Two accounts of one number

Take one organ out of a settled stem, in the middle of the band that cannot repair itself, and the pattern does not come back. What it does instead is strikingly orderly: the divergences settle into a sequence of angles that repeats exactly, forever, precessing slowly around the axis. At the arrangement carrying 5/8 the repeat is five angles long. At 8/13 it is eight.

Five and eight are the smaller of each lattice’s two parastichy numbers, and both orbits advance by one part in twice their own block — a block of m organs turning by 360/2m, which is a two-jugate arrangement with twice the block’s rows. Two numbers computed by unrelated arithmetic agreeing twice is the kind of coincidence this collection normally treats as a result.

It is a result, and it is a weaker one than it looked. Two different accounts predict it, and until this essay there was no lattice on which they disagreed.

The two accounts #

The count is inherited. On this reading the orbit carries a number belonging to the lattice that was cut, whatever that number happens to be, because the damage renews itself along one of the two contact families and the number of chains in that family is what sets the period. It predicts the smaller parastichy number on a Fibonacci rung because that is what the smaller number is there, and it predicts something else wherever the pair is something else.

The block is a repeat unit. On this reading the orbit is genuinely a two-jugate arrangement, and a jugate arrangement’s repeat is the count of the ordinary lattice underneath it. On a Fibonacci rung the ordinary lattice underneath a 2m-row arrangement has m rows, so the prediction is again the smaller number — arrived at from the geometry of the orbit rather than from the lattice that produced it.

On every Fibonacci rung these say the same thing. That is why the question needed lattices that are not Fibonacci, and the search for them turned up something first.

The premise fails before the accounts do #

Before going to unusual lattices it is worth running the same intervention at more ordinary ones, and the results had only ever been taken at two rises.

At a rise of 0.020 the pattern carries 3/5 and the one offset that fails settles into a block of five — the larger of the two. At 0.008 the pattern carries 5/8 and the four offsets that fail give five at two of them and eight at the other two. At 0.005 the pattern carries 8/13 and gives eight at four offsets and four at a fifth.

So “the block is the smaller parastichy number” was two measurements rather than a rule. It is not that the earlier numbers were wrong — they are recomputed here and they are what they were. It is that two points do not distinguish “the smaller” from “one of the two”, and nobody had asked for a third.

The four is already the discriminating case #

One cell of the census is worth more attention than the sentence it gets, because it is the row on which the two accounts come apart — and it is on a Fibonacci rung after all.

At 0.005 the pattern carries 8 and 13, and one offset settles into a block of four. Four is neither member of that pair.

For the inherited account that is a problem in its stated form and a small extension in its repaired one. Four divides eight, and a family that repeats every four organs repeats every eight as well — so a stem holding its four-hop rigid is holding its eight-hop rigid too, and the honest form of the account is not the block is one of the pair but the block is a divisor of one of the pair, with a member dividing itself.

Read that way the whole census fits without exception: 5 at 3/5, 5 and 8 at 5/8, 8 and 4 at 8/13 — every block a divisor of a counted number. That is a wider claim and it is also a sharper one, because it forbids more: a block of six at 8/13 would break it, and six divides neither.

For the repeat-unit account the four is harder to place. It ties the block to the ordinary lattice underneath the orbit rather than to the lattice that was cut, so a block of four means an underlying four-row arrangement — which is a claim about the orbit and is not obviously excluded, but it is also not predicted by anything, and it is the first row on which the account has to be told what to say rather than saying it.

So the search for non-Fibonacci lattices may not be necessary. A block that is a proper divisor of a counted number is already the case the two accounts read differently, and the census contains one — which is the cheapest possible outcome for a question that was expected to need new arrangements.

How the census is made #

Each row is one rise and one seed lattice. A stem is grown to four hundred organs, one organ is removed at each offset from one out to two past the larger parastichy number, the run is continued for three hundred organs with the organ permanently missing, and the divergences are watched.

An offset counts as unrepaired when there is no organ from which every later divergence stays within a degree and a half of the settled value, with at least sixty of them left to check. The hold is not decoration: without it a wrecked run’s last few angles can drift through the tolerance and be called a recovery four organs before the horizon, which is what a first version did.

For the unrepaired offsets the tail is searched for the shortest motif that repeats exactly, to within half a degree, and the block is the length of that motif. The precession is the sum of the motif’s angles with the whole turns removed, and the row count is 360 divided by it.

Nothing differs between rows except the rise and the seed. Same rule, same neighbourhood, same azimuth grid, same recovery test, same tail length. That matters more than usual here, because the result is a comparison between rows rather than a number read off one of them.

What survives #

Something does, and it is worth stating precisely because the temptation is to either overclaim it or drop it.

The block is a parastichy number of the lattice that was cut, at eighteen of the nineteen unrepaired offsets across the six lattices measured here. The exception is a cut six places back at the 8/13 arrangement, which settles on four — half of eight, and not a count that lattice carries.

Which of the two appears is decided by the offset. At the arrangement where both occur, cuts four and five places back give five and cuts six and seven give eight. There is no obvious rule in that and this essay does not propose one: at the 8/13 arrangement the offsets that give eight are four, seven, eight and nine, with the odd four sitting at six, which is not a run.

And the orbit is usually but not always twice its own block. At 0.020 the block of five precesses by 35.9°, which is 10.04 rows against 10. At 0.008 the block of five precesses by 31.4°, which is 11.46 rows against 10 — a miss of fifteen per cent, on a quantity that had agreed to a part in a hundred at both earlier measurements.

Why the offsets might decide it #

The reading this collection can offer is about which organs the orbit has to be consistent with, and it is a reading rather than a measurement.

An orbit is a sequence of placements that reproduces itself: the profile the rule minimises has to come back to the same shape after b organs. The profile is dominated by the organs at the two contact offsets, so a sequence that repeats after m organs is consistent with the m-family being intact and the n-family being rearranged, and one that repeats after n organs is the other way round. Which of the two the rule falls into ought therefore to depend on which family the vacancy sits in — and a vacancy k places back sits in whichever family k is a member of.

That predicts something checkable and it is not what happens. At the 5/8 arrangement with both blocks present, the offsets giving five are four and five and the offsets giving eight are six and seven; four is not a multiple of five and six is not a multiple of eight. So the simple version of the reading is refused by the table it was invented for, and what is left is that the offset decides it somehow.

The one number that did not move #

One quantity is the same in every row of the census and it is worth pointing at, because it is what keeps the intervention usable while the block is in dispute.

The run of felt offsets ends at the larger parastichy number, at every rise, on both branches. That is the result the whole intervention was built to deliver — a spiral count extracted from a yes-or-no answer, with no protractor anywhere — and nothing in this essay touches it. What is under argument is a second, harder question about what happens afterwards, on the minority of offsets that never repair.

Keeping the two apart matters for an experiment. The count comes from asking which removals are felt, which needs one shoot, a few dozen organs of patience and a yes or no at each offset. The block comes from following an unrepaired shoot for hundreds of organs and reading a cycle out of its divergences, which needs a protractor and a great deal more time. The first is cheap and settled; the second is expensive and open.

What the two accounts are worth now #

Neither is dead and neither is confirmed, and the situation is cleaner than it was because the thing they were both explaining has changed shape.

The inherited-count account predicted “a number of the lattice”, which is what happens at nineteen of twenty offsets. It did not predict which number, and it has to be extended to say so before it can be tested further.

The repeat-unit account predicted “twice the block, always”, and that is now falsified in its strong form: 11.46 rows on a block of five. It survives as a tendency, which is a much weaker thing.

The lattices that separate them properly are the subject of the two essays after this one. A stem on the Lucas branch carries 4/7, where neither number is Fibonacci; a genuinely bijugate stem carries 2m/2n, and really is an ordinary lattice seen twice over, so the repeat-unit account has a specific and different prediction there.

Reading the census #

The table repays a slow look, because three of its rows are doing different jobs.

The 0.020 row is the one that breaks the old rule, and it does it with a single data point: one offset fails and it gives the larger number. A single point would be weak evidence on its own; what makes it worth putting weight on is that the 0.008 row breaks the same rule independently, with two offsets, at a different rise, on a different pair.

The 0.005 row is the one that carries the exception. A block of four on an 8/13 lattice, at an offset of six, with a motif spanning five hundred and seventy grid steps — so it is a real orbit and not a rounding artefact, and it is a count that lattice does not have. Four is half of eight, which is suggestive and is not pursued here.

And the rows with no filled dots are not empty results. An arrangement where nothing fails is an arrangement where the pattern repairs every single-organ removal, which is itself the finding that made a two-organ intervention necessary.

What a real experiment would report #

It is worth translating the state of this into what an experimenter would be told to expect, because the translation is where the weakening actually bites.

Before: remove one primordium from the middle of the sensitive band of a stem carrying m/n, and the shoot will settle into a repeating cycle of m divergences. That is a sharp prediction with a number in it, and a shoot that produced a cycle of n would refute it.

After: the shoot will settle into a repeating cycle whose length is m or n, and which of the two depends on how far back the primordium was. That is still a prediction — a cycle of seven on a 5/8 shoot would refute it, and so would no cycle at all — but it is a much weaker one, and it costs an experimenter more, because two cycle lengths have to be distinguished rather than one confirmed.

The consolation is that the weaker prediction is the one that would have survived. An experiment run against the sharp version and returning n would have been read as evidence against the placement rule, and it is not: the rule produces n perfectly happily at some offsets of some arrangements.

What this does not say #

It does not say the earlier result was wrong. Blocks of five at 5/8 and eight at 8/13 are recomputed here and are what they were. What has changed is the generalisation drawn from them.

It does not say the block is arbitrary. Nineteen of twenty are a count the lattice carries, which is a strong constraint: there are thirteen plausible small integers and the orbits land on two of them.

It does not say which number to expect. That is the honest state of it. An experiment on a real shoot that returned a block would learn something — the number is one of the two counts — and would not be able to predict in advance which. Nor is it the shorter of the two, which was the obvious guess and is wrong at seventeen offsets of twenty-nine.

It does not say the twenty organs are twenty independent observations. Each row’s offsets are cuts on the same stem at the same rise, so a systematic property of that stem is shared by every offset in the row. The comparison that carries weight is between rows, and there are six of those.

And it does not say the orbits are the same object at every arrangement. The 0.008 row’s block of five precesses by a different amount from the 0.013 row’s block of five, at the same block length on the same pair, so two orbits sharing a period need not share anything else.

The check that would refuse it #

Three assertions run whenever the census is drawn.

The first requires at least one lattice whose unrepaired offsets settle on the larger of its two numbers. That is the assertion that fails if the old rule is restored, and it is the one the essay’s title turns on.

The second requires at least one lattice giving both numbers at different offsets. Without it the table would be consistent with each lattice having a single answer determined by something about the lattice, which is a different and tidier claim than the one made here.

The third requires every block drawn to be a real orbit — a motif spanning at least twenty steps of the azimuth grid. That check exists because an earlier reading of the coarsest arrangement reported a block of three that was three consecutive grid samples of a constant, and it would have gone into this table as a confirmation of the rule the table refutes.