Phyllotaxis
the spirals are counted, not admired
Stems and cones

Where the survivors meet

At a matched pair the two stems keep exactly the counted numbers their two pairs have in common — the 5 where 3/5 meets 5/8, the 8 where 5/8 meets 8/13, the 7 where 4/7 meets 7/11, and nothing at all where 3/5 meets 8/13. Four rows, including the empty one.
13 min read 17 figures Counted, not admiredState the claim, state the test

Worth reading first: The angle the ladder returns to · The organ that was taken away · Counting the spirals.

The four matched pairs were built to answer one question — whether the settled divergence decides which family a removal leaves standing — and they answered it in the negative. Every one of them handed over a second thing on the way, and it was not looked for.

At each pair, the two stems keep exactly the counted numbers their two pairs have in common. Not roughly, not usually. Four rows, and the fourth is the one where they have none in common and keep none in common.

Shared counted numbers against shared survivors. One row per matched pair, over both branches. The third column is the counted numbers the two rungs have in common and the fourth is the families both stems leave standing; on every row the two are the same set. The row whose rungs share no counted number is the one whose stems share no survivor, which is what makes this a claim about an intersection rather than a restatement that a survivor is usually a contact family. four rows, and the empty case is one of them.
Fig. 1 Each row is a matched pair: the counted numbers the two rungs share, against the families both stems leave standing.

The four rows

The 3/5 rung and the 5/8 rung share the 5. Cut at 137.266°, the 3/5 stem keeps {5} and the 5/8 stem keeps {5, 8}; the two sets meet in {5}.

The 5/8 rung and the 8/13 rung share the 8. Cut at 137.844°, the 5/8 stem keeps {5, 8} and the 8/13 stem keeps {4, 8}; the two meet in {8}.

The Lucas 4/7 rung and the 7/11 rung share the 7. Cut at 99.273°, the 4/7 stem keeps {7} and the 7/11 stem keeps {7, 11}; the two meet in {7}.

And the 3/5 rung and the 8/13 rung share nothing. Cut at 137.85°, the 3/5 stem keeps {5} and the 8/13 stem keeps {4, 8}; the two meet in nothing.

The same angle, two rungs, two answers. Each block is one matched pair: two rises whose stems settle on the same divergence and whose counters return different pairs. Under each is the family every wrecked cut leaves standing. On three of the four pairs both rises wreck at some offset, and on every one of those the two stems keep different families — so the divergence, which is held, is not what decides the survivor. The two stems keep exactly the counted numbers their two pairs share, including the pair that shares none and keeps none.
Fig. 2 The same four comparisons before the intersection is taken, with the offsets each stem wrecks at.

Why the empty row is the important one

A claim of the form “the survivors are the shared counted numbers” could be manufactured by a much weaker fact. If a survivor were simply usually a counted number of its own stem, then two stems that share a counted number would often both keep it, and the claim would follow from an existing result rather than adding to it.

What that weaker fact cannot produce is the empty case. It says nothing about two stems whose pairs are disjoint; they could each keep a counted number and happen to keep the same one anyway — a 3/5 stem keeping its 3 and an 8/13 stem keeping a 3 that it does not count. The prediction here is that the intersection is exactly the shared set, so where the shared set is empty the intersection must be empty too.

It is. The 3/5 stem keeps only the 5; the 8/13 stem keeps the 4 and the 8; and 5 is not among {4, 8}. The claim survives the one row on which it could most easily have failed silently.

Which lag survives, at every lattice and every offset. A row for each of twelve lattices and a column for each offset an organ is removed at, with the lag whose hop survived written in the cell. 30 cells never repair. The lag left standing is one of the two contact families at 29 of them, and at 17 it is the second shortest step on the lattice rather than the shortest, which is what refuses the reading that a rule holds its nearest neighbours. The ringed cells are the five the offset rule gets wrong. Two lattices carry no cells at all because no single removal wrecks them.
Fig. 3 Which surviving lags are contact families of their own stem and which are not, which is the weaker fact this claim is not a restatement of.

The 4 is the row that could have gone wrong

At the 8/13 stem, one offset — six places back — keeps the 4, and 4 is not one of its counted numbers. That is not an anomaly of this essay’s stems; it is a standing result of the census, where some wrecked stems keep a lag ranked well down the length ordering.

The 4 is what makes both the third row and the fourth row informative. In the third row the two stems keep {5, 8} and {4, 8}, and the presence of the 4 in one set and not the other is why the intersection is the single number 8 rather than a set of two. In the fourth row the 4 is one of the two things the 8/13 stem keeps, and the prediction requires that it not be matched by anything on the 3/5 side. It is not.

If the survivors had been nothing but contact families, both of those rows would have come out the same way for a duller reason. The 4 is the row where the prediction is doing work.

The hops of a 5/8 lattice, shortest first — golden, rise 0.010Every lattice hop at this rise, ordered by how long its step is across the surface of the stem, with the two that a wrecked stem here leaves standing picked out. The two shortest are the contact families the counter returns, 5 and 8, and they differ in length by a factor of 1.076. The lags left standing after a removal are 5 and 8, sitting at rank 2 and 1 in this order, so the family the rule holds is a short step but not always the shortest one.85133161021181122624629lag, in organs — ordered by the length of its stepstep length, in turns of the stemkept: lag 5, lag 8golden, rise 0.010 · pair 5/8 · offsets that wreck: 4, 5, 6, 7, 8generated from a stated rule, not drawn to look right
Fig. 4 The length ranking of a stem’s lags, on which a surviving 4 sits well below the two contact families.

What four rows can carry

Four is a small number and the claim is stated as four rows. What makes it worth asserting rather than mentioning is its shape.

A claim about an intersection has to get two things right at once: it must include everything the two sets share and exclude everything they do not. On a row with one shared number that is two conditions, not one — the shared number must be kept by both stems, and every other family either stem keeps must not be kept by the other. Across the four rows that is eleven separate conditions, of which the eleven that could have failed did not.

Counted that way it is not four coincidences. It is also not eleven independent ones — the survivors are not independent of each other — and no probability is being claimed. What is being claimed is that the pattern is stated in the one form where a single exception would show.

Shared counted numbers against shared survivors. One row per matched pair, over both branches. The third column is the counted numbers the two rungs have in common and the fourth is the families both stems leave standing; on every row the two are the same set. The row whose rungs share no counted number is the one whose stems share no survivor, which is what makes this a claim about an intersection rather than a restatement that a survivor is usually a contact family. five rows, and the empty case is one of them.
Fig. 5 Every matched pair on both branches, with the coarse pair that produces no comparison listed rather than dropped.

What the pattern is not saying

It is not saying that a survivor is always a shared counted number. Each stem keeps families the other does not — the 8 on a 5/8 stem where the other is 3/5, the 4 on an 8/13 stem where the other is 5/8 — and those are not shared and are not predicted by anything here.

It is not saying anything about which offset keeps which family. The sets are assembled across all the offsets a stem wrecks at, and the two stems of a pair wreck at mostly different offsets. A statement about the offset needs a census that varies the offset, which exists and is a different measurement.

And it is not an account of the mechanism. It is a regularity in what four pairs of stems did, standing in the position an account would have to explain.

The next organ moves for the last 13, and for no others. One row per organ removed, counted back from the tip of a stem at a rise of 0.005 whose counted pair is 8 and 13. Removing any of the last 13 moves the next organ by 2.6° to 167.6°; removing an older one moves it by at most 0.47°, which is under the azimuth grid. The boundary is at 13, and 13 is the larger parastichy number — so the experiment counts the spirals without measuring an angle.
Fig. 6 The offsets a fine stem wrecks at, which is the axis a matched pair does not hold in common.

The reading it does suggest

Two stems at one angle, sharing a counted number, keep that number. Two stems at one angle sharing none keep none. That is what would happen if the survivor were a property of the contact structure — which family of chains the removed organ belonged to and which the tip is spaced against — rather than of the arrangement’s angle.

That reading is already the direction the collection’s other results point. Removing an organ costs the family it did not belong to, and the side of the tip the removed organ sat on reproduces the published reading and extends it. Both are statements about chains. So is this.

What this adds is that the statement survives holding the angle constant, which neither of the others tested.

Which side of the tip the removed organ was on. Each row is one wrecked cut. The centre line is the azimuth the next organ would have taken; the two open marks are where the two contact families leave it, which are always on opposite sides because that is what makes them the two nearest neighbours of a lattice point. The filled mark is the organ that was removed. Reading the lost-member rule as a question about which side rather than about which chain makes it answerable on all 30 cuts instead of 9, reproduces the published reading on every one of those 9, and is right on 22 of 30 overall.
Fig. 7 The sidedness reading, which is the other statement about chains rather than about angles.

Where a fifth row would come from

Two places, and both are out of reach for a reason worth recording.

The first is the coarse end. The 2/3 and 3/5 rungs match at 139.297°, and neither stem produces a survivor: the coarsest rungs cannot be wrecked by a single removal. A fifth row would need a multi-organ cut, and a multi-organ cut is a different experiment with its own results about what survives.

The second is the fine end. The ladder stops at 0.0040 because below it a stem does not settle onto a lattice at any run length, so there is no 13/21 rung to pair the 8/13 with. The four rows are the four the ladder holds.

The settled divergence down the golden branch. Every rise from 0.07 down to 0.00482, plotted against the divergence the rule settles on, with each rung drawn in its own stroke and the branch's limit angle marked. The curve does not slide: it turns three times in four rungs, climbing across one and falling across the next, so a value it takes on one rung it takes again on another. That is what makes a matched pair possible — two rises, different counted pairs, one angle — and it is the whole reason the design exists on this branch. The widest excursions from the limit angle, coarse rung first, are 3.195°, 3.352°, 0.961°, 0.422°.
Fig. 8 The whole curve, with the two ends beyond which no further matched pair exists.

A check that had to be written carefully

The assertion in the library scores the claim as a set comparison, and there is a way to write that which always passes. Comparing the intersection of the survivors against the intersection of the pairs, and reporting agreement whenever both are empty, would pass on any row where either stem kept nothing — and three of the Lucas 4/7 stem’s four wrecked cuts keep nothing at all.

So the rows scored are the rows where both stems keep something, nulls are stripped before the intersection is taken rather than folded into it, and the existence of a row with an empty shared set is asserted separately. Without that last line a table of three rows that all share a number would satisfy the check and the empty case would never have been tested.

Every stem that never repaired, and the lag it kept. The 19 offsets across six lattices at which a single removal leaves a stem that never returns to its divergence. For each one: which organ was removed, the period of the block of angles the stem settles into, the lag whose hop survived the cut unchanged, and how many whole turns the stem gains over one period of that lag. The block and the surviving lag are the same number in every row. The marked row is the one whose survivor is not a parastichy number of the lattice that was cut — a hop 6.8 times the length of a contact hop, which no census would report and which the rule held rigid all the same.
Fig. 9 The census with every cut listed, including the ones that keep nothing, which is what a set comparison has to strip.

What a specimen could be asked

Nothing directly — this is a statement about two stems grown at rises chosen to share an angle, and a plant is one stem. But it turns into a specimen question one step along.

If the surviving family is set by the contact structure, then two plants counted 5 and 8 should respond to the same removal the same way whatever their divergences are, and two plants counted 5/8 and 8/13 should respond differently even if their divergences agree. The second half is testable in the field in a way the first is not, because the count is the measurement a specimen supplies and the angle is not.

That is a version of the ablation this collection has already specified, with one column added: record the counted pair, not the estimated angle.

The spiral counts four different divergence angles produce. Fibonacci counts come from one angle. The Lucas angle — which the same dynamical model reaches on a different branch — gives 47 and 76, and neither number is a Fibonacci number.
Fig. 10 What a counter returns across a range of angles, which is the reading a field ablation would record.

What it costs to have found this by accident

Nothing, and that is worth a sentence because it could have cost something. The regularity was read off a table built for another purpose, which is exactly the situation in which a pattern gets over-claimed: the rows were not chosen to test it, no threshold was set in advance, and there are four of them.

The protection is that the claim has no free parameter. It does not say the intersection is usually the shared set, or shares most of it; it says the two sets are equal, on every row, and one row predicts an empty set. There is nothing to tune and nothing to choose, so the only way to make it come out right is for it to be right.

Which rungs of the golden branch share a divergence. One row per pair of rungs. A pair whose divergence ranges overlap has a rise on each rung where the rule settles on the same angle; a pair whose ranges do not overlap has none, whatever the search. On this branch four of six pairs match, and three of those match to 0.0000° — the same value of a quantity read on a grid of 1,536 azimuths. The rises differ by factors of 1.48 to 5.09, so the design holds one angle while changing everything the rise controls.
Fig. 11 The pairs as the search found them, fixed before anything was cut.

The three rows that share a number, read separately

The three positive rows are not three copies of one observation. They involve three different shared numbers — 5, 8 and 7 — on two branches and at three different angles, and in each case the shared number is the larger member of one pair and the smaller member of the other.

That last detail is the one worth pinning. The 5 is the larger of 3/5 and the smaller of 5/8. The 8 is the larger of 5/8 and the smaller of 8/13. The 7 is the larger of 4/7 and the smaller of 7/11. So the shared survivor is never playing the same role on both sides of a comparison, and any account that made “the larger family survives” or “the smaller family survives” the mechanism has to explain how one number can satisfy both descriptions at once on the same row.

It cannot, and that is not a new refutation — both readings were scored and refuted on the census — but it is a second route to the same place, on stems the census never grew.

The ordering changes and the survivor does not. Every offset that wrecks, at every rise of the band, with the family left standing written in the cell. The counted pair is 5 and 8 at all 18 rises and the settled divergence is held to a twentieth of a degree, so the one quantity moving across the columns is which of the two contact steps is the shorter — and it changes hands at the marked rise. The cells do not: the 5 family survives at all 24 wrecked cuts, on both sides. Scored on this band, the reading that a wrecked stem keeps its shortest hop is right 14 times out of 24, for an answer that never changed.
Fig. 12 The families a band’s cuts leave standing, which is the other design’s version of the same table.

Consecutive rungs and the ladder’s arithmetic

The three rows that share a number are all pairs of consecutive rungs, and the row that shares none is the one pair of non-consecutive rungs among the four. That is not a separate finding: consecutive rungs on a Fibonacci ladder share a number by construction, since 5/8 follows 3/5 by dropping the 3 and adding the 8.

It does mean the claim has a much cheaper description than the measurement suggests. Two stems at one angle keep a family in common exactly when their rungs are neighbours on the ladder. Stated that way it sounds like arithmetic, and the arithmetic is only half of it — the sharing of a counted number is arithmetic and the sharing of a survivor is a measurement, and the two agreeing is the content.

Every transition as the rise falls. The pair climbs 1/2 → 2/3 → 3/5 → 5/8 → 8/13. Consecutive transitions are 0.381, 0.382, 0.382 of the previous rise — 1/φ² is 0.3820.
Fig. 13 The ladder whose consecutive rungs share a number by construction, which is where the arithmetic half of the claim comes from.

What would have refuted it

Three things, and none of them is exotic.

A row where the two stems shared a counted number and one of them did not keep it. That is the ordinary way for a prediction of this kind to fail, and it did not happen on any of the three rows where it could.

A row where the two stems kept a family in common that neither pair counted. Nothing forbids that: survivors here are not always contact families, and the 8/13 stem’s 4 is the standing example. Had the 3/5 stem kept a 4 as well, the intersection would have contained a number the pairs do not share and the claim would be gone.

And the empty row coming out non-empty, which is the same failure in its most visible form. The 3/5 stem keeps the 5 and the 8/13 stem keeps the 4 and the 8; one more overlap anywhere and there would be nothing here to write.

A wreck is a whole number of extra turns. For each of the 19 stems that never repair, the slip of its settled divergence multiplied by the lag whose hop survived. Every value lands on a whole number of turns — the horizontal lines — with a largest departure of 2.97 degrees, against divergences that have moved between 0 and 103 degrees. 17 of the 19 close on exactly one turn. So a wrecked stem is the stem it was with one extra turn threaded through every period of the family that survived, which is a dislocation with a stated size rather than damage.
Fig. 14 Every wrecked cut in the census with the family it keeps, which is the population these four pairs are drawn from.

Reading it against the angle result

The two results from these four pairs pull in opposite directions and are the same measurement. The first is that the two stems keep different sets, which is what rules the divergence out. The second is that the part of the sets they share is exactly predictable from the counts.

Both are needed. A pair whose stems kept identical sets would have said the angle might be sufficient after all; a pair whose stems kept sets with no relation to their counts would have left the counted pair as unsupported as the angle. What the four rows give is a difference with a structure in it — different where the counts differ, the same where the counts agree.

Shared counted numbers against shared survivors. One row per matched pair, over both branches. The third column is the counted numbers the two rungs have in common and the fourth is the families both stems leave standing; on every row the two are the same set. The row whose rungs share no counted number is the one whose stems share no survivor, which is what makes this a claim about an intersection rather than a restatement that a survivor is usually a contact family. four rows, and the empty case is one of them.
Fig. 15 The intersection column beside the counted-numbers column, which is the whole claim in two columns.

What the Lucas row costs to include

Three of the Lucas 4/7 stem’s four wrecked cuts keep nothing at all, so the set it contributes to the intersection is built from a single cut.

That is worth flagging rather than burying in a table. A row whose set has one member is a row where the intersection is decided by one measurement, and if that one cut had kept an 11 rather than a 7 the row would have read {} against {7, 11} and the claim would have an exception in it.

It is kept because the alternative — scoring only rows whose sets have several members — would drop the Lucas branch entirely and leave the claim on the golden one. A claim tested on one branch of a two-branch ladder is a claim about that branch.

The same angle, two rungs, two answers. Each block is one matched pair: two rises whose stems settle on the same divergence and whose counters return different pairs. Under each is the family every wrecked cut leaves standing. On one of the one pairs both rises wreck at some offset, and on every one of those the two stems keep different families — so the divergence, which is held, is not what decides the survivor. The two stems keep exactly the counted numbers their two pairs share, including the pair that shares none and keeps none.
Fig. 16 The Lucas comparison, on which one side’s set is built from a single wrecked cut.

The one line

At every matched pair where both stems keep something, the families they both keep are exactly the counted numbers their two pairs share — {5}, {8}, {7}, and on the pair whose rungs share nothing, nothing. Four rows, eleven conditions, no exceptions, and no parameter to have chosen.

The same angle, two rungs, two answers. Each block is one matched pair: two rises whose stems settle on the same divergence and whose counters return different pairs. Under each is the family every wrecked cut leaves standing. On one of the one pairs both rises wreck at some offset, and on every one of those the two stems keep different families — so the divergence, which is held, is not what decides the survivor. The two stems keep exactly the counted numbers their two pairs share, including the pair that shares none and keeps none.
Fig. 17 The Lucas row, where three of four wrecked cuts keep nothing and the fourth keeps the shared 7.

Where this goes next

The same angle, two rungs, two answers. Each block is one matched pair: two rises whose stems settle on the same divergence and whose counters return different pairs. Under each is the family every wrecked cut leaves standing. On three of the four pairs both rises wreck at some offset, and on every one of those the two stems keep different families — so the divergence, which is held, is not what decides the survivor. The two stems keep exactly the counted numbers their two pairs share, including the pair that shares none and keeps none. The angle is not the actor Cut an organ out of two stems that settled on the same divergence and return different counted pairs, and the family left standing is different at every one of the four pairs where both stems wreck. The angle is held to a hundredth of a degree underneath. Which lag survives, at every lattice and every offset. A row for each of twelve lattices and a column for each offset an organ is removed at, with the lag whose hop survived written in the cell. 30 cells never repair. The lag left standing is one of the two contact families at 29 of them, and at 17 it is the second shortest step on the lattice rather than the shortest, which is what refuses the reading that a rule holds its nearest neighbours. The ringed cells are the five the offset rule gets wrong. Two lattices carry no cells at all because no single removal wrecks them. A survivor has to be a neighbour A stem that never repairs after a removal keeps exactly one lattice hop rigid, and nothing predicted which one. Sweep every offset at twelve lattices and the answer narrows sharply: at twenty-nine of thirty the surviving hop is one of the two families a counter returns, and the one exception is a step six times too long to be one. Which rungs of the golden branch share a divergence. One row per pair of rungs. A pair whose divergence ranges overlap has a rise on each rung where the rule settles on the same angle; a pair whose ranges do not overlap has none, whatever the search. On this branch four of six pairs match, and three of those match to 0.0000° — the same value of a quantity read on a grid of 1,536 azimuths. The rises differ by factors of 1.48 to 5.09, so the design holds one angle while changing everything the rise controls. The last of three quantities The pair, the divergence and the step ordering move together when the rise is swept, and for four rounds no result could be attributed to any of them. Two designs later, two are ruled out as sufficient and the third has never been held still — because holding it is what a rung already does.

Shares its objects with

Essays that name at least two of the same things, and that neither author linked.

  • The ordering was not the actor — both name ablation, counting blind, claim testing, falsifiability, lattice offset, matched design, negative result, parastichy pair, rigid hop
  • A step of one organ — both name ablation, claim testing, divergence angle, lattice offset, mechanism, negative result, prediction, rigid hop
  • Both walls of the slot — both name ablation, claim testing, lattice offset, matched design, mechanism, negative result, parastichy pair, prediction
  • The ordering on six bands — both name ablation, claim testing, falsifiability, lattice offset, matched design, negative result, parastichy pair, rigid hop
  • The organ that moved furthest — both name ablation, claim testing, divergence angle, falsifiability, lattice offset, negative result, parastichy pair, rigid hop
  • One level and two exceptions — both name ablation, claim testing, lattice offset, mechanism, negative result, prediction, rigid hop

Named objects

A flat tag is an object no other essay names yet.

AblationCounting blindClaim testingDivergence angleFalsifiabilityLattice offsetMatched designMechanismNegative resultParastichy pairPredictionRigid hopSample sizeSelection effect