The Fibonacci ladder

Take a cylindrical lattice at the golden divergence and lower the rise. Nothing else changes: the divergence stays where it is, no dynamics run, no primordia are placed. One number falls, slowly, and the parastichy pair climbs.

At a rise of half a circumference the shortest offsets are 1 and 2. By 0.12 they are 2 and 3. By 0.045 they are 3 and 5, by 0.018 they are 5 and 8, by 0.007 they are 8 and 13, and by 0.0027 they are 13 and 21.

Each rung is the pair before it with the smaller member replaced by the sum. Which is to say the sequence of counts is 1, 2, 3, 5, 8, 13, 21, and it arrives out of an arithmetic in which the word Fibonacci does not appear.

Why the pair has to climb #

The mechanism is short enough to give in full.

The hop for offset $j$ has two components: it goes $\langle j\delta \rangle$ around the cylinder, where $\langle\cdot\rangle$ is the distance to the nearest whole turn, and $jh$ up it. Its length is the hypotenuse.

The vertical component is proportional to $j$ and to $h$. The angular component is not proportional to anything — it oscillates, dipping low at the offsets where $j\delta$ nearly lands on a whole turn.

So when $h$ is large, the vertical term dominates and small offsets win simply for being small. When $h$ is small, the vertical term is cheap, and the competition is decided by the angular term — which favours the offsets that approximate well, regardless of size.

Lowering the rise therefore hands the contest over from “which offset is smallest” to “which offset approximates best”, gradually, one offset at a time. The ladder is that handover happening in slow motion.

Which offsets approximate best #

The offsets that make $j\delta$ nearly a whole number are exactly the denominators of the best rational approximations to $\delta$ — the convergents of its continued fraction. That is a theorem about numbers, not an observation about lattices, and it is the same theorem the disc essays lean on.

For the golden angle, $\delta = 1/\varphi^2$, whose continued fraction is all 1s. Its convergents have denominators 1, 2, 3, 5, 8, 13, 21, 34 — the Fibonacci numbers. So the ladder for a golden-divergence stem is a Fibonacci ladder, and for a different divergence it is a different ladder.

At the Lucas angle the same argument gives 1, 3, 4, 7, 11, 18. At a divergence with a large partial quotient somewhere in its expansion, the ladder skips: a very good approximation appears early, dominates a wide range of rises, and the intermediate rungs never get a turn.

This is worth stating as a prediction rather than a description, because it is testable before anything is drawn. Choose a divergence, compute its continued fraction, and the sequence of parastichy pairs its stem will show as it grows is determined — as is the fact that some divergences show fewer transitions than others over the same range of rise.

The transitions are geometrically spaced #

The rises at which the pair changes, at the golden divergence, are 0.1246, 0.0476, 0.0182, 0.00695 and 0.00266.

Divide each by the one before: 0.3821, 0.3821, 0.3821, 0.3821.

That number is $1/\varphi^2 = 0.38197$, the golden angle expressed as a fraction of a turn, and it appears here as a ratio between transition rises rather than as an angle. The ladder is geometric, and its common ratio is the same constant the divergence is.

This is not a coincidence and it is not deep. A golden-divergence lattice is self-similar: replacing the pair $(m, n)$ by $(n, m+n)$ and scaling the rise by $1/\varphi^2$ maps the lattice onto itself, because the Fibonacci recurrence and the scaling of $\langle j\delta \rangle$ by $1/\varphi$ are the same statement seen twice. The transitions inherit the scaling.

What it does mean is that the ladder has no end and no beginning. Rungs continue upward as the rise falls, forever, at a fixed ratio — which is why a very large seed head can show 144 and 233, and why there is no largest parastichy number in the model.

Each transition is a genuine ambiguity #

Between two rungs the lattice passes through a state where three offsets have nearly equal hops, and this is not a numerical artefact of crossing a boundary. It is a real property of the lattice at that rise, and it has a specific form: at the transition from $(m,n)$ to $(n, m+n)$, the three offsets $m$, $n$ and $m+n$ are all short at once.

Right at the transition, in fact, all three are equal, and the lattice is exactly equilateral. That configuration turns out to have an exact closed form and gets its own essay.

Near it, a counter has to choose two out of three, and which two it picks depends on details with no information in them. This is the cylindrical version of a limit the disc essays already record: a golden-angle head at one radius has offset 55 at 1.68 mean spacings, 34 at 2.01 and 89 at 2.10, and which pair the eye picks is ambiguous there.

So there is a floor on how precise a parastichy count can be, and it is set by the object rather than by the measurement.

What the lattice is doing while the pair changes #

It helps to watch the three competing offsets rather than the winning two, because the transition then stops looking like a discontinuity.

Well above a transition, the pair $(m,n)$ is dominant and the offset $m+n$ is a distant third — its hop perhaps twice the length of the shorter two. As the rise falls, all three hops shorten, but not at the same rate. The offset $m+n$ is the largest of the three, so its vertical cost falls fastest as $h$ does, and its angular cost is the smallest of the three because it approximates best.

So it catches up. It draws level with $m$ — the worse-approximating of the current pair — and overtakes it. Nothing jumps; a continuous quantity crosses another continuous quantity, and the label attached to “the two shortest” changes because of it.

That is why the transition replaces $m$ rather than $n$. The member that gets displaced is always the one with the poorer approximation, which for consecutive convergents is always the earlier one. The Fibonacci recurrence in the counts is the recurrence in the convergents, seen through the lattice.

It also explains why the sum is always the next rung and never a skipped one. The offset that overtakes is the one whose approximation is next-best after the two already present, and for a continued fraction with all partial quotients equal to 1 that is exactly $m+n$.

Why a plant moves down the ladder #

The rise is internode length divided by apex circumference. Both change during growth, and both change in the direction that lowers the rise: internodes shorten as elongation slows, apices broaden as a shoot matures.

So a growing shoot walks down this ladder, in the direction the figure’s axis runs, for reasons that have nothing to do with pattern and everything to do with ordinary development. The counts rise as a consequence.

That is the phenomenon botanists call rising phyllotaxis, and the model says something specific about it: the transitions are not evenly spaced in time or in size, but geometrically spaced in the rise, so the early ones come fast and the later ones come slowly. A shoot spends a short while at 2 and 3 and a very long while at 21 and 34, because the rise has to fall by a factor of $\varphi^2$ to get from one rung to the next and it is already small.

Which is a prediction about the frequency of observed pairs, and it points the same way as the observation that surveys of real material find high pairs much more often than low ones.

What the ladder is not #

Three things this figure does not show, worth separating out because each is a claim that gets attached to pictures like it.

It is not a dynamical process. No element is placed, nothing settles, and there is no time in the computation. It is a family of static lattices indexed by one parameter, and the “climb” is the reader’s eye moving along that index.

It does not explain the golden angle. The divergence is held fixed at 137.5° throughout, so the ladder cannot be evidence about why that angle rather than another. It shows what a golden-divergence stem does as it grows, not why a plant would have that divergence. The question of where the angle comes from is answered elsewhere, by a rule that settles rather than by a lattice that is drawn.

It is not restricted to Fibonacci. Every divergence has a ladder. The Fibonacci one belongs to the golden angle, the Lucas one from the Lucas angle, and a divergence chosen at random gives a ladder of its own with rungs that are usually not consecutive anything.

That third point is the one most often lost, and it is worth a number: at a rise of 0.008, only about 15% of divergences give a pair of consecutive Fibonacci numbers. The share is measured in its own essay and it is not close to universal.

Reaching further than the dynamics can #

There is one more reason this figure matters, and it is about the limits of the other half of this site.

The dynamical model — place each new element where the repulsion from the existing ones is least — is where the golden angle comes from as an output rather than an input. It is the site’s central figure and it has a stated limit: below a growth parameter of about 0.18 the discrete implementation stops converging, and the settled angle wanders over a hundred degrees however long the run.

That limit is a property of a sampled boundary and a finite window rather than of the subject, and the low-growth end is exactly where the high Fibonacci pairs live. So the model that explains the angle cannot reach the regime where the famous counts occur, which is an awkwardness the essays state rather than hide.

The lattice route has no such limit. There is no boundary to sample and no window to truncate, so the ladder runs to a rise of 0.0027 and could run to 0.0001 as easily. It reaches the Fibonacci ladder that the discrete dynamics cannot, by not being dynamics.

What it gives up in exchange is any account of why the pattern is at that divergence. The two halves are complementary and neither is sufficient, which is the subject of the essay that puts them together.

How many rungs an object can show #

A practical question follows from the geometric spacing, and it has a usable answer.

An object shows a transition only if its rise passes through the transition’s value while there are enough elements to count. On a stem the rise is fixed, so a stem shows no transitions — it sits on one rung, which is the property the whole field is built on. On a cone the rise changes along the axis, slowly, and a cone shows as many transitions as its taper spans. On a disc the rise falls as $1/r^2$, which is fast, and a seed head shows several.

Concretely: the rise has to fall by a factor of $\varphi^2 \approx 2.618$ per rung, and on a disc $h \propto r^{-2}$, so the radius has to grow by a factor of $\varphi \approx 1.618$ per rung. A head whose countable region spans a factor of four in radius therefore shows about three rungs, which is what the three-regime figure finds by counting.

That is a prediction with no fitted quantity in it, and it is falsifiable in an obvious way: a head spanning a factor of four in radius that showed six transitions would refute it. This kind of arithmetic is what the ladder is for — not the pretty sequence, which was known in the 1830s, but the ability to convert a fact about an object’s size into a fact about the counts it will display.

The check that has to be able to fail #

The build asserts that every transition adds the two previous counts, and that assertion would be worthless if the machinery could not produce a ladder that does something else.

It can, and it does whenever the divergence is not the golden angle. A stem at 99.5° gives 1, 3, 4, 7, 11 — every rung the sum of the two before it, but a different sequence entirely. A stem at 77.14° gives 1, 4, 5, 9. A stem at a rational divergence gives a pair sharing a factor and does not climb at all, because the lattice is rows rather than spirals and there is nothing for the third offset to overtake.

So the assertion is written in the form that can distinguish these: it requires that each transition replaces one member of the pair by the sum, without reference to what the members are. That passes for every divergence with a well-behaved continued fraction and fails for a lattice that skips a rung — which happens, at divergences whose expansion has a large partial quotient, and which is checked separately.

Requiring “the counts are Fibonacci numbers” instead would have been easier to write, would have passed on the golden angle, and would have proved nothing at all, since the point of the essay is that the sequence is a consequence rather than an input.

The measurement #

Five transitions, computed from which lattice vectors are shortest, at rises of 0.1246, 0.0476, 0.0182, 0.00695 and 0.00266. Four consecutive ratios, all 0.3821, against $1/\varphi^2 = 0.38197$.

The ladder was not looked for. The rise sweep was written to check something duller — that the cylinder counter returned sensible answers across a range of parameters rather than only at the one rise the first figure used — and the sequence of pairs in its output was the reason to look harder. Computing the transitions properly took an afternoon; noticing that their ratios were all the same number took longer than it should have, because four decimal places of 0.3821 look like noise until one thinks to compare them with a constant that is already all over the site.

Every pair along the way is a pair of consecutive Fibonacci numbers, and the build asserts it — not by checking against a stored list, but by requiring that each transition replaces one member of the pair with the sum of the two. A ladder that drifted, skipped, or repeated would fail that check, and a divergence that is not the golden angle produces exactly such a ladder, which is how the check is known to reject.