TITLE: General formula for the 1; 5; 19; 65; 211 sequence
QUESTION [6 upvotes]: I have got array $1; 5;19; 65; 211$. Can I find general formula for my array?
For example, general formula for array $1; 2; 6; 24; 120$ is $n!$. 
I tried a lot for finding the general formula, but I only found recurrent formula: $a_{n+1} = 5a_{n}-6a_{n-1}$. 
Any help will be much appreciated.

REPLY [3 votes]: The sequence you have found is a generalization of the Fibonacci sequence.
There have been many extensions of the sequence with adjustable (integer) coefficients and different (integer) initial conditions, e.g., $f_n=af_{n-1}+bf_{n-2}$. (You can look up Pell, Jacobsthal, Lucas, Pell-Lucas,  and Jacobsthal-Lucas sequences.) Maynard has extended the analysis to $a,b\in\mathbb{R}$, (Ref: Maynard, P. (2008), “Generalised Binet Formulae,” $Applied \ Probability \ Trust$; available at http://ms.appliedprobability.org/data/files/Articles%2040/40-3-2.pdf.)
We have extended Maynard's analysis to include arbitrary $f_0,f_1\in\mathbb{R}$. It is relatively straightforward to show that
$$f_n=\left(f_1-\frac{af_0}{2}\right) \frac{\alpha^n-\beta^n}{\alpha-\beta}+\frac{f_0}{2} (\alpha^n+\beta^n) $$
where $\alpha,\beta=(a\pm\sqrt{a^2+4b})/2$.
The result is written in this form to underscore that it is the sum of a Fibonacci-type and Lucas-type Binet-like terms. It will also reduce to the standard Fibonacci and Lucas sequences for $a=b=1 \ \text{and} \ f_0=0, f_1=1$.
The analysis follows Maynard almost exactly and I can expand upon this or provide a brief manuscript. This analysis is valid for any $a,b,f_0,f_1\in\mathbb{R}$. Notice that only when $a=b=1$ does the ratio of successive terms approach the golden ratio $\Phi$ for large $n$. I haven't fully explored the limiting ratio for the general case, but I find for positive $a,b$ that the limiting ratio is given by $\alpha$.
After all is said and done, your sequence with $f_0=1, f_1=5, \alpha=3, \text{ and } \beta=2$, comes down to
$$f_n=\frac{5}{2}(3^n-2^n)+\frac{1}{2}(3^n+2^n)=3^{n+1}-2^{n+1},\ \ \ n=0,1,2,3...\\
\lim_{n\to \infty}\frac{f_{n+1}}{f_n}=3
$$
which is essentially the same the result $(3^n-2^n)$ noted earlier on this page. However, now you have the tools to tackle many additional problems of this type.
Disclosure: This post is was derived largely from a previous answer of mine in Decimal Fibonacci Number?