For what set of integers $\{i\}$ does modified Binet's formula for Fibonacci sequence (substituting that integer i for 5) equal an integer series?
$$B_{i,n} = \frac{1}{\sqrt i}\left(\left(\frac{1+\sqrt i}{2}\right)^n-\left(\frac{1-\sqrt i}{2}\right)^n\right)$$
So far I see that for $i$ = 1, 5, 9, 13, 17, ... e.g. every 4th integer, the sequence is all integers.
$B_{i,0} = 0$, $B_{i,1} = 1$ and $B_{i, n+1} = B_{i,n} + \frac{i-1}{4}B_{i,n-1}$ for $n \geqslant 1$, with $B_{i,2} =1$ and $B_{i,3} = 1+\frac{i-1}{4}$.
That implies that $B_{i,n}$ is an integer for every $n$ if and only if $i\equiv 1 \pmod{4}$.
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 $a=b=1 \ \text{and} \ f_0=0, f_1=1$.
In the end, what I'm pointing out here is that if you start with an integer sequence, you are lead to the result $\alpha,\beta=(a\pm\sqrt{a^2+4b})/2$ and you can see the factor of 4 plainly there.
Disclosure: this post is derived largely from a previous one: Decimal Fibonacci Number?.
