Timeline for The digit sum: $s(na)=s(nb)$
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 20, 2012 at 21:29 | comment | added | fedja | Done. Is it better now? | |
| Apr 20, 2012 at 20:58 | comment | added | fedja | OK, I'll expand a bit :) | |
| Apr 20, 2012 at 20:03 | comment | added | Seva | @fedja: wrong guess, I still don't get it. Exactly how have you defined your random variables, and how they are related to the story? What is $X_1$, say? What is $X_{25}$? | |
| Apr 20, 2012 at 19:25 | comment | added | fedja | It must be fairly slow if you believe the law of large numbers: after all we have two sums of $\log n$ weakly dependent random variables (the digits in the product), so $1/\log n$ is the best to hope for and $1/\sqrt{\log n}$ is the natural thing to expect. I guess Fedor answered Seva's concerns. What are yours? | |
| Apr 20, 2012 at 19:10 | comment | added | Aaron Meyerowitz | OK, maybe I do see it, but the decay is surprisingly slow. | |
| Apr 20, 2012 at 18:52 | history | answered | Aaron Meyerowitz | CC BY-SA 3.0 |