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### Limie's blog

By Limie, history, 6 weeks ago,

How to quickly find the value of each item in the following sequence.

$a_n=\lfloor (1+\sqrt{3})^n \rfloor$

Calculated in the sense of taking a mode.

• +1

 » 6 weeks ago, # |   0 Auto comment: topic has been updated by Limie (previous revision, new revision, compare).
 » 6 weeks ago, # |   0 Do you want to calculate all the values from 1 to n? or just for specific large n values? It looks like you got this from the formula a_n = 2*(a_(n-1)+a_(n-2)). If you want all values from 1 to n that'l take at least O(n) anyway so then use the recurrence
•  » » 6 weeks ago, # ^ |   0 I only want to calculate one values.
•  » » » 6 weeks ago, # ^ | ← Rev. 2 →   0 then given the fact we know its equal to this recurence, we can use matrix exponentiation to calculate values in O(logn) here's a post that explains about this a bit more.
•  » » » » 6 weeks ago, # ^ |   0 but $\lfloor (1+\sqrt{3})^n \rfloor$ is not equal to the formula a_n = 2*(a_(n-1)+a_(n-2)).
•  » » » » » 6 weeks ago, # ^ |   0 True, i miscalculated, but here you can find the coefficients of a recurrence with the 4 previous elements. https://oeis.org/A080041
•  » » » » » » 6 weeks ago, # ^ |   0 why a(n) = A080040(n) — (1+(-1)^n)/2？
•  » » » » » » » 6 weeks ago, # ^ |   0 No idea, but there are coefficiebts writenn for the recurrnece. Which is a_n = 2a_(n-1)+3a_(n-2)-2(a_(n-3)+a_(n-4))
 » 6 weeks ago, # | ← Rev. 2 →   0 But it's fast enough, since it's exponiental, it will only be calculted in log(n). Or is the problem of float relative error, if yes can you provide the equation that have this number as solution?
•  » » 6 weeks ago, # ^ |   0 Sorry,I need to calculated in the sense of taking a mode.
 » 6 weeks ago, # |   0 Auto comment: topic has been updated by Limie (previous revision, new revision, compare).