Timeline for Closed-form for the number of partitions of $n$ avoiding the partition $(4,3,1)$
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 16, 2023 at 6:13 | vote | accept | Notamathematician | ||
| Dec 14, 2023 at 14:55 | answer | added | Joachim König | timeline score: 6 | |
| Dec 14, 2023 at 14:30 | comment | added | LSpice | It is best not to editorialise in the title, so I have edited out the description of the closed form as amazing (without meaning to render any judgement on the amazing-ness myself). | |
| Dec 14, 2023 at 14:30 | history | edited | LSpice | CC BY-SA 4.0 |
Removed editorialising in title
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| Dec 14, 2023 at 7:48 | comment | added | მამუკა ჯიბლაძე | Remark: your $b(n)$ might be also written as$$\frac{n^2}4+\frac{7+(-1)^n}8+\sum_{i=2}^{\left\lfloor\frac n3\right\rfloor}\sum_{j=i}^{n-2i}\left\lfloor\frac ji\right\rfloor$$ | |
| Dec 14, 2023 at 7:32 | comment | added | Joachim König | The ``$(4,3,1)$-avoiding" condition means equivalently that, after removing all instances of the maximum entry from the partition, one gets either the empty set, or a collection whose maximum and minimum differ by at most 1. So the relevant partitions are exactly those with $\le 2$ distinct parts, together with those of the form $[i,\dots, i, j,\dots, j, j-1,\dots, j-1]$ (with $i>j>1$). Surely that should help getting some explicit enumeration formula. | |
| Dec 14, 2023 at 4:42 | comment | added | Notamathematician | @FabiusWiesner, thank you for comment! Typo was corrected. | |
| Dec 14, 2023 at 4:41 | history | edited | Notamathematician | CC BY-SA 4.0 |
edited body
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| Dec 13, 2023 at 21:50 | comment | added | Fabius Wiesner | The denominator is $i+1$ not $j+1$. | |
| Dec 13, 2023 at 15:44 | comment | added | Notamathematician | @AndreaMarino, thank you for comment! My conjecture is based on attempts to obtain a simple closed form for ordinary partitions. I was trying to find a pattern in the indexes of partitions in the composition table. In an attempt to generate them recursively, I came up with one model that gave me values in the form of a triangle. So it might be called empirical evidence only. | |
| Dec 13, 2023 at 15:34 | comment | added | Andrea Marino | is your conjecture based on empirical evidence or conceptual reasoning? Do you have a similar characterization of associated Ferrers boards? | |
| Dec 13, 2023 at 13:52 | history | asked | Notamathematician | CC BY-SA 4.0 |