Phyllotaxis
the spirals are counted, not admired
Series

Sample size — the series

3 essays on one idea, from the one that introduces it to the one that assumes the rest.
  1. 14 specimens separate 14.7% from 50%. The exact binomial power against sample size, for a one-sided test at 5 per cent. It is a staircase rather than a curve because the decision rule is a whole number of specimens: at 14 the cut sits at 5 and the power is 91.0 per cent. A normal approximation smooths that staircase away and reports a different answer.

    How many plants would it take

    Fourteen specimens separate the geometry's Fibonacci share from a coin weighted to a half. Four separate it from what a grown history gives. One fir cone measured at three rings settles whether its transitions are spaced as a cone's or an ogive's. The sample sizes are small, and that is the uncomfortable part.

    part 5 · wrong
  2. How many specimens the census needs, counted at four different pairs, as the closing error spreads. The exact one-sided binomial sample size, at five per cent and ninety per cent power, separating the geometry's Fibonacci share from a census of grown plants, every specimen scored as read. Counted at 13/21: 4, 4, 5, 6, 14 specimens; counted at 21/34: 4, 5, 9, 17, 43 specimens; counted at 34/55: 5, 10, 27, 54, 276 specimens; counted at 55/89: 8, 32, 104, 449, none specimens, at spreads of 1.8°, 3.6°, 5.4°, 7.2°, 10.8°. With every count right the answer is four.

    The census wants a low count

    Four specimens separate the geometry's Fibonacci share of 14.7 per cent from the ninety per cent a grown history gives — if every count is right. Counted with a closing error spread over 7.2°, the same census needs six specimens counted at 13/21, fifty-four at 34/55 and 449 at 55/89, because the geometry's own pairs are all small enough that no closing error under 11° moves them, while a grown plant counted high loses its Fibonacci reading first. Counted at 55/89 with a spread of 9.83° the census reads plants as less Fibonacci than random angles. The count that pins the divergence best is the one a census should avoid.

    part 6 · wrong
  3. How many counts the census spends when an announced reading is counted again, against counting 13/21 once. Each specimen is counted with a closing error spread over 7.2°, and a reading whose counts share a factor is counted again, up to the number of counts on the horizontal axis; a specimen whose every count announces itself is set aside. Plotted is the census's cost in counts — kept specimens needed times counts spent per kept specimen — solid when the plants are grown and counted at the stated pair, dashed when they are the geometry's. The flat line is 13/21 counted once and every reading scored: 6 counts. 34/55: 29.5, 24.4, 20.7, 21.1, 21.5, 22.6 counts on grown plants, cheapest at 2; 55/89: 71.3, 58.5, 51.7, 52.5, 53.2, 55.3 counts on grown plants, cheapest at 3.

    Counting it again

    A reading whose two counts share a factor says the count went wrong, and the specimen is still there to be counted again. Counted afresh, the reading kept is exactly one reading conditioned on not announcing itself — the second chance a silent error gets is matched by the second chance a right reading gets — so a recount changes which specimens a census keeps, not what a kept reading says. At 34/55 with closing errors spread over 7.2° it takes the census from fifteen kept specimens to ten and from about thirty counts to twenty-one, and against scoring every reading it turns 449 counts at 55/89 into 52. It never makes a high count as cheap as counting 13/21 once.

    part 7 · wrong

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