A count that can be wrong by one

A published count is a pair of numbers, and what a count is worth measured how much a pair pins down the divergence angle of the plant it came from: a band $221°/mn$ wide, so that 2 and 3 leave thirty-nine degrees open and 34 and 55 a little over a tenth of one. Every step up the Fibonacci pairs is worth a factor of $\varphi^2$.

That essay ended on a condition that undid part of its own recommendation. The width is what a correct count leaves open. High counts are worth a thousandfold more than low ones, and high counts are exactly where the spirals are most numerous and most nearly parallel, which is where a person tracing them by eye — following each family round without an index — is most likely to be one out. So the question is what a report is worth when either count may be wrong by one.

The neighbours a miscount produces #

A report of $(m, n)$ read with a tolerance of one allows eight other pairs: either count up or down by one, or both. Some of those are not pairs an ordinary lattice can show at all. If the two counts share a factor, the pattern would be whorled, several primordia arriving at once, and a report of 34 and 56 is a report of a pattern that arrives two at a time rather than a slightly wrong 34 and 55.

The rest share no factor, and each of those is the two shortest families of some lattice at some divergence angle. The same scan that measures a report’s band finds where: one band each. The question becomes where those bands fall relative to the true one.

Low counts blur, high counts split #

For 2/3 the answer is the worst possible. One of its neighbours is 1/2, and 1/2 is the two shortest families at every angle from twenty degrees to a hundred and eighty: a 2/3 that might be wrong by one says nothing about the angle at all. For 5/8 the tolerant report allows 35.2 degrees in five pieces spread over 102. For 13/21 it allows 2.4 degrees in three pieces spread over 84, and for 34/55 it allows 0.354 degrees in three pieces spread over 28.

A miscount moves the report #

At 34/55 the two neighbours that share no factor are 33/56 and 35/54, and their bands sit at 109.23° and 113.27°, each about an eighth of a degree wide: as narrow as the true band at 137.52°, and twenty-four and twenty-eight degrees from it. At 21/34 they are 20/33 at 54.37° and 22/35 at 82.13°.

So a high count that is wrong is not a blurred version of the right answer. It is a different sharp answer somewhere else. Neither wrong band at either pair contains the golden angle, and nothing about the band a wrong 34/55 allows would look less confident than the band a right one allows.

Why the wrong bands sit where they do #

A pair $(m, n)$ is the two shortest families when both $m$ turns of the divergence and $n$ turns land close to a whole number of turns, which means the angle is close to $360°\,a/m$ and to $360°\,b/n$ for some whole numbers $a$ and $b$ at once. The band lives where those two fractions nearly agree.

For 34/55 they are $360° \times 13/34 = 137.65°$ and $360° \times 21/55 = 137.45°$, and the band is between them. For 33/56 the nearest agreement is $360° \times 10/33 = 109.09°$ and $360° \times 17/56 = 109.29°$, and its band is at 109.23°. For 35/54 it is $360° \times 11/35 = 113.14°$ and $360° \times 17/54 = 113.33°$, with the band at 113.27°. Changing a count by one changes which fractions agree, and the agreement it lands on has no reason to be anywhere near the golden angle’s.

Why a low pair’s neighbour covers everything #

The reason 2/3 is worthless under a tolerance of one is its neighbour 1/2. When the rise between successive primordia is large, only the first two offsets are short enough to be the two shortest families at all, whatever the angle, so 1/2 is what every lattice shows at a coarse enough rise. A report of 1 and 2 is a report that the lattice is coarse, and it carries nothing about the divergence.

A low count is therefore fragile in a different way from a high one. A wrong high count is confidently somewhere else; a wrong low count can be a count that was never able to say anything, and a tolerance of one admits that possibility.

The width falls, the span does not #

Read with a tolerance of one, the total width a report allows falls from 160 degrees at 2/3 to 0.354 at 34/55, more than two orders of magnitude. The span of its candidates falls from 160 degrees to 84 by 13/21 and stays there until 34/55, where it drops to 28. The candidates sit where their own lattices live, and a pair one count away from a Fibonacci pair belongs to a lattice nowhere near the golden angle.

That is why a single number stops describing the report. Its measure is still small. Its span is not.

What the tolerance costs, in bits #

Counted as information — the logarithm of the 160-degree range over the width a report allows — 2/3 carries 2.04 bits when both counts are right and none when either may be wrong. 8/13 carries 6.24 and 3.92, and loses 2.32. 34/55 carries 10.40 and 8.82, and loses 1.58, within a hundredth of $\log_2 3$.

That last number is not a coincidence. At 34/55 the three candidates are equally narrow, so the tolerant report is the exact report plus the question of which of three it is, and a three-way choice is exactly $\log_2 3$ bits. On the Lucas pairs the cost is larger: 18/29 loses 2.80 bits, because it has six coprime neighbours rather than two.

A second estimate picks the candidate #

A choice among three sharp candidates tens of degrees apart is a cheap choice to make. Any other estimate of the angle, however rough, that lands nearer the true band than the nearest wrong one settles it.

For 2/3 and 3/5 no second estimate helps, because the wrong bands overlap the right one. For 5/8 it has to be good to 3.1 degrees and for 8/13 to 5.8. For 13/21 and 21/34 an estimate good to 27 degrees suffices, and for 34/55 one good to 12. That is an estimate a person could make from a photograph with a protractor, or from the angle a round trip through the counts returns at a lower count.

Which miscounts announce themselves #

The neighbours that share a factor are the other half of the story, because a miscount that produces one can be caught. A report of 34 and 56 is a whorled report, and a counter who knows the plant is not whorled knows the count is wrong.

How many of the eight share a factor depends on the pair, and not smoothly. At 5/8 three do; at 8/13 four; from 13/21 to 34/55 six do. On the Lucas pairs it runs the other way: at 11/18 and 18/29 only two of the eight share a factor, and six miscounts would pass as ordinary reports.

Two pairs in six cannot be miscounted silently #

The sharper question is about single miscounts — one count wrong by one, the other right — since those are far more likely than both counts wrong at once. For each of the first twenty-four Fibonacci pairs from 2/3, the number of single miscounts that share no factor is two, except at two pairs in every six, where it is none.

The protected pairs are 21/34, 34/55, 377/610, 610/987, 6765/10946, 10946/17711 and 121393/196418. At 34/55 the reason can be read off the numbers: 33 is 3 × 11 and 35 is 5 × 7, while 21 is 3 × 7 and 55 is 5 × 11, so miscounting either count up or down by one lands on a number sharing a prime with the other. None of the thirteen Lucas pairs from 4/7 is protected.

The same factors, six pairs later #

The pattern is not special to 34 and 55. At 377/610 and 610/987, the next protected pairs, 609 is $3 \times 7 \times 29$ and 611 is $13 \times 47$, while 377 is $13 \times 29$ and 987 is $3 \times 7 \times 47$. Miscounting 610 by one in either direction lands on a number sharing a prime with both of its neighbours in the sequence, exactly as miscounting 34 did.

Fibonacci numbers either side of a Fibonacci number factor through its neighbours’ factors in a cycle, and the cycle has length six. The pairs measured here show it holding for twenty-four consecutive pairs and do not prove it for all of them; the identities that would prove it are standard results about Fibonacci and Lucas numbers, and nothing in the argument above depends on more than the twenty-four checked.

What protection is worth #

With each count off by one with a chance of one in twenty, a report of 8/13 or 13/21 is silently wrong about one time in twenty — 4.9 and 4.8 per cent. A report of 21/34 or 34/55 is silently wrong 0.13 per cent of the time. At an error chance of one in a hundred the two figures are one per cent and 0.005 per cent. On the protected pairs a silent miscount needs both counts wrong at once, so its chance goes as the square of the error. It is exactly half that square: both counts must move, and of the four ways two errors of one can combine, two land on the pairs that share no factor, 33/56 and 35/54. At a twentieth the square is a quarter of one per cent and half of it is the 0.13 per cent above.

That is a fact about arithmetic rather than about plants, and it happens to favour exactly the high counts that a census of spiral pairs most needs and is most likely to get wrong.

A second annulus catches the rest #

The earlier essay suggested counting two adjacent annuli rather than one, because the counts change with radius and consecutive counts on a golden stem are consecutive Fibonacci pairs. That check has a precise form: counts $(a, b)$ in one annulus and $(c, d)$ in the next continue each other when $c = b$ and $d = a + b$. Every one of the four counts appears in that relation, so no single miscount can pass it.

Reading 8/13 with 13/21, a wrong pair of readings passes with a chance of 0.11 per cent at an error of one in twenty, against 4.9 per cent from 8/13 alone. Reading 13/21 with 21/34, it passes 0.003 per cent of the time. Reading 21/34 with 34/55, no combination of errors of one passes at all.

What the check costs #

The second annulus is not free, and the cost is worth naming. A reading of two annuli has four counts rather than two, so it is more often flagged: at an error of one in twenty, the two annuli are right and consistent 81 per cent of the time, where one annulus is right 90 per cent of the time. Most of the difference is readings that were right in one annulus and caught an error in the other.

That is the price of a check rather than a failure of it. A flagged reading is a reading to redo, and a reading that passes wrongly is a published number that is confidently wrong — which a count carries no error argues is the worse outcome, and which at high counts the tolerant bands above show is worse still.

What miscounts do to a census #

A survey that counts many heads and tallies their pairs is where these errors accumulate, and they accumulate unevenly. With each count wrong by one a twentieth of the time, a survey of golden heads counted at 13/21 records 90.3 per cent of them correctly, 5.0 per cent as pairs sharing a factor and 4.8 per cent as ordinary pairs that are not Fibonacci.

The same survey counted at 21/34 or 34/55 records 90.3 per cent correctly, 9.6 per cent as pairs sharing a factor and 0.13 per cent as ordinary pairs that are not Fibonacci. So a census taken high on the golden heads almost never manufactures a non-Fibonacci ordinary pair — and manufactures a whorled plant about one head in ten. A census that tallies whorled patterns without checking them against the plant’s own arrangement would find far more of them than there are, and the excess would be a signature of the counting rather than of the plants.

What this changes about counting further out #

The recommendation to count high survives, and it changes shape. A high count is worth a thousandfold more when right. When it may be wrong by one, it is worth somewhat less, and the whole of the loss is a choice among a few distant candidates that any rough second estimate resolves. So the practical recommendation is to count high and record one more thing: either a rough angle, or the count in the next annulus.

What it does not survive is a report of a low pair that might be wrong. A 2/3 read with a tolerance of one is consistent with every angle there is, and no second estimate short of measuring the angle directly recovers anything from it.

What this does not say #

It does not say how often counters are wrong by one, or whether errors in the two counts are independent. A person who loses track of one spiral family is likely to lose track in the same way on the other, which correlates the errors, and correlated errors are exactly the case the square law for protected pairs does not cover. It does not treat miscounts of two or more.

Nor does it say that a count’s error is equally likely up or down. Counting the spirals found an earlier version of the counter used on these heads returning the two smallest offsets rather than the two shortest families, a systematic error that is not an error of one at all, and a scheme that catches errors of one says nothing about it.

The claim, reduced #

A report read with a tolerance of one allows the bands of its coprime neighbours as well as its own. For low pairs they swallow the true band; for high pairs they are as narrow and tens of degrees away, so the tolerance costs a choice among a few candidates — $\log_2 3$ bits at 34/55 — that a rough second estimate resolves. Two Fibonacci pairs in every six, 21/34 and 34/55 among them, admit no silent single miscount, and two annuli whose pairs must follow on admit no silent single miscount anywhere.

What would withdraw it #

A coprime neighbour of a high Fibonacci pair whose band overlaps the true one. A tolerant report at 34/55 losing more than the cost of choosing among its candidates. A protected pair with a single miscount that shares no factor. A single miscount that passes the two-annulus relation. Each is checked whenever the bands are measured.

Still open: a miscount by more than one #

Every tolerance here is one. A counter tracing a high family by eye is as likely to drift by two as by one, and the neighbours at a distance of two include pairs such as 32/55 and 34/57 whose arithmetic is different: some share factors with the true pair’s counts that no neighbour at one does, and the bands they add may fall closer to the truth.

The measurement is the same tolerant reading at a tolerance of two and three: how many candidates each high report then allows, whether they stay separated by more than a rough estimate can resolve, and whether any Fibonacci pair remains protected once a single count can drift by two.