Timeline for Algorithm for main diagonal of integer coefficients associated with Schroeder numbers
Current License: CC BY-SA 4.0
7 events
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| Sep 17 at 13:23 | history | edited | Peter Taylor | CC BY-SA 4.0 |
cases environment
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| Sep 17 at 11:17 | comment | added | Notamathematician | @PeterTaylor, $T_q(n,0)=1$ is also works (because of the symmetry of the recurrence), isn't it? Could you please edit the question to indicate the cases? | |
| Sep 17 at 11:13 | history | edited | Notamathematician | CC BY-SA 4.0 |
deleted 3 characters in body
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| Sep 17 at 10:38 | comment | added | joro |
You have a lot of conjectures about a(n)=b(n). For me interesting cases are when one is significantly faster than the other. You may need to work modulo prime to avoid extremely large integers. The process can be automated and interrupt very large running times. Very rough test is log(time(a(N)))/log(time(b(N))). If you wrap your pari in sage, you can do alarm(seconds).
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| Sep 17 at 9:38 | comment | added | Peter Taylor |
@joro, they're both quadratic (well, cubic when you take into account that the point is for $q$ to be a variable and $T_q$ to be a symbolic polynomial in $q$). OP, I think you're missing a base case. $T_q(0, k) = 1$ seems to work. And the notation is somewhat ambiguous about priority of the cases. Using the cases environment would allow it to be expressed more clearly.
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| Sep 17 at 8:53 | comment | added | joro | What is the complexity of both approaches (possibly defined modulo some large prime to avoid big integers)? Looks like $T(n,n)$ is computable in time exponential in $n$ because of the branches, right? | |
| Sep 17 at 8:27 | history | asked | Notamathematician | CC BY-SA 4.0 |