Timeline for Which numbers are sums of finite numbers of reciprocal squares?
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| yesterday | comment | added | Michael Lugo | The paper gives $1/2 = 1/2^2 + 1/3^2 + 1/4^2 + 1/5^2 + 1/6^2 + 1/15^2 + 1/18^2 + 1/36^2 + 1/60^2 + 1/180^2$, which feels like it has some sort of structure but is not the result of the greedy algorithm. The $1/2$ case has come up here at math.SE before (math.stackexchange.com/questions/4126592/…) and relates to problem 152 of Project Euler: projecteuler.net/problem=152 | |
| Jun 29 at 8:52 | audit | First answers | |||
| Jun 29 at 9:03 | |||||
| Jun 29 at 7:41 | comment | added | gnasher729 | To see why I find it amazing: Just for fun, try writing 1/2 as the sum of the smallest number of squares of different reciprocals, and that’s about the most trivial case. Writing an algorithm for this is likely to be challenging. | |
| Jun 29 at 0:03 | comment | added | Arthur | Without having even looked at the actual paper, if they say it's not difficult to obtain representations of specific rationals, then my immediate assumption from just looking at the language in that sentence is that the greedy algorithm works, or something very close to it. | |
| Jun 28 at 18:36 | comment | added | Michael Lugo | I didn't know this before, and honestly the result kind of surprises me. Thank you for pointing me to finding this. | |
| Jun 28 at 17:43 | comment | added | templatetypedef | Amazing! Thanks so much for sharing this. | |
| Jun 28 at 17:43 | vote | accept | templatetypedef | ||
| Jun 28 at 17:24 | comment | added | gnasher729 | That's absolutely astonishing. | |
| Jun 28 at 16:15 | comment | added | Michael Lugo | I meant $[\pi^2/6 - 1, 1)$ - fixed. | |
| Jun 28 at 16:14 | history | edited | Michael Lugo | CC BY-SA 4.0 |
added 4 characters in body
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| Jun 28 at 15:37 | history | answered | Michael Lugo | CC BY-SA 4.0 |