I have the following recurrence given:
$$a_{0}=1$$ $$a_{1}=1$$ $$a_{n}=3a_{n-2}+3a_{n-1}$$
Why is that equal to something like this?:
$$a_{n}=3a_{n-2}+3a_{n-1}-2[n=1]+[n=0 ]$$
What are those brackets?
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4If I recall correctly, these are also known as Iverson brackets, $[n=1]$ evaluates to $1$ if $n = 1$, and to $0$ if $n \neq 1$. Here, it could be simply done with the Kronecker delta, $a_n = 3a_{n-2} + 3a_{n-1} - 2\delta_{1n} + \delta_{0n}$ – Daniel Fischer Sep 18 '13 at 20:40
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Argh. :) You beat my answer, @DanielFischer. :) – apnorton Sep 18 '13 at 20:41
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@anorton I bet it was the link that held you up ;) – Daniel Fischer Sep 18 '13 at 20:43
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So...these Iverson brackets don't do anything here? Because for any n>1 Kronecker delta gives 0. If so, then is something like $a_{n}=3a_{n-2}+3a_{n-1}+[n=0]+[n=1 ]$ also equal to this recurrence? – khernik Sep 18 '13 at 20:49
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To use Iverson brackets you'll need to assume that $a_n=0$ for $n<0$. This is no clearer than the original formulation. – lhf Sep 18 '13 at 21:00
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These are most likely Iverson Brackets. If the Boolean expression inside the square bracket is true, then the bracket evaluates to $1$. Otherwise, the bracket evaluates to $0$.
apnorton
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