Over the past few months I have been investigating one the generalizations of the Fibonacci numbers, called the Generalized Fibonacci Numbers (GFNs).
The GFNs are just like the regular Fibonacci numbers except that the sequences of GFNs have arbitrary initial values.
The GFNs obey the same type of recurrence relation as the Fibonacci numbers, namely: $G_n=G_{n-1}+G_{n-2}$ but can also be computed by $G_n=aF_{n-2}+bF_{n-1}$ where $G_1=a$ and $G_2+b$ are the initial values.
I am hoping that someone familiar with the GFNs can tell me whether or not certain sums I have been investigating have already been studied. I have been told that they have but so far I haven't found any papers that deal with these particular sums, and I would really like to see what other authors have done.
My Work:
So what I have been investigating are sums and other identities that involve more that one GFN sequence at a time. The identities that I have seen have only dealt with one GFN sequence at a time, or maybe the GFNs and Fibonacci and Lucas numbers.
Such as Identity 13 on page 113 of Fibonacci and Lucas Numbers with Applications by Thomas Koshy:
$\sum_{i=1}^n G_{2i}=G_{2n+1}-a$
The sums I am considering sum through multiple GFN sequences. For instance you might begin with the first term of the Fibonacci sequence which can be written as $G_(1,1)_1$, where $(1,1)$ are $(a,b)$, the first and second terms in the sequence. Then continuing on to the second term of the sequence $G(1,2)$, written as $G(1,2)_2$. Each time we increase $b$ by one and $n$ by one in the notation $G(a,b)_n$.
This can be written as: $\sum_{i=0}^n G(1,1+i)_{i+1}=nF_{n+2}+1$
The general form of this sum is: $\sum_{i=0}^n G(a,b+i)_{i+1}=G(a,b)_{n+3}+nF_{n+2}-F_{n+3}+2-b$
So does anyone know if this sort of thing has already been studied? If so I would love to see the paper/papers which dealt with these sorts of identities.
In all I have around 10 identities that deal with multiple GFN sequences, and I would love to be able to say that they are novel results!
But I realize I am in a well-trodden area, so I will not be surprised if this has already been done.
Your Gibonacci numbers can be written as linear combinations of Fibonacci numbers:
$\begin{align*} G^{a, b}_0 &= a - b, \quad G^{a, b}_1 = a \\ G^{a, b}_{n + 2} &= G^{a, b}_{n + 1} + G^{a, b}_n \end{align*}$
Define the generating function:
$\begin{align*} g^{a, b}(z) &= \sum_{n \ge 0} G^{a, b}_n z^n \end{align*}$
From the recurrence:
$\begin{align*} \sum_{n \ge 0} G^{a, b}_{n + 2} z^n &= \sum_{n \ge 0} G^{a, b}_{n + 1} z^n + \sum_{n \ge 0} G^{a, b}_n z^n \\ \frac{g^{a, b}(z) - (a - b) - a z}{z^2} &= \frac{g^{a, b}(z) - (a - b)}{z} + g^{a, b}(z) \\ g^{a, b}(z) &= \frac{b - a - b z}{1 - z - z^2} \end{align*}$
Remember the generating functions:
$\begin{align*} F(z) &= \sum_{n \ge 0} F_n z^n \\ &= \frac{z}{1 - z - z^2} \\ \frac{F(z) - F_0}{z} &= \sum_{n \ge 0} F_{n + 1} z^n \end{align*}$
Thus:
$\begin{align*} g^{a, b}(z) &= \frac{F(z) (b - a)}{z} - a F(z) \\ G^{a, b}_n &= (b - a) F_{n + 1} - a F_n \end{align*}$
From the generating function (or the explicit expression in terms of Fibonacci numbers and their identities) all sorts of fun expressions can be derived.