Timeline for Closed form for the number of steps required to get $n$ balls in the last box
Current License: CC BY-SA 4.0
17 events
| when toggle format | what | by | license | comment | |
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| Dec 21, 2022 at 13:28 | comment | added | Notamathematician | @PeterTaylor, thank you for suggestion! | |
| Dec 21, 2022 at 12:50 | comment | added | Peter Taylor | It doesn't have to be a single sentence. Splitting it into multiple sentences makes it easier to name things (e.g. $j$) and remove scope ambiguity. Given n balls, all of which are initially in the first of n numbered boxes, a(n-1) is the number of steps of the following process required to move them all the last box. A step consists of first identifying j, the lowest numbered box which has at least one ball. If it has only one ball then move it to box j+1; otherwise move half its balls rounded down to box j+1 and (unless it's the first box) half its balls rounded down to box j-1. | |
| Dec 21, 2022 at 10:05 | comment | added | Notamathematician | @PeterTaylor, thank you for editing A077071! Do you have any ideas how to improve my comment about balls and boxes? | |
| Dec 21, 2022 at 9:47 | comment | added | Peter Taylor | In fact, the result is already in OEIS, hidden away in Ralf Stephan's comment on A077071. | |
| Dec 21, 2022 at 9:03 | comment | added | Notamathematician | @PeterTaylor, ok, done. | |
| Dec 21, 2022 at 9:01 | history | edited | Notamathematician | CC BY-SA 4.0 |
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| Dec 21, 2022 at 8:52 | comment | added | Peter Taylor | Thank you, that's much clearer. Directly from OEIS we have $a(n) = 2n(n+1) - 2\sum_{k=0}^n h(k)$ where $h$ is the Hamming weight, and that seems potentially a more useful starting point than the balls and boxes process. | |
| Dec 21, 2022 at 8:16 | comment | added | Notamathematician | @PeterTaylor, could you please improve the comment in A077071? | |
| Dec 21, 2022 at 7:50 | comment | added | Notamathematician | @kodlu, thank you for comment! Done. | |
| Dec 21, 2022 at 7:49 | comment | added | Notamathematician | @PeterTaylor, do you have any other questions? | |
| Dec 21, 2022 at 7:47 | history | edited | Notamathematician | CC BY-SA 4.0 |
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| Dec 21, 2022 at 3:08 | review | Close votes | |||
| Jan 4, 2023 at 18:36 | |||||
| Dec 21, 2022 at 2:43 | comment | added | kodlu | I think it's off scope to suggest "please write a program in PARI" instead of properly explaining your iteration mathematically and precisely. | |
| Dec 21, 2022 at 0:07 | comment | added | Peter Taylor | In rereading I think there's an implied "select a box" at the start of the step, which I didn't get first time. I would understand "moving to the next box every second ball" to mean "half rounded down" except that rounding up seems necessary to make sufficient progress. But what does "also moving (except for the first box) to the previous box the same number of balls from the lowest-numbered box that has at least one ball" mean e.g. in cases where the lowest numbered non-empty box doesn't have enough balls in it? | |
| Dec 20, 2022 at 17:45 | comment | added | Notamathematician | @PeterTaylor, thank you for comment! Please try to write a prog (for example on PARI) and check the result. If that does not help, just tell me and I will add mine prog which most likely will make things absolutely clear. | |
| Dec 20, 2022 at 12:04 | comment | added | Peter Taylor | The description of the process is extremely confusing. If $d_t(i)$ denotes the number of balls in box $i$ after step $t$ we have $d_0(i) = n[i=0]$ using Iverson brackets; what is $d_{t+1}(i)$ in terms of $d_t$? | |
| Dec 20, 2022 at 11:16 | history | asked | Notamathematician | CC BY-SA 4.0 |