Timeline for Number triangle
Current License: CC BY-SA 4.0
24 events
| when toggle format | what | by | license | comment | |
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| May 5, 2020 at 12:27 | comment | added | DSM | @IlyaBogdanov, I agree with your logic. I think I just gor lucky with 13, 109 and ignored that intricacy. Thanks for that! | |
| May 5, 2020 at 9:45 | comment | added | Ilya Bogdanov | On the other hand,if a doubling works on a specific word, then you can add an arbitrary number of such runs, and it will work as well. | |
| May 5, 2020 at 9:35 | comment | added | Ilya Bogdanov | It does not seem so; one of your four triangles is turned upside-down, and such triangles ate governed by different rules. Check out doubling 111001 | |
| May 5, 2020 at 4:46 | comment | added | DSM | @IlyaBogdanov, thanks for some brilliant observations. Taking some cues from your thoughts, I guess one can construct infinitely many special numbers from one special number. The idea is: if $(a_1,a_2, \cdots. a_n)$ is a special number, then $(a_1,a_2, \cdots. a_n, a_2, \cdots. a_n )$ must also be special. This can be seen easily by pasting 4 triangles of $(a_1,a_2, \cdots. a_n)$, appropriately. I did check this on Python starting with (1,1,0,1). Let me know if there's an error in logic. | |
| May 4, 2020 at 23:45 | comment | added | YCor | My solution easily describes the dimension (when nonzero) $u_n$ of the solutions of size $n$; nevertheless it does not say how the $2^{u_n}$ numbers thus defined by binary expansion distribute in $[2^{n-1},2^n]$. The picture might suggest it tends to a uniform distribution (this ought to be confirmed, or not, by more precise computations). I haven't thought about it so far. | |
| May 4, 2020 at 23:15 | comment | added | Ilya Bogdanov | @YCor, after this all has been done --- isn't your solution beter in a sense that it could provide more uniformity? The OP's graphs look really piecewise-linear --- is it true that your construction confirms that? Does it preserve some order of the proper vectors, or it rather smashes them all around? | |
| May 4, 2020 at 22:41 | comment | added | Pat Devlin | Oh, great! Nice work on the odd case. That’s a satisfying solution. :-) | |
| May 4, 2020 at 20:01 | comment | added | Ilya Bogdanov | Done! Could I ask some of you to cjeck? | |
| May 4, 2020 at 20:01 | history | edited | Ilya Bogdanov | CC BY-SA 4.0 |
Finished the solution
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| May 4, 2020 at 19:43 | history | edited | Ilya Bogdanov | CC BY-SA 4.0 |
All rewritten
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| May 4, 2020 at 19:38 | comment | added | YCor | Actually my argument doesn't involve parity of $n$, and just passes from each solution (allowing 0 on corners) for $n-3$ to two solutions for $n$. It just doesn't say, for given $n$, whether all solutions have 0 on corners, or if half of them have 1 on corners). | |
| May 4, 2020 at 19:38 | comment | added | Pat Devlin | Sounds good. Also. I now see that the dimension of U_k seems to be floor(k/3) when k is not 2 mod 3, and otherwise it seems to be floor(k/3) + 1. In particular, it seems like dim(U_k) = dim(U_{k+3}) - 1 | |
| May 4, 2020 at 19:18 | comment | added | Ilya Bogdanov | @Pat: Wait till I finish editing it in a clearer manner, please. Perhaps I'll also sort it out -- still unsure. | |
| May 4, 2020 at 19:09 | comment | added | Pat Devlin | Ah! I’m seeing it now! If I finish sorting out the case of n odd, then I could either (i) edit your answer to include this (and perhaps to be a bit clearer in parts) or (ii) post a follow-up answer. Whichever you’d prefer | |
| May 4, 2020 at 18:43 | comment | added | Ilya Bogdanov | @Pat Perhaps, it is too sketchy; I will try to fill the details or to rewrite all this. | |
| May 4, 2020 at 18:27 | comment | added | Pat Devlin | This seems better, but your assertion (in both cases) that $b = a + c$ doesn’t look like it’s true to me. I agree that in the case that $n$ is even, then the vector $b$ is completely determined by the vector $a$. And in the odd case, $a$ must be a solution to the case $(n+1)/2$. But this looks like it has good ideas that can be worked into a correct solution. | |
| May 4, 2020 at 15:48 | comment | added | Ilya Bogdanov | I@YCor: Yes, in these terms I now know that $2f(k)\cdot 2f(2k)=2^k$ which seems to agree with this formula. An advantage is that I know that for any even $n$ there are solutions with leading ones... Need to think more about an odd case. | |
| May 4, 2020 at 15:39 | comment | added | YCor | It seems that you haven't still reached the modulo 3 phenomenon. Namely, define $f(3n)=f(3n+2)=2^{n-1}$, $f(3n+1)=2^{n}$. Then the number of solutions, if nonzero, is equal to $f(n)$. (While $2f(n)$ is, for every $n\ge 0$, the number of solutions allowing $0$ on corners.) | |
| May 4, 2020 at 15:26 | comment | added | Ilya Bogdanov | Added some Addendum which makes the things more clear. | |
| May 4, 2020 at 15:25 | history | edited | Ilya Bogdanov | CC BY-SA 4.0 |
added 233 characters in body
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| May 4, 2020 at 15:14 | comment | added | Ilya Bogdanov | @PatDevlin: Thanks for pointing that out! I've made some corrections (and introduced one new idea); seems that it is corrected now. Does a new text agree with the numerical data? | |
| May 4, 2020 at 15:13 | history | edited | Ilya Bogdanov | CC BY-SA 4.0 |
Corrected an error
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| May 4, 2020 at 14:11 | comment | added | Pat Devlin | There is an error in your reasoning for the even case. For instance, $n=4$ and $n=6$ and $n=8$ all have the same number of solutions (all have 2 solutions if we require the top left entry to be 1). $n=10$ has 8 solutions (if we require top left entry to be 1). | |
| May 3, 2020 at 21:03 | history | answered | Ilya Bogdanov | CC BY-SA 4.0 |