Timeline for What is this sequence counting?
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 5, 2019 at 16:41 | comment | added | Max Alekseyev | I've added the sequence to the OEIS as A324368. | |
| Sep 1, 2019 at 20:17 | answer | added | Max Alekseyev | timeline score: 3 | |
| Sep 1, 2019 at 7:00 | answer | added | Bjørn Kjos-Hanssen | timeline score: 3 | |
| Sep 1, 2019 at 5:50 | comment | added | TheTwistedSector | @GerhardPaseman Yes, the recursion process I am working with does involve adding 2. Actually, at a "level" $n$ I am solving for $P(n)$ unknowns and there are $P(n-1)$ linear equations from the next lower level and $P(n-2)$ linear equations from the next-to-next lower level. The redundancies occur because the system gets over-constrained for $n\geq5$ . | |
| Sep 1, 2019 at 5:29 | comment | added | Gerhard Paseman | There is a way of generating all partitions of N from all partitions of N-1, assuming they are presented as ordered tuples of positive integers: add 1 in at most two ways. If a partition of N-1 does not have a unique smallest member (say it starts 3,3,5, ...), then prepend a 1 (to make it 1,3,3,5, ,...), otherwise prepend and then for a new partition add 1 to the smallest number. I wonder if your process involves "adding 2". Gerhard "Just Trying To Brainstorm Here" Paseman, 2019.08.31. | |
| Sep 1, 2019 at 4:06 | comment | added | TheTwistedSector | @GerhardPaseman Could you please elaborate a bit more or give an example? I didn't quite follow what you meant by smallest class. | |
| Sep 1, 2019 at 3:37 | comment | added | Gerhard Paseman | Take a partition of n elements and remove 1 element from its smallest class. (If this gives an empty class, throw that away.) This gives P(n-1) results with (P(n)-P(n-1)) duplicates. Maybe you are removing something like two elements? Gerhard "Trying To Add An Idea" Paseman, 2019.08.31. | |
| Sep 1, 2019 at 3:20 | comment | added | Noam D. Elkies | By Euler's pentagonal-number theorem, the denominator $\prod_{n=1}^\infty (1-q^n)$ is $1 - q - q^2 + q^5 + q^7 - q^{12} - q^{15} + + - - \cdots$, so your sequence is $P(n-5) + P(n-7) - P(n-12) - P(n-15) + + - - \cdots$. Beyond that I don't know. | |
| Sep 1, 2019 at 2:50 | review | Close votes | |||
| Sep 6, 2019 at 17:19 | |||||
| Sep 1, 2019 at 1:38 | history | asked | TheTwistedSector | CC BY-SA 4.0 |