Timeline for Numerical evidence that $\pi$ is not normal in base two [closed]
Current License: CC BY-SA 4.0
20 events
| when toggle format | what | by | license | comment | |
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| Mar 11, 2020 at 11:29 | comment | added | joro | @WillSawin Many thanks for the debugging :) I wrote a lot of wrong stuff trying to "rationalize" the bug... | |
| Mar 7, 2020 at 3:43 | comment | added | Will Sawin | I also just want to point out that when I said the numbers should add up to 10000, I obviously meant plus or minus 1. (or even plus or minus 2 depending on how you count the digits before the "decimal" point) so the 110 is not a counterexample. | |
| Mar 7, 2020 at 0:22 | history | closed |
Mark Wildon R.P. Steven Landsburg Wojowu Tobias Fritz |
Not suitable for this site | |
| Mar 6, 2020 at 21:51 | comment | added | მამუკა ჯიბლაძე | A place to test sage code: https://sagecell.sagemath.org/ | |
| Mar 6, 2020 at 17:52 | comment | added | Mark Wildon | I'm voting to close this question as off-topic because the comment strongly suggests the numerical data is just wrong. | |
| Mar 6, 2020 at 17:10 | review | Close votes | |||
| Mar 7, 2020 at 0:22 | |||||
| Mar 6, 2020 at 16:26 | comment | added | joro | @მამუკაჯიბლაძე Many thanks :) | |
| Mar 6, 2020 at 16:04 | comment | added | მამუკა ჯიბლაძე |
For example, "00000000".count('00') gives 4
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| Mar 6, 2020 at 15:55 | comment | added | Will Sawin | e.g. see line 230 of github.com/python/cpython/blob/master/Objects/stringlib/… | |
| Mar 6, 2020 at 15:46 | answer | added | Wojowu | timeline score: 6 | |
| Mar 6, 2020 at 15:45 | comment | added | Will Sawin | From the data, we can tell with high confidence, without looking in the spec, that count does not allow overlaps. | |
| Mar 6, 2020 at 15:37 | comment | added | joro | @WillSawin I edited with the program, are there bugs in my program? | |
| Mar 6, 2020 at 15:37 | comment | added | joro | @Wojowu I edited with the program, are there bugs in my program? | |
| Mar 6, 2020 at 15:36 | history | edited | joro | CC BY-SA 4.0 |
added 360 characters in body
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| Mar 6, 2020 at 15:31 | comment | added | Will Sawin | @მამუკაჯიბლაძე This has to be it - if you every time you see $11$ you delete it and move on to the next one, you're essentially counting occurrences of $011$, $01111$, $0111111, \dots$, which gives $1/2^3 + 1/2^5 + 1/2^7 + \dots = 1/6$. | |
| Mar 6, 2020 at 15:27 | comment | added | Wojowu | The numbers I get are $2510,2505,2505,2480$ (give or take 1 on each). Could you perhaps share the code you have used? | |
| Mar 6, 2020 at 15:26 | comment | added | მამუკა ჯიბლაძე | You are probably counting substrings without overlaps? | |
| Mar 6, 2020 at 15:17 | comment | added | Wojowu | They should still add up to 10000 (give or take a few): the first 10000 bits contain 9999 consecutive bit pairs | |
| Mar 6, 2020 at 15:03 | comment | added | Will Sawin | How could this happen for random integers? Also, shouldn't your numbers add up to 10000? | |
| Mar 6, 2020 at 14:56 | history | asked | joro | CC BY-SA 4.0 |