Abstract
It is easy to assume that a visitor working many flowers pollinates better. This paper counts that, on one plant, the total and the share of pollen from other plants point opposite ways. No new theorem or law is claimed.
First, the total from other plants grows: if $0.3$ of the body’s pollen is left at each flower, the total after $k$ flowers is $1-(1-0.3)^k$: $0.971752$ at $10$ flowers, $3.239175$ times that at $1$ flower (Section 2).
Second, and this is the core. Even so, the share from other plants falls: of the pollen delivered, the share from other plants is $1.000000$ with $1$ flower, $0.629630$ with $2$, $\mathbf{0.271977}$ with $5$ and $\mathbf{0.125670}$ with $10$ (Section 2).
Third, pollen from the same plant grows faster: pollen from other plants never exceeds the $1$ brought in, but pollen from the same plant is picked up again at every flower and reaches $\mathbf{6.760825}$ at $10$ flowers (Section 2).
Fourth, later flowers receive little from other plants: pollen from other plants reaching flower $j$ is $0.3\times0.7^{j-1}$; the tenth flower receives $0.012106$, a share of $\mathbf{0.012458}$ of what it receives (Section 3).
Fifth, per-flower amounts and totals were obtained separately and match: per-flower pollen from other plants matched the formula within $10^{-12}$ for all ten flowers, and summing per-flower amounts returned the totals within $10^{-12}$ (Section 3).
Sixth, the separator is whether one asks about the total pollen delivered or the share within it: the total rises with more flowers and the share falls. The same visit points opposite ways depending on which is asked (Section 4).
Keywords: pollination, pollen, geitonogamy, outcross pollen, share, model.
Scope (what is not claimed)
No new theorem or law is claimed: that pollen moves between flowers of the same plant (geitonogamy), and that pollen on a visitor’s body is left a little at each flower and runs down, are both known in pollination studies. No real values are claimed: the share left, $0.3$, and the amount picked up, $1$, are model values matching no plant and no visitor. Mechanisms are not included: pollen lifetime, limits on how much a stigma accepts, rejection of the plant’s own pollen, and visitors grooming themselves are not included. Fruit set is not discussed: whether delivered pollen becomes seed, or how many fruits form, is not treated. Relation to earlier papers: no earlier paper takes pollination as its subject. Paper 864 showed that higher up the chain there is less of the substance, yet it is more concentrated. This paper stands beside it. There a fixed share per level shrank the living mass in the denominator and the concentration rose. Here a factor $0.7$ per flower shrinks the pollen from other plants while pollen from the same plant is added, and the share falls. What is added is laying out totals and shares by the number of flowers worked, measuring that the share falls to $0.125670$ while the total grows $3.239175$-fold, following what each flower receives up to the tenth, and placing the separator on total delivered versus the share within it.
1. Notation and the Question
Setting: a visitor arrives at a plant carrying $1$ unit of pollen from other plants. At each flower it leaves a share $d$ of each kind of pollen on its body, then picks up $p$ of the plant’s own pollen and moves on. With $a_j$ the pollen from other plants and $b_j$ the pollen from the same plant received by flower $j$:
$$a_j=d(1-d)^{j-1},\qquad \sum_{j=1}^{k}a_j=1-(1-d)^{k},\qquad \text{share from other plants}=\frac{\sum_{j} a_j}{\sum_{j} a_j+\sum_{j} b_j}$$We set $d=0.3$ and $p=1$. $b_j$ was not written as a formula but counted flower by flower from the same-plant pollen left on the body.
Question: does working more flowers on one plant deliver pollen from other plants better?
2. Number of Flowers Worked
Computation 2.1: the number $k$ of flowers worked in a row on one plant was varied and the pollen left was counted.
| $k$ | from other plants | from same plant | share from other plants |
|---|---|---|---|
| $1$ | $0.300000$ | $0.000000$ | $1.000000$ |
| $2$ | $0.510000$ | $0.300000$ | $0.629630$ |
| $3$ | $0.657000$ | $0.810000$ | $0.447853$ |
| $5$ | $0.831930$ | $2.226900$ | $0.271977$ |
| $10$ | $0.971752$ | $6.760825$ | $0.125670$ |
Observation 2.2 this is the core. The more flowers worked, the smaller the share from other plants. The share is $1.000000$ with $1$ flower, $0.271977$ with $5$ and $0.125670$ with $10$: from the second flower on, the visitor starts carrying pollen from the same plant.
Observation 2.3 the total grows, but has a ceiling. The total from other plants is $0.971752$ at $10$ flowers, $3.239175$ times that at $1$, and matched $1-(1-d)^k$ within $10^{-12}$. It never exceeds the $1$ brought in. Pollen from the same plant is picked up again and again, has no ceiling, and reaches $6.760825$ at $10$ flowers.
3. What Each Flower Receives
Computation 3.1: for ten flowers in a row, what each flower receives was laid out.
| flower | from other plants | from same plant | share at this flower |
|---|---|---|---|
| $1$ | $0.300000$ | $0.000000$ | $1.000000$ |
| $2$ | $0.210000$ | $0.300000$ | $0.411765$ |
| $3$ | $0.147000$ | $0.510000$ | $0.223744$ |
| $5$ | $0.072030$ | $0.759900$ | $0.086582$ |
| $10$ | $0.012106$ | $0.959646$ | $0.012458$ |
Observation 3.2 later flowers receive little from other plants. The tenth flower receives $0.012106$ from other plants against $0.959646$ from the same plant, a share of $0.012458$: per-flower pollen from other plants matched $d(1-d)^{j-1}$ within $10^{-12}$ for all ten flowers, and summing per-flower amounts returned the $k=10$ row of Computation 2.1 within $10^{-12}$.
4. The Separator
Observation 4.1 the separator is whether one asks about the total pollen delivered or the share within it. Asked as a total, more pollen from other plants arrives the more flowers are worked: asked as a share, it is buried in pollen from the same plant the more flowers are worked. The same numbers of the same visit point opposite ways depending on which is asked.
Observation 4.2 placed beside Paper 864. 864 showed that a fixed share per level leaves less substance higher up, yet at a higher concentration. There the denominator shrank and the ratio rose. Here the numerator is multiplied by $0.7$ per flower while pollen from the same plant is added to the denominator, and the ratio falls.
Observation 4.3 what is not said. Real plants and visitors, pollen lifetime, stigma limits, rejection of the plant’s own pollen and fruit set are not treated. What is shown is that working more flowers raises the total and lowers the share.
5. Census and Closing
- total is $1-(1-d)^k$ (Section 1): capped by the $1$ brought in.
- share $1.000000$ with $1$ flower (Computation 2.1): only from other plants.
- $0.629630$ with $2$ (Computation 2.1): same-plant pollen begins to mix in.
- $\mathbf{0.271977}$ with $5$ (Computation 2.1): still falling.
- $\mathbf{0.125670}$ with $10$ (Computation 2.1): the core.
- total $0.971752$ at $10$ (Observation 2.3): grows $3.239175$-fold.
- same-plant pollen $\mathbf{6.760825}$ (Computation 2.1): no ceiling.
- share $\mathbf{0.012458}$ at the tenth flower (Computation 3.1): later flowers receive little.
- per-flower and totals within $10^{-12}$ (Observation 3.2): independent.
- total delivered, or share within it (Observation 4.1): the separator; Paper 864.
Closing: The more flowers a visitor works on one plant, the smaller the share of pollen from other plants. If $0.3$ of the body’s pollen is left at each flower and $1$ unit of the plant’s own pollen is picked up, the share is $0.629630$ with $2$ flowers, $0.271977$ with $5$ and $0.125670$ with $10$: from the second flower on the visitor carries the plant’s own pollen, which is picked up again and again and reaches $6.760825$ at $10$ flowers. The total from other plants grows to $0.971752$ but never exceeds the $1$ brought in: of the pollen the tenth flower receives, only $0.012458$ is from other plants. Placed among the earlier papers: no earlier paper takes pollination as its subject; beside Paper 864, where a fixed share per level shrinks the denominator, we placed the side where a fixed share per flower shrinks the numerator and the denominator grows. To be honest: the share left and the amount picked up are model values, and pollen lifetime, stigma limits and rejection of the plant’s own pollen are not included.
References
- Willmer, P. (2011). Pollination and Floral Ecology. Princeton University Press.
- Yamagishi, Y. (2026). Higher Up the Chain There Is Less of the Substance, Yet It Is More Concentrated [Paper 864]. YOI: https://www.yamagishiyuuki.com/yoi/864
Reference
Yamagishi, Y. (2026). The More Flowers a Visitor Works on One Plant, the Smaller the Share of Pollen from Other Plants [Paper 866]. YOI: https://www.yamagishiyuuki.com/yoi/866
ORCID: 0009-0006-8587-3450
PDF (English) · PDF (Japanese) · Date proof of the English PDF (OpenTimestamps .ots) · SHA256SUMS
The record page for this paper (in Japanese, with the PDFs in each language and the date proof): YOI 866. This paper is a preprint and has not been peer reviewed.