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I have a recursive formula for a sequence O: $ O_n = O_{n-1} + O_{n-2} + F_{n-1}$ where $F_n$ is the n-th Fibonacci number, $O_1 = 1$ and $O_2 = 2$.

After playing around with it, I found a new formula that might be easier to convert to the form I search: $ O_n = F_{n-3} * O_1 + F_{n-2} * O_2 + \sum_{k=2}^{n-1} F_{n-1-k} * F_k$.

Now what I am searching for is an explicit formula for $O_n$ that doesn't include a summation.

I also tried filling in Binet's formula and simplify, to no avail.

Here's a similar post, but the math is too hard for me, so I can't transform it to fit my problem.

An interesting property I found is that $\lim_{x\to\infty} \frac{O_x}{O_{x-1}}$ is equal to the golden ratio.

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  • $\begingroup$ Try a term like $nF_n$ $\endgroup$ – Empy2 Jul 5 '20 at 14:27
  • $\begingroup$ This is A029907. There is some discussion on the linked page, in particular you can find the claim that $O_n=\frac {(n+4)F_n+2nF_{n-1}}5$ which should be easy to verify. $\endgroup$ – lulu Jul 5 '20 at 14:27
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    $\begingroup$ As another approach, it follows immediately from your definition that $O_n$ satisfies $O_n=2O_{n-1}+O_{n-2}-2O_{n-3}-O_{n-4}$ and the associated characteristic polynomial factors as $(x^2-x-1)^2$. $\endgroup$ – lulu Jul 5 '20 at 14:34
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I calculated a few terms and got $(O_n)=(1,2,4,8,15,28,51...)$.

I looked this up in OEIS and found this formula, which could be made explicit with Binet's formula:

$O_n=\dfrac{(n+4)F_n+2nF_{n-1}}5.$

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