Timeline for When is it possible to arrive from $a$ to $b$ by this procedure?
Current License: CC BY-SA 3.0
23 events
| when toggle format | what | by | license | comment | |
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| Sep 27, 2017 at 0:21 | comment | added | user114642 | But do we have if we count zero-th step? | |
| Sep 26, 2017 at 23:14 | comment | added | user114642 | @LSpice Do we have an example where number of steps does not divide neither $a$ nor $b$? Edit: Yes, we have $9 \to 12 \to 15$. | |
| Sep 26, 2017 at 22:30 | comment | added | user114642 | I do not know does GCD always must divide at least one intermediate sequence. @LSpice | |
| Sep 26, 2017 at 22:07 | comment | added | user114642 | @LSpice Ah, I did not read carefully your question. Then it depends on the way we reach $b$, obviously. But $k(a,b)$ is here worth studying, especially multi-valued definition of it. | |
| Sep 26, 2017 at 21:40 | comment | added | LSpice | But $(12, 20)$ is not an example where no connecting sequence has all terms divisible by $\gcd(a, b)$, because there is a connecting sequence $12 \to 18 \to 20$ (or just $12 \to 14 \to 16 \to 18 \to 20$). | |
| Sep 26, 2017 at 20:56 | answer | added | 8163264128 | timeline score: 3 | |
| Sep 26, 2017 at 19:45 | answer | added | Gerhard Paseman | timeline score: 1 | |
| Sep 26, 2017 at 19:43 | comment | added | user114642 | $$12 \to 15 \to 18 \to 20$$ | |
| Sep 26, 2017 at 19:41 | comment | added | LSpice | Sorry; I guess I meant to append to my comment "… and, if so, what is it?" | |
| Sep 26, 2017 at 19:41 | comment | added | user114642 | @LSpice Yes sir, I do know. | |
| S Sep 26, 2017 at 19:14 | history | suggested | Glorfindel | CC BY-SA 3.0 |
grammar corrections
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| Sep 26, 2017 at 19:10 | comment | added | user114642 | @LSpice After going through your description of the procedure I think that you completely understood what I meant to say in the text. You can define $k(a,b)$ to be minimal or to be multi-valued (since there, as shown, can exist $a$ and $b$ for which there is more than one way to reach $b$ by starting from $a$.Both definitions lead to, hopefully, fruitfull study, because I think that $k(a,b)$ is very interesting. Why define it in one way when we can define it in at least two ways and so study two different functions. | |
| Sep 26, 2017 at 19:04 | review | Suggested edits | |||
| S Sep 26, 2017 at 19:14 | |||||
| Sep 26, 2017 at 19:01 | comment | added | LSpice | Do you know of an example where $a$ is reachable from $b$, but no 'connecting sequence' (in the hopefully obvious sense) has all its terms divisible by $\gcd(a, b)$? | |
| Sep 26, 2017 at 18:57 | comment | added | user114642 | @LSpice I added description of the procedure in other words under the question raised in the text, you will not face any difficulties or ambiguities when reading it. | |
| Sep 26, 2017 at 18:55 | comment | added | LSpice | Does the following shorter description capture your procedure correctly? "For natural numbers $a$ and $b$ with $a < b$, we say that $b$ is one-step reachable from $a$ if $b - a$ is a divisor of $a$ other than $1$. We say that $b$ is $n$-step reachable from $a$ if there is a sequence $a = a_0, a_1, \dotsc, a_n = b$ in which $a_{i + 1}$ is one-step reachable from $a_i$ for all $0 \le i < n$." Then maybe you want to define $k(a, b)$ to be the set of all $n$ for which $b$ is $n$-step reachable from $a$, or to be the minimum of that set? | |
| Sep 26, 2017 at 18:44 | history | edited | user114642 | CC BY-SA 3.0 |
expained in more detail
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| Sep 26, 2017 at 18:36 | history | edited | user114642 | CC BY-SA 3.0 |
added 19 characters in body
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| Sep 26, 2017 at 18:29 | history | edited | user114642 | CC BY-SA 3.0 |
added 19 characters in body
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| Sep 26, 2017 at 18:27 | comment | added | user114642 | @SylvainJULIEN No, since such an example does not exist. | |
| Sep 26, 2017 at 18:10 | comment | added | Sylvain JULIEN | Do you have an example of a prime $ b $ reachable by some $ a<b $? | |
| Sep 26, 2017 at 18:03 | review | Close votes | |||
| Sep 26, 2017 at 20:57 | |||||
| Sep 26, 2017 at 17:35 | history | asked | user114642 | CC BY-SA 3.0 |