Decimal Fibonacci Number?

Your question is valid, but incomplete insofar as the Fibonacci sequence requires two initial conditions. In other words, you must provide $f_0$ and $f_1$ before proceeding with $f_n = f_{n - 1} + f_{n - 2}$.

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 $$\alpha, \beta = \frac{a \pm\sqrt{a^2 + 4b}}{2}.$$

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 $\alpha$.

If you say THE Fibonacci sequence, the answer is no, since the Fibonacci sequence is defined by $F_0=1=F_1$ and $F_{n+2}=F_{n+1}+F_n$ for all $n\geq 0$.

However, if you only specify the recurrence relation $F_{n+2}=F_{n+1}+F_n$ for all $n\geq 0$ and demand that $(F_0,F_1)\neq (0,0)$ there are many similarities between both. For example, the limit $\lim_{n\to\infty}\frac{F_{n+1}}{F_n}$ is independent of $F_0$ and $F_1\in \mathbb{R}$.

YES, Fibonacci is: just adding the previous two numbers together. example: 0.25 + 0.5 =0.75 0.5 + 0.75 =1.35 0.75 + 1.35 = 2.1, ect.