Sequence — the series

The memory of a divergence sequence, at 0.75° of scatter. With no noise at all the lag-one correlation is 0.54: the rule corrects itself, so a lattice arrives with a memory in it. Matched at the same recorded scatter, placement noise leaves -0.10, jostle noise leaves 0.66, field noise leaves 0.47. The band is ±0.13, which is what an uncorrelated sequence of this length gives.

The sequence has a memory

Every measurement this collection has made of a stem's divergence angles throws the order away. A spread is invariant to shuffling. Put the angles back in order and there is a large correlation between one and the next — 0.54 with no noise at all — which is the rule correcting itself, and which nothing had looked at.

part 1 · cylinder

What the sequence sees that the scatter cannot. Each point is an ensemble at one amplitude, placed at the scatter it produces. A stem at three quarters of a degree of scatter has a lag-one correlation near zero if its noise arrived after the primordium was placed, and near 0.7 if it arrived before — and no measurement of the scatter can tell those apart. The separation closes above about a degree, because what the other two kinds preserve is the correlation of a lattice.

What one angle says about the next

A tenth of a degree of placement noise moves a stem's divergence scatter from 0.50° to 0.62°, which nobody would report. It takes the correlation between consecutive angles from 0.54 to below zero. The other two kinds of noise, at scatters where no measurement can separate them, leave it at 0.6.

part 2 · cylinder

What the experiment costs, in internodes. The combined sampling band of two autocorrelations falls as one over the root of the sequence length. The difference to be resolved is 0.76 — between noise that arrives before the primordium is placed and noise that arrives after — so the count needed is 56 internodes on a single stem. Every other open question in this collection is priced in tens of specimens.

The test a plant could settle

Every other open question in this collection is priced in tens of specimens, and one of them in a hundred and sixty. This one is priced in internodes on a single stem, and the number is fifty-six — because it is a statistic of one sequence rather than a share of a population.

part 3 · wrong

The order of the angles carries the count. Three stems, each held at a fixed rise so the pattern sits on one rung of the ladder. At a rise of 0.032 the positions count 3 and 5 spirals and the angles peak at 3; At a rise of 0.013 the positions count 5 and 8 spirals and the angles peak at 5; At a rise of 0.005 the positions count 8 and 13 spirals and the angles peak at 8. Each panel marks the peak and its multiples; the pale strip is what an uncorrelated sequence of this length gives.

The order carries the count

Take the divergence angles off a stem, throw away every coordinate, and autocorrelate what is left. The result is periodic at the smaller parastichy number — peaks at it and at every multiple of it. A list of angles, with no picture and no position in it, carries the spiral count.

part 4 · cylinder

The memory belongs to the rise, not to the lattice. The lag-one correlation of a noiseless rising stem, against how fast it climbs the ladder. Below about sixty nodes per rung it is negative; above it, 0.54, 0.74, 0.58, 0.58, 0.55 — flat across a fivefold change in rate. The horizontal line is the same rule with the rise held FIXED, where the correlation is -0.68. So the +0.74 the earlier work called the sequence's own memory is the pattern chasing an equilibrium that is moving under it.

The memory was the rise

The earlier work measured a lag-one correlation of 0.54 in a noiseless divergence sequence and called it the sequence's own memory. Hold the rise fixed and there is no sequence at all — every angle identical — and under a disturbance the correlation is negative. The 0.54 belongs to the pattern chasing an equilibrium that is moving under it.

part 5 · cylinder

The measurement is limited by the protractor, not by the plant. The peak falls as the reading error grows, and it falls by an arithmetic factor with nothing fitted: a position error enters two consecutive divergences with opposite signs, adding variance at every lag while the pattern's signal sits at one. At a quarter of a degree the readout is right on all 5 runs; at half a degree on 2; at a degree on 1. Below the dashed floor the peak is the largest of thirty noisy numbers rather than a measurement.

What the protractor has to be

The readout that names the parastichy number costs sixty internodes, which is cheap. It also needs every organ's position measured to better than a quarter of a degree, which is not — and the requirement follows from arithmetic rather than from care, so no amount of averaging relaxes it.

part 6 · wrong

Two combs, at a rise of 0.005. The autocorrelation of 760 divergence angles from one stem held at a rise of 0.005. The filled teeth are the lags at multiples of 8; the open teeth are the second comb, at the same spacing offset by 5. Reading the spacing off the first and the offset off the second gives the pair 8 and 13, which is what the position counter reports for the same stem — from angles alone, with no coordinate anywhere in the calculation.

The second comb

The autocorrelation of a divergence sequence has peaks at the smaller parastichy number and at every multiple of it. It also has a second set of peaks, at the same spacing, offset by the difference of the pair — so a list of angles with no coordinate in it returns both numbers rather than one.

part 7 · cylinder

Two combs, at a rise of 0.013. The autocorrelation of 760 divergence angles from one stem held at a rise of 0.013. The filled teeth are the lags at multiples of 5; the open teeth are the second comb, at the same spacing offset by 3. Reading the spacing off the first and the offset off the second gives the pair 5 and 8, which is what the position counter reports for the same stem — from angles alone, with no coordinate anywhere in the calculation.

A harmonic is a step taken twice

The spectrum contains the larger parastichy number, their sum, and echoes of the smaller one, and no ranking of peak heights separates them. What separates them is arithmetic: a harmonic is a multiple of the spacing and a family is not, and the two kinds sit in different residue classes.

part 8 · cylinder

Both statistics, on the same stems, at a rise of 0.005. Five seeded stems at each disturbance, held at a fixed rise. Bars are how many returned the pair the position counter finds; open portions are refusals. The pair comes out from 0.1 to 0.25, and across that whole range the lag-one correlation of the same sequences is -0.33, -0.58, -0.59 — decisive, negative and flat. There is no trade between the two: one stem supplies both. Below the window the sequence has locked onto the sampling grid and is a cycle rather than a sample; above it there is no lattice left, at 116° of scatter.

Two readings from one stem

Three note left with the work in a row have recorded that the two statistics of a divergence sequence want opposite plants — one quiet, one disturbed. Measured on the same stems they do not. The conflict was in the interpretation of a sign, and the window in which both are readable is wide.

part 9 · cylinder