Timeline for When is it possible to arrive from $a$ to $b$ by this procedure?
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Sep 27, 2017 at 14:27 | comment | added | Gerhard Paseman | Suppose a and b are multiples of 6. There are at least $3^{(b-a)/6}$ paths to take involving any number from $(b-a)/6$ steps all the way up to $(b-a)/2$ steps. In general it is not clear what the utility is of knowing possible values of k(a,b). Gerhard "Unsure How To Use It" Paseman, 2017.09.27. | |
| Sep 27, 2017 at 7:41 | comment | added | user114642 | Is there any hope to determine generally number of values of k, (since for some choices of a and b (almost all, probably) there is more than one way to arrive from a to b).? k is obviously multi-valued almost always. | |
| Sep 27, 2017 at 7:33 | comment | added | Gerhard Paseman | It does not need to be. Consider a and b squares of primes which differ by two. K can be 4 or 5, depending on how you count it. Gerhard "There Are Many Other Examples" Paseman, 2017.09.27. | |
| Sep 27, 2017 at 7:31 | comment | added | user114642 | Sorry, it is not always the case that it is so. | |
| Sep 27, 2017 at 7:26 | comment | added | user114642 | Why $k(a,b)$ seems to be always divisor either of $a$ or $b$ or both? | |
| Sep 26, 2017 at 21:13 | comment | added | Gerhard Paseman | Further, one can always take a step to reach an even number, so an upper bound for minimal k(a,b) is 2 log_2 b . Gerhard "And Lower Bound Of Zero" Paseman, 2017.09.26. | |
| Sep 26, 2017 at 21:09 | history | edited | Gerhard Paseman | CC BY-SA 3.0 |
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| Sep 26, 2017 at 20:00 | history | edited | Gerhard Paseman | CC BY-SA 3.0 |
added 26 characters in body
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| Sep 26, 2017 at 19:58 | comment | added | Gerhard Paseman | Of course, if gcd(a,b) is bigger than 1, then it is possible, especially if b-a is the least prime factor of both b and a. Gerhard "Also Leaves That To Reader" Paseman, 2017.09.26. | |
| Sep 26, 2017 at 19:45 | history | answered | Gerhard Paseman | CC BY-SA 3.0 |