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How can we solve the non-homogeneous recurrence relation $$a_n = 3a_{n-2}+2a_{n-3}+f(n),$$ where $$f(n) = \begin{cases} -2^{k+1}\binom{2k+2}{k} & n=3k, \\ 2^{k+1} \binom{2k+2}{k+1} & n=3k+1, \\ 2^{k+1}\binom{2k+3}{k+1} & n=3k+2, \\ \end{cases}$$ and $a_1=10$, $a_2=10$, and $a_3=14$?

I can solve the associated homogeneous recurrence $a_n=3a_{n-2}+2a_{n-3}$, but I don't know how to find a particular solution.

Adam Lowrance
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  • Let the parts be $f_1, f_2, f_3$. Can you find a particular solution for each of them? If so, combine it with $ \delta_3(n) f_1 + \delta_3(n-1) f_2 + \delta_3(n-2) f_3$, where $\delta_3 (x) = 1 $ iff $3 \mid x$, 0 otherwise. You can use the cube roots of unity if you don't want it to be too ugly / obviously piecewise IE $\delta_3 (n) = 1/3 ( 1^n + \omega^n + \bar{\omega}^n)$, and then combine coefficients of $ \omega$ and see if it gives something nice. – Calvin Lin Jun 22 '23 at 17:42

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