Timeline for Finding a new level-$7$ pi formula using the relation $j_{7A}(\tau) = \left(\sqrt{j_{7B}(\tau)}+\frac{7}{\sqrt{j_{7B}(\tau)}}\right)^2$
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 19, 2023 at 9:29 | comment | added | Tito Piezas III | I used Zagier's six 3-term recurrences to find a new cfrac for $\zeta(3)$ (I hope). Kindly see this MO post. | |
| May 17, 2023 at 18:28 | comment | added | Somos | That is a very good MO question. Some of this is known in theory. For example, the solutions of linear with constant coefficient recursions are known to be sums of simple exponentials with overall exponential growth rates. | |
| May 17, 2023 at 17:49 | comment | added | Gerald Edgar | Interesting. If we take $k$ large enough, can we get $n=0$? (Recurrence with constant coefficients has a more complete theory.) Or, on the other hand, if we take $n$ large enough can we get $k=3$? (Three-term recurrences are associated with continued fractions.) | |
| May 17, 2023 at 17:19 | comment | added | Somos | @TitoPiezasIII The Fibonacci sequence is a good example. There are very many recurrence relations that it satisfies. The defining relation $F_{n+2}=F_{n+1}+F_n$ is one but there is also $F_{n+3}=2F_{n+1}+F_n.$ | |
| May 17, 2023 at 14:57 | comment | added | Tito Piezas III | Somos remarked that an integer sequence may have more than one recurrence. Let $R(k,n)$ be an $k$-term recurrence with polynomial coefficients of deg-$n$. So the second sequence has at least three, namely $R(6,4)$, $R(8,3)$ by Somos and $R(5,6)$ by G.Edgar? I always assumed that, like the Fibonacci sequence, there can only be one recurrence relation. | |
| May 17, 2023 at 12:32 | history | edited | Gerald Edgar | CC BY-SA 4.0 |
added 644 characters in body
|
| May 17, 2023 at 12:27 | history | edited | Gerald Edgar | CC BY-SA 4.0 |
added 644 characters in body
|
| May 17, 2023 at 11:51 | history | answered | Gerald Edgar | CC BY-SA 4.0 |