This post contains many images, most of them 480x360 Julia images that link to larger sizes. Detailed statistics for a particular image can be had by clicking it; there are too many to list all the stats in this post. The slowest was
Magnetic Force at nearly 10 hours.
Christmas Balls itself only took half an hour; besides
Magnetic Force,
Vergence was slower. Those two Julias required aggressive antialiasing and the minimum iterations in
Magnetic Force is 718.
The Matchmaker Mandelbrot
bplane with
a = 0. This is the unsoupedup fractal that Mandelbrot himself studied at one time.
One layer is a greyscale gradient smoothediteration layer. Four more layers color particular subsets of the buds by filtering; for instance the red layer filters out the green and blue components of the underlying greyscale layer. All of the filtering layers color disjoint subsets of the buds, and together they color them all save for the "insideout" one, radius 1 centered on infinity.
The grey area corresponds to zero being a stable fixed point and the only attractor. The red areas have a pair of equalperiod attracting cycles and this period is odd; blue areas have a pair of equalperiod attracting cycles and this period is even. Yellow areas have a single attractor whose period is not divisible by four. Green areas have a single attractor whose period is divisible by four.
The black area (no attractors at all) is perforated with swarms of minibrots. Minibrot cardioids come in every color except grey.
There are patterns in what colors occur where that are explained below and illustrated by numerous Julia sets.
The grey zone corresponds to the
bvalues of
for which zero attracts. It is easy to prove mathematically that zero is a fixed point and that it attracts for 
b > 1; 0 goes to 0/
b = 0 so the first part is obvious, and the rest follows with the use of some elementary calculus: the derivative of
f with respect to
z is
, which at zero is 1/
b, indicating stability for 
b > 1.
The colors of the other areas reflect which of four possible types of dynamics occur when zero is not an attractor. Black points have no attractors.
The critical points are at 1 and 1 (easily inferred from the derivative's numerator) and
. Since negating
z negates
f(
z,
b), the critical points have symmetric orbits, exchanged by a 180 degree rotation. It follows that if one goes to zero both of them do, so in the grey zone zero is the only attractor.
It also follows that if there are two attractors, they are rotated images of one another, whereas if there is only one attractor, every point in it has its negative in it as well, and if the attractor isn't just the origin, the cycle is even in length.
Below is a Julia set from the grey zone; with only a single attractor and that a fixed point, the Julia sets out here are dusts.

Final Frontier 
The red buds have a cycle of odd length, and therefore have two attractors.
Since negating
b negates
f(
z,
b), when you negate
b, you negate the odd iterates but not the even iterates:
f(
z,
b) = 
f(
z,
b),
f(
f(
z,
b),
b) = 
f(
f(
z,
b),
b) = 
f(
f(
z,
b),
b) = (
f(
f(
z,
b),
b) =
f(
f(
z,
b),
b) and then it repeats.
If we have an odd length attracting cycle
a_{1},
a_{2},
a_{3}, ... then 
a_{1}, 
a_{2}, 
a_{3}, ... is the other attractor. If we then negate
b, the iteration starting at
a_{1} now goes
a_{1}, 
a_{2},
a_{3}, ... and eventually hits the last element of the attractor unnegated, following which 
a_{1},
not a_{1}. So
a_{1} is now a member of an evenlength cycle that consists of all the points of both of the old attractors.
Furthermore, this cycle is attracting: the derivative of
f, given above, is an even function whose magnitude isn't altered by negating
z or
b, so even when it is applied over the whole cycle using the chain rule the same magnitudes get multiplied and only the sign may change. If those magnitudes were diminishing before, they still are. Or just note that the orbits of 1 and 1 are shuffled with each other but otherwise unchanged, so they still converge on the same set of points. It's just that those points form a single attractor instead of two.
So the buds 180 degrees opposite red buds have a single attracting evenlength cycle; for a given
bvalue in one of these buds this attractor contains the same points as the pair of oddlength cycles you get with the point 
b in the opposite red bud. These buds I've called "splitter" buds, because negating
b splits the attractor into two attractors. They are colored yellow in the image of the Mandelbrot set at the top of this post. Here are some illustrative Julia sets, one from the edge of the largest yellow bud and one with that one's
b negated (image flipped horizontally):
 
Golden  Space Frontier 
The left image is all one attracting basin, colored with the same golden gradient throughout, unlike the right image, which has two distinct basins (red and darkpurplish).
Whenever there's only one attractor, it's an evenlength cycle invariant under negation, but it isn't always a "splitter". In particular, a "splitter"'s cycle length has to be congruent to 2 modulo 4, since its length is double an odd number. So, what happens if a cycle whose length is divisible by four is subjected to negation of alternating points? The result ultimately depends on how points and their negatives are distributed in an evenlength cycle.
In fact, if
f^{k}(
a) = 
a, then since negating the input negates the output,
f^{k}(
a) =
a. If a cycle contains
a and 
a, then, they are halfway around the cycle from one another. Therefore if the cycle length is not divisible by 4, so
k is odd, negating alternate iterates causes
f^{k}(
a) =
a and therefore produces two separate odd cycles (so all such attractors are splitters, the converse of a result obtained above). If, however, the cycle length is divisible by four and so
k is even,
f^{k}(
a) is still 
a, and the cycle remains unchanged in length and set of member points. However, if the cycle originally went
a_{1},
a_{2},
a_{3}, ... 
a_{1}, 
a_{2}, 
a_{3}, ... it now goes
a_{1}, 
a_{2},
a_{3}, ..., 
a_{1},
a_{2}, 
a_{3}, ...; it has been rearranged. I call these attractors "shufflers", because negating
b shuffles the points in the attractor but doesn't change the period or membership. "Shuffler" buds therefore lie opposite other "shuffler" buds under a 180 degree rotation of the Mandelbrot set. I have colored these buds green in the Mandelbrot image at the top of this post.
The following Julia set images demonstrate a "shuffler" at work: the top images show the Julia set from a tiny green bud close to the attachment point of the giant red bud to the grey outer zone, with smoothed iterations coloring, and its opposite number; the bottom images show the same with phase coloring instead. Though they look identical under smoothed iterations coloring, the phase coloring clearly shows how the order of visitation of components of the immediate attracting basin is shuffled.
Finally, it is possible to have two disjoint evenlength attracting cycles. Unlike the odd case, negating every other point won't make the cycles join into a single doublelength cycle. Instead, the two attractors will exchange an alternating subset of their members. These I call "mixer" buds, because the points of two attractors get mixed by negating
b. Mixer buds are colored blue in the Mandelbrot image at the top of this post, and they are always opposite other mixer buds.
The following two Julia sets show the basins of two opposite mixer buds (second image flipped horizontally):
 
Interlock I  Interlock II 
Dark blue is one basin and light blue is the other in both cases. Note how some Fatou components belong to different basins after
b is negated.
Bifurcations of these attractors have various predictable patterns. Where a smaller bud buds off a larger bud the associated attracting cycle has its period multiplied by a factor as each point in the attractor becomes a group of points. For example, if we had a 2cycle
x,
y and it trifurcates we get a 6cycle
x_{1},
y_{1},
x_{2},
y_{2},
x_{3},
y_{3}, with
x_{1},
x_{2}, and
x_{3} near
x and
y_{1},
y_{2}, and
y_{3} near
y.
Where there are two symmetric attractors, both must undergo identical bifurcations. Pairs of odd cycles bifurcate into either pairs of odd cycles or pairs of even cycles, so one expects alternate blue and red buds attached to red buds, which is exactly what we get. Similarly, pairs of even cycles only bifurcate into pairs of even cycles, so blue buds have only blue buds attached. The behavior of yellow buds can be predicted from the
bnegation rules and the behaviors of red and blue buds: they have yellow and blue buds where the red ones have red and blue buds.
Green buds, interestingly, have the same behavior again, blue buds at even multipliers and still green at odd ones. This is harder to explain (and prove) but consider any evenlength selfsymmetric cycle at all,
q,
r,
s, ... 
q, 
r, 
s, ... bifurcating with any multiplier. The first part of the larger cycle will be
q_{1},
r_{1},
s_{1}, ..., (
q)
_{1}, (
r)
_{1}, (
s)
_{1}, ... with (
q)
_{1} not necessarily the same thing as (
q_{1}). But it WILL be (
q_{x}) for some
x.
In fact, the cluster of new attracting points about members of the old cycle must slowly make one revolution as one completes a circuit around the old cycle, so that after (
z)
_{1} (say) comes
q_{2}. So half way around (at 
q) it has rotated halfway, and (
q)
_{1} should be
close to (
q_{1}) (since the two near180degree rotations should about cancel). If the multiplier is 2, it has to
be (
q_{1}) as (
q_{2}) is nowhere nearby, so the cycle breaks into two: after (
z)
_{1} comes
q_{1} again rather than
q_{2}, so the
x_{2}s form a separate cycle. If the multiplier is higher it
cannot be (
q_{1}) as the number of separate cycles would be more than the number of available critical points. It must therefore be (
q_{2}) or (
q_{N}) with N the multiplier, i.e. there is a shift of either 1 or 1 modulo
N on reaching the "negative half" of the cycle. Which means there is a shift of 2 or 2 modulo
N on each complete cycle. Which means that if 2 and
N are not relatively prime (that is, if
N is even) the cycle returns to its starting point after exhausting only half the points, and again the attractor has split in two.
In particular, this predicts not only the same behavior for yellow buds noted above (the evenmultiplier smaller buds are blue), but the observed behavior for green ones as well. (With these, all attractor periods in the whole bifurcation "family tree" will be divisible by four.)
It also predicts that the attractors for blue buds attached to yellow or green ones will be interleaved with each other, which is exactly the behavior demonstrated by some stunning Julia set images I've generated.
The above pair,
Interlock I and
Interlock II, are from blue buds attached to, respectively, the big red bud and the big yellow bud; in the yellow case, the 2cycle has become a pair of interleaved 4cycles, as indicated by alternate light and dark petals (light blue is one attracting basin, dark blue the other). In the red case, two separate attracting fixed points bifurcated into 4cycles, which remained clustered.
What happens if the bifurcation is by a factor of three instead is illustrated below:
 
Lava Shore  Froth 
The two fixed points become two threecycles (left); the 2cycle opposite becomes a single 6cycle (right; all one color indicates single attracting basin).
Below is a Julia set from a "mixer" bud attached to a "shuffler" bud  in fact, the same "shuffler" bud that produced
Vergence, which is reproduced to its left below:
 
Vergence  Magnetic Force 
Again, note that when an even cycle splits the basins become interleaved as tightly as possible. A 1cycle split into 32 (
Vergence), and then this split into two 32cycles. Each point of the original 32cycle split in two, but as per the model above, one wound up belonging to each of the new attractors, so instead of the 32 petals becoming 64 with one group of 32 and then a second group of 32, or even with two interleaved groups of 32,
each petal became a chain of beads belonging alternately to the two attractors. The largest beads in each petal contain attractor points, and the attractors form two concentric rings of 32 points each.
The Julia sets of
f(
z,
b), in general, have a number of symmetries, some deriving from the same reasoning as the above. For one thing, repelling periodic points are still repelling periodic points after negating
b or
z, so the Julia set itself, as the closure of this set, is unchanged under both symmetries. In particular, it is always 2fold rotationally symmetric about the origin.
f(1/
z,
b) =
f(
z,
b) so it is also inversionsymmetric. The usual conjugacy symmetry applies, such that conjugating
b mirrors the Julia set in the
yaxis. (Which the twofold symmetry makes equivalent into mirroring in the
xaxis; the mirrored images above were actually mirrored by negating the imaginary part of
b). Negating the real part of
b amounts to negating
b and then conjugating, so it shuffles shufflers, mixes mixers, swaps splitters with pairs of odd cycles, and mirrors in the
yaxis all at once. The Mset is therefore
yaxis symmetric, and
xaxis symmetric except for red and yellow buds exchanging, which reasoning is borne out by the image at the top of this post.
The biggest red bud corresponds to two symmetric attracting fixed points; in the corresponding big yellow bud one gets single 2cycles. The biggest two green buds (top and bottom of set) have single 4cycles. The black area (no attractors at all) is perforated with swarms of minibrots. Minibrot cardioids come in every color except grey. 180degree rotation (and
xaxis mirroring) again must swap red and yellow and fix green and blue. The Julia sets in these are foams surrounding "bubble" basins, as demonstrated with our final pair of Julia images:
 
Gold Nuggets  Dark Matter 
As usual, each image link leads to a 1024x768 image, with a 2048x1536 image available by following one more link from there, and a 2048x1536 lossless PNG available upon request.
All images are, as usual, freely redistributable and usable subject to the Creative Commons Attribution license, version 3.0.