Timeline for Generalizing Ramanujan's and the Chudnovskys' 1/pi formula (Part 1)
Current License: CC BY-SA 4.0
17 events
| when toggle format | what | by | license | comment | |
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| Sep 11, 2019 at 11:27 | history | edited | Tito Piezas III | CC BY-SA 4.0 |
Modified title
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| S Aug 31, 2019 at 7:00 | history | bounty ended | CommunityBot | ||
| S Aug 31, 2019 at 7:00 | history | notice removed | CommunityBot | ||
| Aug 24, 2019 at 12:09 | history | edited | Tito Piezas III | CC BY-SA 4.0 |
Small typo
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| Aug 23, 2019 at 5:08 | history | edited | Tito Piezas III | CC BY-SA 4.0 |
Some comment
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| S Aug 23, 2019 at 5:06 | history | bounty started | Tito Piezas III | ||
| S Aug 23, 2019 at 5:06 | history | notice added | Tito Piezas III | Draw attention | |
| Aug 21, 2019 at 15:38 | history | edited | Tito Piezas III | CC BY-SA 4.0 |
Phrasing
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| Aug 21, 2019 at 15:04 | comment | added | Tito Piezas III | @GerryMyerson: That's a very fortunate coincidence. | |
| Aug 21, 2019 at 13:27 | comment | added | Gerry Myerson | Wadim Zudilin gave a talk today about series like these. He may be able to help you. | |
| Aug 21, 2019 at 6:55 | history | edited | Tito Piezas III | CC BY-SA 4.0 |
deleted 14 characters in body
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| Aug 21, 2019 at 6:15 | comment | added | Tito Piezas III | @Wolfgang: Actually, $p=2$ and $p=3$ work for ANY arbitrarily large $d$ yielding $C_d$ as an algebraic number of degree $n$. It's just that when $d$ has small class number $h(d) = 1$ (Heegner numbers) or $h(d)=2$ that $C_d$ also has conveniently small degree $n$. However, there doesn't seem to be any pi formula known using $p=4$ (or $p=1$ for that matter). | |
| Aug 21, 2019 at 6:10 | history | edited | Tito Piezas III | CC BY-SA 4.0 |
Some details
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| Aug 21, 2019 at 5:46 | comment | added | Wolfgang | If p=3 already mounts to the last Heegner number 163, can we expect more for a bigger p? | |
| Aug 21, 2019 at 5:07 | history | edited | Tito Piezas III | CC BY-SA 4.0 |
added 8 characters in body
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| Aug 21, 2019 at 5:05 | comment | added | Tito Piezas III | Note that, as pointed out by L.Miller, we have $6\times1448 = 8688 = A_{163}$ (without the root of unity) in this MO post. | |
| Aug 21, 2019 at 4:59 | history | asked | Tito Piezas III | CC BY-SA 4.0 |