Timeline for A conjecture about certain values of the Fabius function
Current License: CC BY-SA 3.0
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| Feb 25, 2017 at 20:37 | history | edited | Vladimir Reshetnikov | CC BY-SA 3.0 |
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| Feb 25, 2017 at 19:30 | vote | accept | Vladimir Reshetnikov | ||
| Feb 22, 2017 at 8:58 | answer | added | juan | timeline score: 14 | |
| Feb 11, 2017 at 0:52 | history | edited | Vladimir Reshetnikov | CC BY-SA 3.0 |
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| Feb 8, 2017 at 6:42 | history | edited | Vladimir Reshetnikov | CC BY-SA 3.0 |
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| Feb 8, 2017 at 6:11 | history | edited | Vladimir Reshetnikov | CC BY-SA 3.0 |
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| Feb 8, 2017 at 6:06 | history | edited | Vladimir Reshetnikov | CC BY-SA 3.0 |
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| Feb 8, 2017 at 5:07 | history | edited | Vladimir Reshetnikov | CC BY-SA 3.0 |
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| Feb 8, 2017 at 1:05 | history | edited | Vladimir Reshetnikov | CC BY-SA 3.0 |
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| Feb 8, 2017 at 0:56 | history | edited | Vladimir Reshetnikov | CC BY-SA 3.0 |
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| Feb 7, 2017 at 23:58 | history | edited | Vladimir Reshetnikov | CC BY-SA 3.0 |
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| Feb 7, 2017 at 23:39 | comment | added | Kevin Buzzard | My initial thought was to prove that your "fudge factor" used when going from $F$ to $a$ would be more than enough to make each term in the defining sum integral, but it's more delicate than that (as you probably knew). For example the $2^{-n(n+1)/2}/n!$ term when multiplied by the fudge factor is not in general an integer; there are powers of 2 in the denominator. So one has to think harder. | |
| Feb 7, 2017 at 23:16 | history | asked | Vladimir Reshetnikov | CC BY-SA 3.0 |