Most of the Fibonacci share was one family read twice

How often is it Fibonacci asked a question that surveys of plants never quite had, by asking the geometry. Pick a divergence at random, fix a rise, ask what pair a lattice at that divergence carries. At a rise of 0.12, 67.5 per cent of the circle gives consecutive Fibonacci numbers. At 0.03, 31.1 per cent. At 0.008, 14.7 per cent. The popular claim was stated as a near-universal, and the geometry says it fades with fineness. That number has been quoted here ever since. What “whorled” was hiding then opened the bucket that held a further 35 per cent, and found every pair in it to be $k$ and $2k$.

Both essays are careful about what they were measuring, and it should be said directly what a pair meant in them: the two offsets with the shortest hop on the cylinder. The measure was exact, in the sense that the calculation does what the sentence says. The trouble is one of definition. A person counting spirals on a stem counts families, chains running in one direction that between them reach every organ. Two directions are needed for a pair. And the two shortest hops on a lattice need not be two directions.

This came to light by accident, while looking at a reading flag that a sequence that was a reading had found being thrown away. Trying to restrict the census to pairs whose two families wind opposite ways, the whorled bucket disappeared altogether and the Fibonacci share fell by two thirds. It was not a matter of winding. On a lattice with its offsets near a small fraction, the second-shortest hop is often simply the shortest hop taken twice.

One family, counted twice #

A divergence of exactly a quarter turn makes the case as cleanly as possible, and the dial’s first stop shows it. The organs lie in four vertical rows, climbing with a rise of 0.008 per organ. The shortest hop is 4 organs, straight up one row. The next shortest is 8, also straight up the same row, twice as far. The census calls the lattice 4/8. It then notices a factor in common and puts it in the whorled bucket.

But 8 is not a second family. It runs in the same direction as 4 and it visits a subset of the same organs. Nobody looking at that stem would say it carries two families at those counts. What the lattice actually has is the rows, four of them, and the one family that gets from one row to the next, the single spiral passing through every organ in order. That is hop 1, which is long by comparison, at a quarter of the circumference. The lattice’s two generators are 1 and 4, and any cell of this pattern has those on its edges.

At 120 degrees the census reads 3/6, and the lattice has 1 and 3. At 135 degrees it reads 3/8, and that is genuinely two directions, which the lattice agrees with. At the golden angle it reads 5/8, and that is two families, as it should be. So in many places the two readings agree exactly and no number changes. The question is how much of the circle lies where they do not.

The reading that asks for two families #

The test for two families is exact. Offsets $a$ and $b$ generate a lattice, every organ reachable from every other by steps of them, precisely when $|a\,e_b - b\,e_a| = 1$, where $e_j$ is how far $j$ turns of the divergence fall from a whole turn. One over root two relied on that identity to compute a cell. So the pair is the shortest hop, and the shortest hop that completes a cell with it. That is the reduced basis of the lattice, and it can be found by Gauss reduction from any basis, starting with one organ’s step and a whole turn at no height.

Such a pair can never share a factor. A common factor would mean that the two families reach only every $k$-th organ, and could not together reach them all. So the whorled bucket does not exist in this reading, by construction, on a stem whose organs arrive one at a time. That is the right answer. A stem of single organs cannot be whorled, and the essay that opened the bucket made the same point in its last section, about what the census could not contain. It is a pattern with $k$ rows, which is something else.

The census is otherwise exactly the same: thirty-six hundred divergences over half a turn, at each of fifteen rises from 0.12 to 0.001, each read by both rules, classified by whether its pair is consecutive Fibonacci numbers, consecutive Lucas numbers, or something else.

What the share becomes #

At a rise of 0.12 the two shortest hops read Fibonacci on 67.5 per cent of the circle and the lattice’s two families on 32.4 per cent. At 0.03 they are 31.1 and 14.3. At 0.008, 14.7 and 3.2. At 0.001, 4.7 and 0.3. So across the range, the share by families is about a half down to a tenth of the share as published, falling away faster at the finer rises.

The missing part has a simple name. The pair 1/2 is Fibonacci, the first pair of the sequence, and on the two-hop rule a divergence with a small residual reads 1/2 at any rise. The first hop is a small sideways step, up a shallow spiral, and the second is the same step taken twice. It is one family, read twice, in the same way that 4/8 is. But it does not share a factor. So it went straight into the Fibonacci bucket and nobody could see it. At a rise of 0.12, 56.6 per cent of the circle reads 1/2, out of the 67.5 the census reported. At 0.008 it is 12.7 per cent, out of the 14.7. By the time the headline figure is at its smallest, almost nine parts in ten of it are this one pair.

Taking 1/2 out by itself leaves 10.9 per cent at 0.12, 7.3 at 0.03, 1.9 at 0.008 and 0.2 at 0.001. The lattice’s own reading comes out somewhat above those, since some of these same divergences have a true Fibonacci basis at their second family when the first is read properly. The figure shows all three, and it is plain which way the correction goes. Every reading that asks for two families says that the fraction of the circle reading as Fibonacci is far smaller than the census said.

How much of the circle was affected #

Together the two kinds of reading that were really one family fill a remarkably steady part of the circle. At 0.12, 56.6 per cent reads 1/2 and 4.9 shares a factor. At 0.03, 23.8 and 24.6. At 0.008, 12.7 and 35.4. At 0.001, 4.5 and 43.7. The total, 61, 48, 48, 48 per cent, is close to half at every rise beyond the coarsest. As the rise falls the two-hop reading moves from one form of this error to another, with no change in the size of the mistake.

The figure below says the same thing from the other side. There is a reason it should be about half. The two shortest vectors of a lattice are collinear when the second-shortest is the double of the first, which happens when the shortest vector is so much shorter than its nearest independent one that twice its length is still less. It is a statement about the aspect of the cell. For a lattice with a random shape, a long thin cell is about as common as a nearly square one, and half of all lattices are thin enough for this. There is an old result here that the shape of a random lattice has a stated distribution, and nothing below depends on it, though it would account for the steady half.

Read directly, the share of divergences whose two readings are one pair lies between 51 and 55 per cent at every rise. So about half of what the census has ever reported is a correct reading, and it needs no change at all. The other half is a pair that a person looking at the pattern could not count.

Those two accounts fit together exactly, at the fine rises, and that is a useful check on the claim that one family read twice is the whole story. At 0.008 the readings disagree on 48.1 per cent of the circle, and 12.7 per cent reading 1/2 plus 35.4 per cent sharing a factor comes to 48.1. At 0.03 they disagree on 48.2 and the two kinds come to 48.4. At 0.001 they disagree on 48.2, and the two kinds come to 48.2. So there is no third way for the two rules to differ. Whenever the two shortest hops are not the lattice’s two families, the reason is that the second is a repeat of the first. Only at the coarsest rises, where a basis of 1/2 is sometimes what the lattice really does have, do the totals part, and at 0.12 a fifth of the circle reads 1/2 with a genuine lattice behind it.

Fibonacci against Lucas #

The most interesting consequence is on a comparison nobody was making. By hops, at a rise of 0.008, 14.7 per cent of the circle is Fibonacci and 1.5 per cent Lucas, a lead of about ten to one. At 0.002 it is 6.8 to 0.6. That lead appeared to be something like a fact about the geometry: Fibonacci pairs took up ten times as much room as Lucas pairs do. It has been among the things that survive for a long time in one form or another.

By families they run close together at every rise. At 0.03, 14.3 and 14.7. At 0.008, 3.2 and 3.0. At 0.002, 0.8 and 1.1. There is no lead. In fact there is no preference at all: a random divergence is about as likely to carry a Lucas pair as a Fibonacci pair at any rise. So the geometry has no bias in the room it gives these two sequences, and whatever favours Fibonacci among real plants cannot be read out of the circle. That supports what has been said here from the start, that Fibonacci is a branch and not a law, rather more directly than anything measured until now.

It also fits with what a count is worth: a pair of counts $m$ and $n$ leaves open a band of angles about $221°/mn$ wide. That formula says nothing about which sequence a pair belongs to, and it ought to give similar room to similar products. The band for 21/34 and the band for 18/29 are near each other in size, and the census is now saying what that implies.

Everything is rare #

What sits in place of the buckets is a very long tail. The lattice’s families read 116 different pairs at a rise of 0.008, the commonest of them, 6/7, on 3.3 per cent of the circle, and 828 at 0.001. The share of no named sequence at all rises from 36 per cent at 0.12 through 71 per cent at 0.03 to 94 per cent at 0.008 and 99 per cent at 0.001. So the honest answer to the question of how often a random divergence reads as Fibonacci is that at fine rises almost nothing reads as anything in particular, and each sequence has its small share of a very large number of other pairs.

That changes what the original essay’s phrase meant. “Collapses with scale” is still right, but the Fibonacci figure was falling from a false height. The right picture is one in which all named pairs were always a small part of the circle, except at the very coarsest rises.

What this corrects, and what it does not #

The earlier essays’ arithmetic was right about what it counted. A census of the two shortest hops is what they measured and they said so. The conclusion that the share falls with fineness stands, and its direction is stronger than ever. The essay on the whorled bucket concluded that the generous reading “admits 35% of the circle on a technicality”, and that was right too, and in a sharper way than it knew: that 35 per cent was not a pattern with multiple rows being read up to jugacy. It was a single family being counted twice, and the census’s whole bucket is a side effect of the rule.

What has to change is any use of 14.7 per cent as a measure of how much of the circle reads as Fibonacci at a rise of 0.008. The figure is 3.2. Similarly, 67.5 per cent at the coarsest rise is 32.4, and the fact that about two thirds of the circle was Fibonacci at a coarse rise was never true in a sense a botanist would recognise. Anything that used the census as a prior should be read with that in mind, and the survey specification, which compares a hoped-for plant census against the geometry’s share, gets a lower null and so more power for the same specimens, which is the only good news in this essay.

The cylinder’s counter, the two-shortest rule itself, is also not wrong on a real stem. Settled stems, on the essays’ evidence, sit at angles where the two readings agree: at the golden and the Lucas angle they are one pair, and on the settling table only five runs in three hundred and eighty-four came up with an odd reading of this kind. So this is a limit on a census of all divergences and no problem with counting settled stems.

What this does not establish #

The classification still has only the two named sequences and everything else. At the coarsest rises several pairs, like 1/1 and 1/3, are very short and it can hardly matter what they are called. A pair of families has been taken to be the reduced basis, and other definitions of a counted pair are possible: a person might call the two most visible families, and those can differ from the two shortest on a real photograph. The measure on the circle is uniform, as it has been throughout, and whether that is a reasonable prior at all is a separate question. And none of this has been compared with any real plant.

What would overturn it #

A person counting two families off the stem at 90 degrees who came back with 4 and 8 would mean the definition here is wrong. A census of a larger sample at which Fibonacci and Lucas families separated again at some rise would remove the most striking claim here. A divergence whose reduced basis was not coprime would be a defect in the computation. It has been checked at several hundred placements, and none has been found.

Still open: what the census says of the settled stems #

A uniform measure on the circle was always a stand-in. Stems settle, and the destinations they settle on occupy a much smaller set of angles than the circle does. The next measurement re-reads the settling table’s three hundred and eighty-four runs and the destinations reached from forty starting angles by the same two rules, weighted by how many starting angles arrive at each. It asks how far apart the rules are on the angles a rule actually reaches, what Fibonacci’s share is among destinations rather than among angles, and whether the lead that has vanished from the circle returns once the rule is allowed a say.