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I have the sequence

$a_0=0$, $a_1=3$, $a_2=0$, $a_3=23$,

and

$a_n=6a_{n-2} + 8a_{n-3} + 3a_{n-4}$ for $n\ge 4$

and I have to find the formula for the generating function $A(t)=\sum_{n=0}^\infty a_n t^n$ and find a formula for $a_n$.

So far I have found

$A(t)=(26-18t^3) / (1-6t^2-8t^3-3t^4)$

but when I use this I get a formula for $a_n$ that does not work.

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    $\begingroup$ Note that the leading term in the numerator should not be $26$ ... it should be $3t$. Besides this the auxuillary equation factorises nicely and might be an easier route to a formula for $a_n$. $\endgroup$ Feb 25 '20 at 0:10
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The recurrence relation and initial conditions imply $$A(t)-0 t^0-3 t^1-0 t^2-23 t^3=6t^2 (A(t)-3t^1) + 8t^3 A(t) + 3t^4 A(t).$$ Solving for $A(t)$ yields $$A(t)=\frac{3t+5t^3}{1-6t^2-8t^3-3t^4}=\frac{1/2}{1-3 t} - \frac{3/2}{1+t} + \frac{3}{(1+t)^2} - \frac{2}{(1+t)^3},$$ which immediately implies that $$a_n=\frac{1}{2}\cdot 3^n - \frac{3}{2}(-1)^n + 3\binom{n+1}{1}(-1)^n - 2\binom{n+2}{2}(-1)^n=\frac{3^n-(-1)^n (2 n^2 + 1)}{2}.$$

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