Timeline for Binary weight of shifted integers
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
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| Dec 12, 2017 at 4:53 | answer | added | fedja | timeline score: 2 | |
| Dec 11, 2017 at 20:55 | comment | added | fedja | OK. As soon as I have some free time for typing more than 2 lines. Maybe in the evening :-) | |
| Dec 11, 2017 at 6:40 | comment | added | Shahrooz | Dear @fedja, would you please give your proof here? I am curious to see your proof! | |
| Dec 10, 2017 at 17:33 | comment | added | fedja | I'm not that sure yet. It is easy to show that regular patterns enjoy nice properties due to their regularity (judging from the tone of your remarks, it looks like we both know the proof of that part now) , but to show that irregular ones do not enjoy those properties even accidentally is often much harder. At least I do not see it yet in this particular case :-) | |
| Dec 10, 2017 at 13:17 | comment | added | Shahrooz | Yes, this is the case dear fedja... | |
| Dec 10, 2017 at 8:29 | comment | added | fedja | OK, then all numbers $n=2^s-1$ create all even overlaps indeed. Not sure about the other way yet, but we'll see... | |
| Dec 10, 2017 at 6:01 | history | edited | Shahrooz | CC BY-SA 3.0 |
Related problem added
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| Dec 10, 2017 at 5:29 | history | edited | Shahrooz | CC BY-SA 3.0 |
corrected the bound for the number $n$
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| Dec 10, 2017 at 5:26 | comment | added | Shahrooz | The first natural number that is of the form $2^s-1$ is $3$ and the claim is true for $n\geq 3$. Anyway, you are right and I must mention in the question that $n\geq 3$. Thanks for these important trivial cases:) | |
| Dec 10, 2017 at 3:52 | comment | added | fedja | Well, then my eyesight is even worse than I thought because $n=2$ (which is not of the form $2^s-1$) gives me trouble already: $0110$ and $1001$ have an even overlap each with itself and between them. What do you claim, really? | |
| Dec 9, 2017 at 17:34 | comment | added | Shahrooz | @fedja: Please see the note that I added. I forgot to mention that I use the uniform representation, which explained in the question now. | |
| Dec 9, 2017 at 17:32 | comment | added | Shahrooz | @AaronMeyerowitz: yes the claim is as you said. | |
| Dec 9, 2017 at 17:31 | history | edited | Shahrooz | CC BY-SA 3.0 |
Added some explenation
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| Dec 9, 2017 at 17:00 | comment | added | fedja | Take $n=3=2^2-1$. I have $X_2(3)=11011$, $X^1_2(3)=11110$. So it looks (to me) like the intersection has $3$ elements (1st,2nd, and 4th positions). Am I daltonic, or what? | |
| Dec 9, 2017 at 16:31 | comment | added | Aaron Meyerowitz | You mean that the set of $n$ such that for all $k$ and $m$ the "intersection" is even is exactly the set of $n$ of the form $2^s-1.$ | |
| Dec 9, 2017 at 13:55 | history | asked | Shahrooz | CC BY-SA 3.0 |