# Need formula for sequence related to Lucas/Fibonacci numbers

I am trying to get a formula for the nth term of the following sequence:

2, 14, 22, 58, 266, 398, 1042, 4774, 7142, 18698, 85666, 128158, 335522,...

It's not in OEIS and as far as I can tell differencing (up to 15th order) does not yield any insight. However, dividing each term by $F_{(2n-1)}+F_{(2n+1)}$ (where $F_u$ is the $u$th Fibonacci number) yields the following:

0.666666etc., 2, 1.22222etc., 1.234042553191489, 2.162601626016260, 1.236024844720497, 1.236061684460260, 2.163117353874036, 1.236067843544478, 1.236067957955972, 2.163118955634673, 1.236067977083774, 1.236067977439093, 2.163118960609133, 1.236067977498497,...

which for even terms is converging to $$\frac{7}{2\phi}$$ and $$\frac{2}{\phi}$$ for odd terms ($\phi$ being the golden ratio). In rational form this sequence is:

Numerators: 2, 2, 11, 58, 266, 199, 1042, 4774, 3571, 18698, 85666, 64079, 335522,...

Denominators: 3, 1, 9, 47, 123, 161, 843, 2207, 2889, 15127, 39603, 51841, 271443,...

The numerators of the new sequence are the same as the terms in the original sequence except every third term is half the term in the original (except for the second term, which one would expect to be 14 but is instead 2).

How can I use all this info to come up with a formula for the nth term of the original sequence?

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Maple's gfun package says the generating function could be

$$g(t) = 2\;\frac { \left( t+1 \right) \left( {t}^{4}+6\,{t}^{3}+5\,{t}^{2}+6 \,t+1 \right) }{t^6 - 18 t^3 + 1}$$

A formula for the terms (starting at $k=0$) is

$$a_k = \sum_r \frac{5 (1-r)}{2 r^{k+1}} - \sum_s \frac{3 + 8 s + 4 s^2 + s^3}{8 s^{k+1}}$$ where the first sum is over the roots $r$ of $r^2 - 3 r + 1$ and the second is over the roots $s$ of $s^4 + 3 s^3 + 8 s^2 + 3 s + 1$.

Note that $r^2 - 3 r + 1 = (r - 1)^2 - (r-1) - 1$, so the roots of $r^2 - 3 r + 1$ are $1+\phi$ and $1 - 1/\phi$. This seems to be how your sequence is related to the Fibonacci numbers.

EDIT: There is more. The first sum $$\sum_r \frac{5 (1-r)}{2 r^{k+1}} = \frac{15}{2} F_{2k} + \frac{5}{2} F_{2k-1}$$

The second sum

$$\sum_s \frac{3 + 8 s + 4 s^2 + s^3}{8 s^{k+1}} = \cos(2 \pi (k-1)/3) (3 F_{2k} + F_{2k-1})$$

where $\cos(2 \pi (k-1)/3) = 1$ if $k \equiv 1 \mod 3$ and $-1/2$ otherwise. So

$$a_k = \left(\frac{5}{2} + \cos\left(\frac{2\pi (k-1)}{3}\right)\right)\left(3 F_{2k} + F_{2k-1}\right) = \cases{\dfrac{21}{2} F_{2k} + \dfrac{7}{2} F_{2k-1} & if k \equiv 1 \mod 3\cr 6 F_{2k} + 2 F_{2k-1} & otherwise\cr}$$

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