The census wants a low count

How many plants would it take sized the one experiment the frequency question needs. Score each specimen as Fibonacci or not by its counted pair; under the geometry’s census 14.7 per cent of divergence angles give a consecutive Fibonacci pair, under a grown history about ninety per cent do, and four specimens tell the two apart at five per cent and ninety per cent power. Fourteen tell the geometry from a coin weighted to a half. The answer was small, and the essay’s point was that smallness makes the missing survey a matter of recording the right fields rather than of expense.

Every specimen in that arithmetic was counted right. Two counts that slip together has since measured the counting error most likely to occur — a counter closing the circle a few degrees early or late, which multiplies both counts by the same factor — and found that it leaves a high count right much less often than a low one. This essay carries that error through the census, on both sides of the comparison, and asks what the four specimens become.

The experiment as it was sized #

The null is the geometry’s census at a rise of 0.008: every divergence angle from nought to 180°, each read for the two shortest families, the pairs tallied. It holds forty-eight pairs, and its Fibonacci share is 14.67 per cent. That share is almost entirely two pairs — 1/2, at 12.7 per cent, and 5/8, at 1.9 — because most angles at that rise give either a coarse pair or one that shares a factor, and the largest count anywhere in the census is 16.

The alternative is stated here as ninety per cent of specimens showing one Fibonacci pair, the pair they are counted at, and the other ten per cent drawn from the null’s non-Fibonacci pairs in proportion. The test is one-sided and exact, with the cut placed where the chance of a false positive first falls under five per cent. With every count right it needs four specimens, the cut at three, and it achieves 94.8 per cent power.

One closing error for every specimen #

Each specimen is counted with its own closing error, drawn from a normal distribution about zero with a stated spread. A spread of 3.6° is a hundredth of a turn: a careful counter, perhaps. A spread of 10.8° is a counter who is not marking the starting spiral at all. Nothing on a head says which of these a real counter is, and the spread is swept rather than chosen.

A reading is scored Fibonacci when it is a consecutive Fibonacci pair, whatever pair it was meant to be. Every probability is an exact integral over the intervals of closing error on which the rounded pair holds, as in the essay that measured the error, and every sample size is the exact binomial one from the essay that sized the census. Nothing is simulated.

The null does not move #

The first result is on the side where it might have been expected to matter most, and it is that the error does almost nothing there.

A count of $c$ spirals changes once the counter overruns or falls short by half a spacing of its family, which is $180/c$ degrees. The largest count in the null’s census is 16, so no closing error under 11.25° changes any of its readings at all, and its largest share, 1/2, needs an error of ninety degrees. Even at a spread of 10.8°, where 8.8 per cent of the null’s specimens are read as some other pair, the Fibonacci share moves from 14.67 per cent to 14.69. The low counts that make up the null are the robust ones, and what little they gain or lose is exchanged among non-Fibonacci pairs.

The alternative’s counts are the other end of the same curve. A grown plant counted at 34/55 changes its 55 at 3.27° and its 34 at 5.29°; at 55/89 the 89 changes at 2.02°. The error that leaves the null untouched is working on the alternative from the first few degrees.

The alternative does #

Counted at 13/21, the grown plants read 88.4 per cent Fibonacci at a spread of 3.6°, 69.0 at 7.2°, and 51.5 at 10.8°. Counted at 34/55, they read 57.3, 31.6 and 21.4. Counted at 55/89, 38.3, 19.9 and 13.4 per cent.

The reason is the one the closing-error essay found for a single report, now averaged over a population: a high count has more spirals per degree, so a given closing error changes it more often, and nearly every change takes the reading off the consecutive Fibonacci pairs. At 34/55 almost none of those changed readings is silent — they share a factor and would be recorded as whorled — but the census as sized scores them simply as not Fibonacci, and that is all the scoring rule can do with them.

A census that reverses itself #

The lowest line in the figure crosses the dashed one. Counted at 55/89, the census of grown plants reads below the geometry’s own 14.7 per cent from a closing spread of 9.83° onward, and counted at 34/55 from a spread of 15.8°. Past those spreads a survey of plants that really are ninety per cent Fibonacci would find them less Fibonacci than randomly chosen divergence angles, and a one-sided test in the direction the question poses would have nothing to say. The essay that designed the test defended the one-sided form on the ground that a result in the other direction would refute the counting before it refuted anything else. That was right, and here is a counting error that produces exactly that result.

Counted at 21/34 or 13/21, the census does not reverse at any spread up to 21.6°, which is sixteen per cent of a turn and further than anyone who is counting at all could miss by.

Where the reversal falls #

The two spreads at which the census reverses, 9.83° and 15.8°, are not an accident of the grid, and they can be predicted from one line.

A grown plant counted at a Fibonacci pair reads a consecutive Fibonacci pair almost only while its reading is exact — the readings a closing error produces near the truth share a factor or pass as some other lattice, and very rarely land on a different consecutive Fibonacci pair. The reading is exact while the closing error stays inside half a spacing of the larger family, $1/(2n)$ of a turn. So the alternative’s Fibonacci share is close to ninety per cent times the chance a normal error falls inside that interval, and it meets the null’s 14.7 per cent when that chance falls to about a sixth.

That happens when the interval is 0.2057 of the spread, which puts the reversal at $180/(0.2057\,n)$ degrees. For $n = 89$ that is 9.83°, exactly the crossing measured; for $n = 55$ it is 15.9°, within one per cent of the 15.8° measured, the difference being the few readings of 34/55’s neighbourhood that happen to be Fibonacci. For 21/34 the same line gives 25.7° and for 13/21 41.7°, both beyond the range read, which is why neither reversed there.

The reversal spread is inversely proportional to the larger count. Every step up the Fibonacci sequence divides the closing error a census can tolerate by the golden ratio, and there is no count so high that the census gains anything from it.

What the error costs in specimens #

The sample sizes follow from the shares, and they spread apart quickly. At a spread of 3.6° the census needs four specimens counted at 13/21, five at 21/34, ten at 34/55 and thirty-two at 55/89. At 7.2° it needs six, seventeen, fifty-four and four hundred and forty-nine. At 10.8° it needs fourteen, forty-three and two hundred and seventy-six, and counted at 55/89 there is no sample size at all, since the two shares have crossed.

The two ends of that spread at 7.2° are worth seeing as the power curves they come from.

Counted at 13/21 the grown plants read 69 per cent Fibonacci, and six specimens reach 92 per cent power with the cut at three. The staircase is barely longer than the error-free one; a census counted low hardly notices a closing error of this size. The cost is two specimens.

Counted at 34/55 they read 31.6 per cent, and the staircase runs to fifty-four specimens before it clears ninety per cent. This is the regime the original sizing called the expensive one — separating two shares that differ by seventeen points rather than seventy-five — reached not because plants are less Fibonacci than supposed but because of how they were counted.

The weaker comparison suffers more #

The comparison the earlier sizing called the more interesting one — the geometry against a coin weighted to a half, the question to ask if plants are only often Fibonacci rather than nearly always — needed fourteen specimens with every count right, and it is hit harder, because it starts with less room.

With a closing spread of 3.6° it needs seventeen specimens counted at 13/21, fifty-three at 34/55 and 289 at 55/89. At 7.2° it needs thirty-two at 13/21 and 1,420 at 34/55, and counted at 55/89 the grown plants read 11.1 per cent Fibonacci, below the geometry, so there is no answer. A half is a smaller effect than ninety per cent, and a closing error multiplies the Fibonacci part of either alternative by the same factor, so the smaller one falls through the null first. A survey designed around the fourteen-specimen question and counted high is not a slightly more expensive survey. It is one that may not be able to answer the question at all.

The count that pins the angle is the one to avoid #

This reverses a recommendation that runs through this whole line of reading. What a count is worth found that a reported pair pins the divergence angle to about $221°/mn$, so a count of 34 and 55 fixes it to a tenth of a degree while 13 and 21 leaves eight tenths open, and the advice that followed was to count high. For the angle that advice stands. For the census it is wrong.

The arithmetic is short. A count of 13 and 21 leaves $221/273$, about 0.81°, of divergence open; a count of 34 and 55 leaves $221/1870$, about 0.12°. Both are far finer than anything the census needs: the golden angle and the Lucas angle are thirty-eight degrees apart, and even 0.81° separates them forty-seven times over. What the higher count buys is a seventh of the uncertainty about an angle the census never looks at, and what it costs, at a closing spread of 7.2°, is forty-eight specimens.

The census asks a coarser question than the angle does. It needs to know only whether a specimen’s pair is consecutive Fibonacci, and a low count answers that as well as a high one. What a high count adds is precision about the angle, which the census throws away, and what it costs is exposure to the commonest counting error, which the census keeps. So a survey that wants both answers from one specimen should count twice: a low annulus for the census and a high one for the angle, which is also the two-annulus record the independent-error readings recommended, used here for a different reason.

It is also a point about how published counts were chosen. A counted pair is a statement about an annulus, and the annuli people report are the ones with high, photogenic counts. If those counts carry closing errors of a few degrees, the published Fibonacci share of such specimens is lower than the plants’ own — the opposite of the selection bias that essay was worried about, and not a correction for it.

Setting aside what announces itself #

The closing-error essay found that at 34/55 nearly every wrong reading shares a factor, and so announces itself. The census as sized scores those as not Fibonacci. A census that set them aside instead — recording them as whorled, as the census’s own bucket did — would lose the specimens but keep its shares honest.

Set aside that way, the census counted at 34/55 needs four specimens kept at a spread of 3.6°, against ten scored; fifteen at 7.2°, against fifty-four; and fifty at 10.8°, against 276. Counted at 55/89 it needs six, twenty-eight and 304, where scoring every reading needed thirty-two, 449 and none. The null moves too, since a third of its angles give pairs that share a factor and those are set aside as well: its Fibonacci share among the specimens kept is 22.7 per cent rather than 14.7.

The price is paid in specimens counted rather than specimens kept. At 34/55 and a spread of 3.6° only 63.7 per cent of the grown plants yield a kept reading, so the four kept specimens cost about six counted; at 7.2° about half are kept. And it rescues only the readings that announce themselves. The silent ones — 13/21’s in particular, whose first wrong readings are silent — are kept and scored as not Fibonacci whatever policy is used, which is why setting aside helps the high counts much more than the low ones.

A third of the grown plants in the whorled bucket #

One consequence of the dropping policy deserves a sentence of its own, because it touches a number already published here. The geometry’s census puts 35.4 per cent of divergence angles into pairs that share a factor, and opening that bucket found it full of genuinely jugate pairs, $k$ and $2k$.

A census of grown plants counted at 34/55 with a closing spread of 3.6° puts 36.3 per cent of its specimens into the same bucket, for a reason that has nothing to do with jugacy: the counter closed the circle a few degrees off and read 34/54 or 33/54. Counted at 55/89 the share is 55.3 per cent; at 13/21 it is 4.2. So a whorled share near a third, found in a survey of heads counted high, is not by itself evidence that the heads resemble the geometry’s census. It is what a careful counter produces from plants that are ninety per cent Fibonacci, and the pattern that arrives two at a time — the rotational symmetry of the youngest whorl — is the only field that tells the two apart.

What this sizing leaves out #

It carries one error model, the closing error, and not the independent miscount the earlier readings used. An independent miscount of one on the null’s low pairs is a large relative error — 1/2 read as 1/3 — and would move the null in a way a closing error does not. That needs its own model and is not attempted here.

It fixes the alternative at ninety per cent of specimens on one stated pair. A real population counted at a fixed fraction of the radius would show a spread of pairs, and its exposure to the closing error would be an average over them.

And the spread of closing errors is swept, not measured. How far a person counting spirals by eye misses the starting spiral is a property of the person and the head, and the sample sizes above should be read as a function of it rather than as a design.

Three readings that would undo it #

A closing error under 11.25° that changes a count in the geometry’s census. A census of grown plants counted at 13/21 or 21/34 that reads below the geometry’s share at any spread up to 21.6°. A spread of closing errors at which a census counted at 55/89 needs fewer specimens than one counted at 13/21. Any of them would mean the arithmetic is wrong rather than merely incomplete, since each is a direct consequence of $180/c$.

Still open: a recount rather than a set-aside #

Setting aside an announced reading throws the specimen away. A counter could instead count it again, with a fresh closing error, and keep the second reading if it does not announce itself. That changes the arithmetic in two directions at once: it recovers specimens, and it gives a silent error a second chance to be drawn. The measurement is the census under a recount policy — how many counts a specimen costs on average, how many specimens the census then needs, and whether counting at 34/55 with recounts ever becomes cheaper, in counts rather than specimens, than counting at 13/21 once.