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A question
By Limie, history, 5 months ago, In English

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.

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5 months ago, # |
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5 months ago, # |
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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

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    5 months ago, # ^ |
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    I only want to calculate one values.

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      5 months ago, # ^ |
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      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.

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        5 months ago, # ^ |
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        but $$$\lfloor (1+\sqrt{3})^n \rfloor$$$ is not equal to the formula a_n = 2*(a_(n-1)+a_(n-2)).

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          5 months ago, # ^ |
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          True, i miscalculated, but here you can find the coefficients of a recurrence with the 4 previous elements. https://oeis.org/A080041

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            5 months ago, # ^ |
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            why a(n) = A080040(n) — (1+(-1)^n)/2？

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              5 months ago, # ^ |
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              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))

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5 months ago, # |
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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?

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    5 months ago, # ^ |
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    Sorry,I need to calculated in the sense of taking a mode.

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5 months ago, # |
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