I'm trying to discover the general formula of a series defined with recursion:
$$ a_1 = 2, a_2 = 3, a_3 = 4 $$ and $$ a_n = a_{n-1} + a_{n-3} $$
It looks like Fibonacci, but the starting points are different, and the rule is different.
I tried to compare to Fibonacci, but got nothing.
Anyone can think how can I reach the general formula (depending only on 'n')? Some material on how to reach Fibonacci's general formula might help (saw that a long time ago and can't find it, just found proof that the formula is valid, given we have the formula, which does not help me)
Regards
Letting $a_0=1$ extends the relation.
If $r_1,r_2,r_3$ are the three distinct roots of $x^3-x^2-1=0$ then the formula will be of the form:
$$a_n=\alpha_1r_1^n+\alpha_2 r_2^n +\alpha_3r_3^n$$
You can solve for $\alpha_i$ by looking at the equations:
$$1 = a_0 =\alpha_1 + \alpha_2 + \alpha_3\\2 = a_1=\alpha_1r_1+\alpha_2r_2+\alpha_3r_3\\3 = a_2=\alpha_1r_1^2+\alpha_2r_2^2+\alpha_3r_3^2$$
Unfortunately, the cubic formula gives very messy values for $r_1,r_2,r_3$. The solution with the largest modulus is the real solution, $r_1\approx 1.4656$, so, for $n$ large, $a_n\approx \alpha_1r_1^n$. The other roots are $\approx -0.23279\pm0.79255 i$. These have modulus of about $0.826$ so they should pretty quickly contribute very little. For $n$ large enough, then, it will be the nearest integer to $\alpha_1r_1^n$. What "large enough" means depends on the values $\alpha_2,\alpha_3$.