Simple Observations on Ford Circles (Speiser circles
or Haros Circles) with the Harmonic Sequence that Suggest Deeper Connections to
Number Theory
Richard G. Lanzara
These are vast areas that need further exploration. I only
present my humble and simple observations that may help those with a deeper
knowledge of number theory to make future discoveries.
Like pearls on a necklace, there are two parabolas that
intersect the center of the Ford circles (Speiser circles or Haros circles)
that include the Harmonic sequence (Hn) (1/2,1/3,1/4,1/5,…) (y=(1/2)x^2) and
the sequence ((n1)/n) (1/2,2/3,3/4,4/5,5/,…) (y=(1/2)(x1)^2). To this
author’s knowledge, these parabolas were previously overlooked, but appear to
be an essential connection between the Farey sequence, Ford circles and the
Harmonic sequence. They may have deeper connections to number theory that are
beyond this author’s expertise.
Figure 1: Connections between the Ford circles and the two
parabolas (y=(1/2)x^2 and y=(1/2)(x1)^2).
Looking at the tangent lines between the two adjacent Ford
circles (for example at, x=1/2 and 1/3 on the left side and x=1/2 and 2/3 on
the right side), we see that for each pair of Ford circles on each side of the
middle Ford circle, at x=1/2, the tangent lines always converge on the x=1/2 line
(proof not shown).
Figure 2: The tangent lines between the Ford circles meet on
the x=1/2 line.
All the yintercept points where these pairs of tangent
lines meet are on the x=1/2 line, and are given by the following progression of
intercept points: 1/5, 5/7, 11/9, 19/11, 29/13, 41/15, 55/17, 71/19,
89/21, 101/23… These intercept points have the odd numbers as denominators
with the numerators being of the form n+(n+1)^2, which is found in the integer
sequence data base (oeis) as A028387 (a(n)=n+(n+1)^2) and A110331 (related to
the Pell numbers) and A165900, which is also the Fibonacci polynomial n^2 – n
1 (which is the same as n+(n+1)^2 when n+2 is substituted for n in n^2 – n
1).
***Note that these intersections of the tangent lines progress
down the negative side of the yaxis. They all converge at (0,1) or (1,1) for
each side of the xaxis (0,1) respectively.
Figure 3: Showing just the tangent lines between successive
Ford circles that all intercept the x=1/2 line.
Table I: Equations for the paired tangent lines from
the Ford circles surrounding the circle at x= 1/2, (x1/2)^2 + (y1/8)^2 =
(1/8)^2
Equations for
the tangent lines between the two Ford circles

For the Ford
circles at x:

yintercept
at the line x=1/2

y=112x/5

1/2,1/3

1/5

y=12x/57/5

1/2,2/3

1/5

y=124x/7

1/3,1/4

5/7

y=1/7(24x17)

2/3,3/4

5/7

y=140x/9

1/4,1/5

11/9

y=1/9(40x31)

3/4,4/5

11/9

y=160x/11

1/5,1/6

19/11

y=1/11(60x49)

4/5,5/6

19/11

y=184x/13

1/6,1/7

29/13

y=1/13(84x71)

5/6,6/7

29/13

y=1112x/15

1/7,1/8

41/15

y=1/13(112x97)

6/7,7/8

41/15

Checking oeis.org for the general forms for the numerators
and denominators of the tangent line intercepts with the x=1/2 line yields the
following general equation for these intercepts, which has the following limit
for tangent line intercepts equation:
The Appendix lists more details about
the tangent line intercepts, Pythagorean triangles, modified Farey sequences
and modified Ford circles (see Appendix below).
Mapping the Ford
Circles to the Complex Plane
These figures show the before and after mapping of the Ford
circles to the complex plane. The two Ford circles at x=0 and x=1 with both
upper and lower bound parabolas are included to better visualize this mapping. Figure
4: The Ford circles with two bounding parabolas, y=(x^2)/2 and y=1(x^2)/2.
If we allow for a mapping of the Ford circles to the complex
plane by changing x to z, then we have the following: Figure 5: The mapping of
the Ford circles with two bounding parabolas, y=(x^2)/2 and y=1(x^2)/2 to the
complex plane. Upper and lower plots Demonstrate the Mirror Symmetries Through
the Intercepts on the Line y = 1/2.
Figure 6: By translating the yaxis to z = 0.5, the
symmetries around the tangent line intercepts from the Ford circles become evident.
The tangent lines of intersection intersect on the y=1/2
line in the complex plane. Note their mirror symmetry and that these lines
would extend from the Ford circles to symmetrically intersect the points on the
line y = 1/2.
If we consider a picture of an Apollonian gasket, or a Ford
configuration of Apollonian type, or Ferry gasket (not shown), then the Ford
circles at the top of the gasket (the line y=1 in our case), touching the upper
line, will have tangent lines that intersect in a mirror image to these we have
seen for the tangent lines for the bottom Ford circles.
Figure 7: When mapped onto the complex plane, these tangent
line intersections proceed down both the negative and positive sides of the
zaxis at y=1/2.
Figure 8: The symmetries
suggesting an Apollonian packing.
Concluding remarks:
Perhaps Figures 1 and 7 are two dimensional representations
of a threedimensional Apollonian packing (also called Apollonian sphere
packing, which have many beautiful examples on the internet). Since all proper
fractions are underneath the umbrellas of the two parabolas (1/2x^2 and 1/2(x1)^2)
and the unique properties of the Farey sequence insures the orderly inclusion
of all proper fractions, the circle at x=1/2 represents a boundary condition.
One might visualize this as picking up the x=1/2 line (y=1/2 line in the
complex case, Figure 7) with the parabolas forming the sides of this unique
threedimensional shape.
Appendix
Tangent line
intercepts:
Table I: Table of the paired tangent
line intercepts at x=1/2 with the equations for the tangent lines from the Ford
circles surrounding the center circle at x= 1/2, (x1/2)^2 + (y1/8)^2 =
(1/8)^2.
yequations for
the tangent lines

For the Ford
circles at the x values

yintercepts* at
the line x=1/2

(x,y) values for
the centers of the two circles

Perpendicular
equations to the tangent line equations

xintercepts** at
y=0

y=112x/5

1/2,1/3

1/5

(1/3,1/18)
(1/2,1/8)

y=5/12x1/12

5/12

y=12x/57/5

1/2,2/3

1/5

(1/2,1/8)
(2/3,1/18)

y=5/12x+1/3

7/12

y=124x/7

1/3,1/4

5/7

(1/3,1/18)
(1/4,1/32)

y=7/24x1/24

7/24

y=1/7(24x17)

2/3,3/4

5/7

(2/3,1/18)
(3/4,1/32)

y=5/12x+1/3

17/24

y=140x/9

1/4,1/5

11/9

(1/5,1/50)
(1/4,1/32)

y=9/40x1/40

9/40

y=1/9(40x31)

3/4,4/5

11/9

(4/5,1/50)
(3/4,1/32)

y=5/12x+1/3

31/40

y=160x/11

1/5,1/6

19/11

(1/6,1/72)
(1/5,1/50)

y=11/60x1/60

11/60

y=1/11(60x49)

4/5,5/6

19/11

(5/6,1/72)
(4/5,1/50)

y=5/12x+1/3

49/60

y=184x/13

1/6,1/7

29/13

(1/7,1/98)
(1/6,1/72)

y=13/84x1/84

13/84

y=1/13(84x71)

5/6,6/7

29/13

(6/7,1/98)
(5/6,1/72)

y=5/12x+1/3

71/84

y=1112x/15

1/7,1/8

41/15

(1/8,1/128)
(1/7,1/98)

y=15/112x1/112

15/112

y=1/13(112x97)

6/7,7/8

41/15

(7/8,1/128)
(6/7,1/98)

y=5/12x+1/3

97/112

*All the yintercept points where
these pairs of tangent lines meet are on the x=1/2 line, and are given by the
following progression of intercept points: 1/5, 5/7, 11/9, 19/11, 29/13,
41/15, 55/17, 71/19, 89/21, 101/23… These intercept points have the odd
numbers as denominators with the numerators being of the form n+(n+1)^2, which
is found in the integer sequence data base as A028387 (a(n)=n+(n+1)^2) and
A110331 (related to the Pell numbers) and A165900, which is also the Fibonacci
polynomial n^2 – n 1 (which is the same as n+(n+1)^2 when n+2 is substituted
for n in n^2 – n 1).
**Roots: Subtracting the
complimentary equations for y (eg. y=(112x/5)(12x/57/5)) always gives x=1/2
as a root, whereas dividing them gives the value for the first xintercept (eg.
y=(112x/5)/(12x/57/5) gives x= 5/12 as the root).
The general equation
for the tangent line intercepts with the x=1/2 line:
y = ((((n1) + n^2)/(2(n+1) + 1)))
which has roots:
n = 1/f and f,
which are the reciprocal and negative values of the golden
ratio (1/2(1 + sqrt(5)). This appears to connect the Fibonacci sequence (the
ratios of successive terms equal the golden ratio) with the tangent lines
between successive Ford circles.
Pythagorean
triangles:
Another fascinating
property of these tangent lines is that they form groups of Pythagorean
triangles with perfect square rational numbers. If we look at where the
extensions of the two adjacent circles’ radii meet with the tangent lines, we
see that the base of the triangle is formed by the distance between the two
circles centers and the height is two times the distance where the yaxis hits
the lines extended from the two circles’ radii.
The tangent line
intercepts and representative Pythagorean triangles:
It is interesting to
note that the Ford circles have a symmetry around, x=1/2, that appears to
connect their parabolic, circular and tangent relationships with rational
Pythagorean triangles.
Table II: Pythagorean triangles:
Equations for the tangent lines from the Ford circles surrounding the circle at
x= 1/2, (x1/2)^2 + (y1/8)^2 = (1/8)^2, and containing the Harmonic sequence.
yequations for
the tangent lines

xintercept* at
y=0

yintercept at
x=1/2

Rational
Pythagorean triples**

y=112x/5

5/12

1/5

(6/5,1/2,13/10)

y=12x/57/5

7/12

1/5


y=124x/7

7/24

5/7

(12/7,1/2,25/14)

y=1/7(24x17)

17/24

5/7


y=140x/9

9/40

11/9

(20/9,1/2,41/18)

y=1/9(40x31)

31/40

11/9


y=160x/11

11/60

19/11

(30/11,1/2,61/22)

y=1/11(60x49)

49/60

19/11


y=184x/13

13/84

29/13

(42/13,1/2,
85/26)

y=1/13(84x71)

71/84

29/13


y=1112x/15

15/112

41/15

(56/15,1/2,
113/30)

y=1/13(112x97)

97/112

41/15


**In general, the
equation for the Pythagorean triangles for the tangent lines of the Ford
circles for the Harmonic sequence is:
For circle, c1,
where x1 = 1/a gives the position along the xaxis for c1, and c2, where x2 =
1/b
(2/(a+b))^2 +
(1/(ab))^2 = c^2
For example: for the
Ford circles at 1/5 and 1/6, this gives:
(2/11)^2 + (1/30)^2
= (61/330)^2
In lowest terms this
is:
60^2 + 11^2 = 61^2
Each Pythagorean
triangle can be divided into two equivalent triangles where the tangent line
crosses the xaxis. In fact, there is a larger Pythagorean triangle related to
these other three. This larger triangle is formed where the tangent line
crosses the yaxis and where the lowest point of the other triangles meet
(where the tangent line crosses the perpendicular line through the center of the
larger circle not shown).
Modified Farey
sequence and modified Ford circles:
There is much to
discover here, but in the future, I attempt to write more about modified Farey
sequences with their accompanying modified Ford circles and their corresponding
rational Pythagorean triangles in part II.
For the modified
Ford circles:
Modified Farey
Sequence and Modified Ford Circles:
Beginning the Farey
sequence at 0/2 instead of 0/1, creates a modified Farey sequence that starts
as: 0/2, 1/3, 1/1. The next level is 0/2,1/5,1/3,1/2(2/4),1/1. This continues
with the righthand side from 1/2 being the regular Farey sequence, while the
lefthand side between 0/2 and 1/3 generates the reciprocals of all the odd
numbers, while the region between 1/3 and 1/2 reverts into the original Farey
sequence (This also works for the even numbers (not shown)).
In the region of the
reciprocal odd numbers, modified Ford circles can be generated using the
squares as denominators for the yvalues of the circles (y=1/x^2) instead of
twice the squares for the regular Ford circles (y=1/(2(x^2)).
The reciprocals of
all the odd numbers can now be related to this modified Ferry sequence and
their modified Ford circles.
Solving for these
tangent lines to the successive, modified Ford circles (only for the odd
fractions 1/3,1/5,1/7, etc.), gives the following equations for y:
Table III: Equations for the tangent
lines from the modified Ford circles surrounding the circle at x= 1/3,
(x1/3)^2 + (y1/9)^2 = (1/9)^2
yequations for
the tangent lines**

For the Modified
Ford circles at the x values

yintercept at
x=1/2*

y=1/8(415x)

1/3,1/5

7/16

y=1/12(635x)

1/5,1/7

23/24

y=1/16(863x)

1/7,1/9

47/32

y=1/20(1099x)

1/9,1/11

79/40

y=1/24(12143x)

1/11,1/13

119/48

y=1/28(14195x)

1/13,1/15

167/56

y=1/32(16255x)

1/15,1/17

223/64

y=1/36(18323x)

1/17,1/19

287/72

y=1/40(20399x)

1/19,1/21

359/80

y=1/44(22483x)

1/21,1/23

439/88

y=1/48(24575x)

1/23,1/25

527/96

y=1/52(26675x)

1/25,1/27

623/104

y=1/56(28783x)

1/27,1/29

727/112

y=1/60(30899x)

1/29,1/31

839/120

y=1/64(321023x)

1/31,1/33

959/128

*For y=1/2 all x=0.
The numerators are all of the form 4n+3 and prime, except for 119 (7x17) and
287 (7x41) and 527 (17x31) and 623 (7x89) and 959 (7x137) and 1519 (7^2x31).
The general equation: y= (1/(4+4(n1)))(2n(4n^21)x).
**For the
yequations, the numerators for the slope of x in the tangent equations are of
the form 4n^21 (A00466, oeis). The denominators are of the form 8+4n, which
can be seen to also be the simple sum of the denominators of the x values for
the two circles (eg. for c1,c2 with x values, 1/3,1/5; 3+5 =8, the denominator
in the equations for the tangent lines.
For the Pythagorean triangles of the modified odd
Ford circles:
For circle, c1, where x1 = 1/a gives the
position along the xaxis for c1, and c2, where x2 = 1/b
(2/(a+b))^2 + (2/(ab))^2 = c^2
Where the difference between these
Pythagorean triangles and the other Pythagorean triangles for the normal Ford
circles is the numerator 2, in the (2/(ab))^2 term.
For example: for the Ford circles at 1/3 and
1/5, this gives:
(2/8)^2 + (2/15)^2 = c^2
Which gives c=17/60.