/tree) + 1
which means Y/(X+Y) and (X+Y)/X taking N bits low to high.
N=1 1/1 ------ ------ N=2 to N=3 1/2 2/1 / \ / \ N=4 to N=7 2/3 1/3 3/1 3/2 | | | | | | | | N=8 to N=15 3/5 3/4 1/4 2/5 5/2 4/1 4/3 5/3
tree_type => "Bird" 10 | 682 41 38 597 9 | 341 43 45 34 36 298 938 8 | 170 23 16 149 469 7 | 85 20 22 17 19 74 234 59 57 6 | 42 37 117 5 | 21 11 8 18 58 28 31 61 4 | 10 9 29 30 50 3 | 5 4 14 15 25 24 54 2 | 2 7 12 27 52 1 | 1 3 6 13 26 53 106 213 426 853 Y=0 | ---------------------------------------------------- X=0 1 2 3 4 5 6 7 8 9 10
Notice that unlike the other trees N=1,2,5,10,etc in the X=1 vertical of fractions 1/Y is not the row start or end, but instead are on a zigzag through the middle of the tree giving binary N=1010...etc alternate 1 and 0 bits. The integers X/1 in the Y=1 vertical are similar, but N=11010...etc starting the alternation from a 1 in the second highest bit, since those integers are in the right hand half of the tree.
The Bird tree N values are related to the SB tree by inverting every second bit starting from the second after the high 1-bit, ie. xor "001010...". So if N=1abcdefg binary then b,d,f are inverted, ie. an xor with binary 00101010. For example 3/4 in the SB tree is at N=11 = binary 1011. Xor with 0010 for binary 1001 N=9 which is the 3/4 in the Bird tree. The same xor goes back the other way Bird tree to SB tree.
This xoring is a mirroring in the tree, swapping left and right at each level. Only every second bit is inverted because mirroring twice puts it back to the ordinary way on even rows.
tree_type=>"Drib" selects the Drib tree by Ralf Hinze.
tree_type=>"Drib"
http://oeis.org/A162911
It reverses the bits of N in the Bird tree (in a similar way that the SB and CW are bit reversals of each other).
N=1 1/1 ------ ------ N=2 to N=3 1/2 2/1 / \ / \ N=4 to N=7 2/3 3/1 1/3 3/2 | | | | | | | | N=8 to N=15 3/5 5/2 1/4 4/3 3/4 4/1 2/5 5/3
The descendants of each node are
X/Y / \ Y/(X+Y) (X+Y)/X
The endmost fractions of each row are Fibonacci numbers, F(k)/F(k+1) on the left and F(k+1)/F(k) on the right.
tree_type => "Drib" 10 | 682 50 44 852 9 | 426 58 54 40 36 340 683 8 | 170 30 16 212 427 7 | 106 18 22 24 28 84 171 59 51 6 | 42 52 107 5 | 26 14 8 20 43 19 31 55 4 | 10 12 27 23 41 3 | 6 4 11 15 25 17 45 2 | 2 7 9 29 37 1 | 1 3 5 13 21 53 85 213 341 853 Y=0 | ------------------------------------------------------- X=0 1 2 3 4 5 6 7 8 9 10
In each node descent the left Y/(X+Y) < 1 and the right (X+Y)/X > 1, which means even N is above the X=Y diagonal and odd N is below.
Because Drib/Bird are bit reversals like CW/SB are bit reversals, the xor procedure described above which relates Bird<->SB applies to Drib<->CW, but working from the second lowest bit upwards, ie. xor binary "0..01010". For example 4/1 is at N=15 binary 1111 in the CW tree. Xor with 0010 for 1101 N=13 which is 4/1 in the Drib tree.
tree_type=>"L" selects the L-tree by Peter Luschny.
tree_type=>"L"
http://www.oeis.org/wiki/User:Peter_Luschny/SternsDiatomic
It's a row-reversal of the CW tree with a shift to include zero as 0/1.
N=0 0/1 ------ ------ N=1 to N=2 1/2 1/1 / \ / \ N=3 to N=8 2/3 3/2 1/3 2/1 | | | | | | | | N=9 to N=16 3/4 5/3 2/5 5/2 3/5 4/3 1/4 3/1
Notice in the N=9 to N=16 row rationals 3/4 to 1/4 are the same as in the CW tree but read right-to-left.
tree_type => "L" 10 | 1021 37 55 511 9 | 509 45 33 59 47 255 1020 8 | 253 25 19 127 508 7 | 125 21 17 27 23 63 252 44 36 6 | 61 31 124 5 | 29 9 11 15 60 20 24 32 4 | 13 7 28 16 58 3 | 5 3 12 8 26 18 54 2 | 1 4 10 22 46 1 | 0 2 6 14 30 62 126 254 510 1022 2046 Y=0 | ------------------------------------------------------- X=0 1 2 3 4 5 6 7 8 9 10
N=0,2,6,14,30,etc along the row at Y=1 are powers 2^(X+1)-2. N=1,5,13,29,etc in the column at X=1 are similar powers 2^Y-3.
The SB, CW, Bird, Drib, AYT and HCS trees have the same set of rationals in each row, just in different orders. The properties of Stern's diatomic sequence mean that within a row the totals are
row N=2^level to N=2^(level+1)-1 inclusive sum X/Y = (3 * 2^level - 1) / 2 sum X = 3^level sum 1/(X*Y) = 1
For example the SB tree level=2, N=4 to N=7,
sum X/Y = 1/3 + 2/3 + 3/2 + 3/1 = 11/2 = (3*2^2-1)/2 sum X = 1+2+3+3 = 9 = 3^2 sum 1/(X*Y) = 1/(1*3) + 1/(2*3) + 1/(3*2) + 1/(3*1) = 1
Many permutations are conceivable within a row, but the ones here have some relationship to X/Y descendants, tree sub-forms or continued fractions. As an encoding of continued fraction terms by bit runs the combinations are
bit encoding high to low low to high ---------------- ----------- ----------- runs 000 or 111 SB CW alternating 0101 Bird Drib runs 10000 HCS AYT
The runs of alternating 101010 end at a kind of phase shift, where the bit is not the expected alternating 0,1 but instead a doubled 00 or 11.
The Minkowski question mark function is a +/- sum of the quotients in the continued fraction of a real number,
1 1 1 ?(r) = 2 * (1 - ---- + --------- - ------------ + ... ) 2^q0 2^(q0+q1) 2^(q0+q1+q2)
For a rational r the continued fraction is finite and so the sum is rational too. The pattern of + and - in the terms gives runs of bits the same as the N values in the Stern-Brocot tree. The code here can calculate the ? function on a rational r=X/Y using
N = xy_to_n(X,Y) tree_type=>"SB" depth = floor(log2(N)) # row containing N (depth=0 at top) Ndepth = 2^depth # start of row containing N 2*(N-Ndepth) + 1 ?(X/Y) = ---------------- Ndepth
The effect of N-Ndepth is to remove the high 1-bit, and 2*()+1 appends an extra 1-bit at the end. The division /Ndepth scales down from integer N to a fraction.
N = 1abcdef in binary ? = a.bcdef1 binary fraction
For example ?(2/3) is X=2,Y=3 which is N=5 in SB. It has depth=2, Ndepth=2^2=4, and so ?(2/3)=(2*(5-4)+1)/4=3/4. Or in binary N=101 gives Ndepth=100 and N-Ndepth=01 so 2*(N-Ndepth)+1=011 and divide Ndepth=100 for ?=0.11.
In practice this is not a very efficient way to handle the question function, since the bit runs in the N values may become quite large for relatively modest continued fraction terms. (Math::ContinuedFraction may be better, and in particular allows repeating patterns of quadratic irrationals to be represented exactly.)
See "FUNCTIONS" in Math::PlanePath for behaviour common to all path classes.
$path = Math::PlanePath::RationalsTree->new ()
$path = Math::PlanePath::RationalsTree->new (tree_type => $str)
Create and return a new path object. tree_type (a string) can be
tree_type
"SB" Stern-Brocot "CW" Calkin-Wilf "Bird" "Drib" "AYT" Andreev, Yu-Ting "HCS" "L"
$n = $path->n_start()
Return the first N in the path. This is 1 for SB, CW, Bird, Drib and AYT, but 0 for L.
($n_lo, $n_hi) = $path->rect_to_n_range ($x1,$y1, $x2,$y2)
Return a range of N values which occur in a rectangle with corners at $x1,$y1 and $x2,$y2. The range is inclusive.
$x1
$y1
$x2
$y2
For reference, $n_hi can be quite large because within each row there's only one new X/1 integer and 1/Y fraction. So if X=1 or Y=1 is included then roughly $n_hi = 2**max(x,y). If min(x,y) is bigger than 1 then it reduces a little to roughly 2**(max/min + min).
$n_hi
$n_hi = 2**max(x,y)
@n_children = $path->tree_n_children($n)
Return the two children of $n, or an empty list if $n < 1 (ie. before the start of the path).
$n
$n < 1
This is simply 2*$n, 2*$n+1. The children are $n with an extra bit appended, either a 0-bit or a 1-bit.
2*$n, 2*$n+1
$num = $path->tree_n_num_children($n)
Return 2, since every node has two children. If $n<1 (ie. before the start of the path) then return undef.
$n<1
undef
$n_parent = $path->tree_n_parent($n)
Return the parent node of $n. Or return undef if $n <= 1 (the top of the tree).
$n <= 1
This is simply Nparent = floor(N/2), ie. strip the least significant bit from $n (undoing what tree_n_children() appends).
tree_n_children()
$depth = $path->tree_n_to_depth($n)
Return the depth of node $n, or undef if there's no point $n. The top of the tree at N=1 is depth=0, then its children depth=1, etc.
This is simply floor(log2(N)) since the tree has 2 nodes per point. For example N=4 through N=7 are all depth=2.
The L tree starts at N=0 and the calculation becomes floor(log2(N+1)) there.
$n = $path->tree_depth_to_n($depth)
$n = $path->tree_depth_to_n_end($depth)
Return the first or last N at tree level $depth in the path, or undef if nothing at that depth or not a tree. The top of the tree is depth=0.
$depth
The structure of the tree means the first N is at 2**$depth, or for the L tree 2**$depth - 1. The last N is 2**($depth+1)-1, or for the L tree 2**($depth+1).
2**$depth
2**$depth - 1
2**($depth+1)-1
2**($depth+1)
The trees are in Sloane's Online Encyclopedia of Integer Sequences in various forms,
http://oeis.org/A007305 (etc) A007305 SB X numerators, Farey fractions (extra 0,1) A047679 SB Y denominators A007306 SB X+Y sum, Farey 0 to 1 part (extra 1,1) A153036 SB int(X/Y), integer part A002487 CW X and Y, Stern diatomic sequence (extra 0) A070990 CW Y-X diff, Stern diatomic first diffs (less 0) A070871 CW X*Y product A007814 CW int(X/Y), integer part, count trailing 1-bits which is count trailing 0-bits of N+1 A020650 AYT X A020651 AYT Y (Kepler X) A086592 AYT X+Y sum (Kepler denominators) A135523 AYT int(X/Y), integer part, count trailing 0-bits plus 1 extra if N=2^k A071766 HCS Y A071585 HCS X+Y sum (sum giving rationals >= 1) A162909 Bird X A162910 Bird Y A162911 Drib X A162912 Drib Y A174981 L-tree X A002487 L-tree Y, same as CW X,Y, Stern diatomic A000523 tree_n_to_depth(), being floor(log2(n)) A086893 position Fibonacci F[n+1],F[n] in Stern diatomic, CW N of F[n+1]/F[n] Drib N on row Y=1, being X/1 A061547 position Fibonacci F[n],F[n+1] in Stern diatomic, CW N of F[n]/F[n+1] Drib N in column X=1, being 1/Y A081254 Bird N in row Y=1, binary 110101010...10 A000975 Bird N in column X=1, binary 1010..1010 A088696 length of continued fraction SB left half (X/Y<1) A059893 permutation SB<->CW, reverse bits below highest A153153 permutation CW->AYT, reverse and un-Gray A153154 permutation AYT->CW, reverse and Gray code A154437 permutation AYT->Drib, Lamplighter low to high A154438 permutation Drib->AYT, un-Lamplighter low to high A003188 permutation SB->HCS, Gray code shift+xor A006068 permutation HCS->SB, Gray code inverse A154435 permutation HCS->Bird, Lamplighter bit flips A154436 permutation Bird->HCS, Lamplighter variant A054429 permutation SB,CW,Bird,Drib N at transpose Y/X, (mirror binary tree, runs 0b11..11 down to 0b10..00) A004442 permutation AYT N at transpose Y/X, from N=2 onwards (xor 1, ie. flip least significant bit) A063946 permutation HCS N at transpose Y/X, extra initial 0 (xor 2, ie. flip second least significant bit) A054424 permutation DiagonalRationals -> SB A054426 permutation SB -> DiagonalRationals A054425 DiagonalRationals -> SB with 0s at non-coprimes A054427 permutation coprimes -> SB right hand X/Y>1
The sequences marked "extra ..." have one or two extra initial values over what the RationalsTree here gives, but are the same after that. And the Stern first differences "less ..." means it has one less term than what the code here gives.
Math::PlanePath, Math::PlanePath::FractionsTree, Math::PlanePath::CfracDigits, Math::PlanePath::CoprimeColumns, Math::PlanePath::DiagonalRationals, Math::PlanePath::FactorRationals, Math::PlanePath::GcdRationals, Math::PlanePath::PythagoreanTree
Math::NumSeq::SternDiatomic, Math::ContinuedFraction
http://user42.tuxfamily.org/math-planepath/index.html
Copyright 2011, 2012, 2013 Kevin Ryde
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