Timeline for Subsequence such that $c(a(n))=2^n$
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 13, 2023 at 13:07 | comment | added | Alapan Das | Ok. My email is: [email protected] | |
| Jan 13, 2023 at 9:47 | comment | added | Notamathematician | @AlapanDas, could you please tell me your e-mail? I would like to share with you an idea for an article. | |
| Jan 12, 2023 at 15:34 | comment | added | Notamathematician | @AlapanDas, thank you very much! | |
| Jan 12, 2023 at 15:21 | comment | added | Alapan Das | From the recurrence relation of $b(n,k)$ we get, $b(n,k)=2^{k-1}(2n-k)$. Now, if $c(n)$ is comprised of even numbers s.t $m=2^a_1n$, where $n$ is odd, we have $l(m)=a_1+\lfloor \log_2{n} \rfloor$,then $b(m,l(m)+1)=2^{l(m)}(2^{a_1+1}n-l(m)-1)$ Now, $m \mid b(m,l(m)+1) \Rightarrow n \mid l(m)+1$, say, $l(m)=nl-1,l \in \mathbb N$. | |
| Jan 12, 2023 at 14:50 | comment | added | Notamathematician | @AlapanDas, thank you for comment! Very nice observation! Do you have a proof? | |
| Jan 12, 2023 at 9:38 | comment | added | Alapan Das | The sequence $c(m)$ is made of numbers of the form, $m=2^{nl-1-\lfloor \log_2 n \rfloor}n$ where $n$ are all odd natural numbers and $l\in\mathbb N$. This satisfies the condition that $n<\sqrt{2m}$. | |
| Jan 11, 2023 at 17:46 | history | asked | Notamathematician | CC BY-SA 4.0 |