Related papers: On $k$-Fibonacci sums by matrix methods

We show that the product of two consecutive Fibonacci (respectively Lucas) numbers is divisible by the sum of their indices if this sum is a prime number different from 5 and in the form (4r+1)(respectively (4r+3)).

In this article we present a new recurrence formula for a finite sum involving the Fibonacci sequence. Furthermore, we state an algorithm to compute the sum of a power series related to Fibonacci series, without the use of term-by-term…

In this paper, we define Tribonacci and Tribonacci-Lucas matrix sequences and investigate their properties.

The recurrence for the $k$-Fibonacci polynomials is usually iterated upwards to positive values of $n$ only. When the recurrence is iterated downwards to $n<0$, there are indices where the polynomials vanish identically. This fact does not…

We develop a recursive scheme, as well as polynomial forms (polynomials in $n$ of degree $m$), for the evaluation of Ledin and Brousseau's Fibonacci sums of the form $S(m,n,r)=\sum_{k=1}^nk^mF_{k + r}$, $T(m,n,r)=\sum_{k=1}^nk^mL_{k + r}$…

In this paper we use a formula for the $n$-th power of a $2\times2$ matrix $A$ (in terms of the entries in $A$) to derive various combinatorial identities. Three examples of our results follow. 1) We show that if $m$ and $n$ are positive…

We evaluate various binomial sums involving the powers of Fibonacci and Lucas numbers.

The Fibonacci number is the residue of a rational function, from which follows that Fibonacci number summation identities can be derived with the integral representation method, a method also used to derive combinatorial identities. A…

We show that for the classical Fibonacci sequence (Fn) and the Lucas sequence (Ln) the following identity holds for every integer n >= 2: (n-1)Fn equals the sum from k=1 to n-1 of Lk multiplied by F(n-k). Equivalently, this gives a…

Spivey presented a new approach to evaluate combinatorial sums by using finite differences. We present some closed forms for sums involving the binomial coefficients, Fibonacci and Lucas numbers in terms of the falling factorial.

We present numerous interesting, mostly new, results involving the $n$-step Fibonacci numbers and $n$-step Lucas numbers and a generalization. Properties considered include recurrence relations, summation identities, including binomial and…

We consider a family of infinite sums of products of Catalan numbers, indexed by trees. We show that these sums are polynomials in $1/\pi$ with rational coefficients; the proof is effective and provides an algorithm to explicitly compute…

In this paper, we establish several formulae for sums and alternating sums of products of generalized Fibonacci and Lucas numbers. In particular, we recover and extend all results of Z. Cerin and Z. Cerin & G. M. Gianella, more easily.

In this paper, for the generalized Fibonacci sequence $\left\{W_n\left(a,b,p,q\right)\right\}$, by using elementary methods and techniques, we give the asymptotic estimation values of…

At this paper, we derive some relationships between permanents of one type of lower-Hessenberg matrix and the Fibonacci and Lucas numbers and their sums.

In this paper, we introduce three new classes of binomial sums involving Fibonacci (Lucas) numbers and weighted binomial coefficients.

This paper considers the properties of Tribonacci numbers on identities, matrices, and determinants. In the first front part, we obtain several symmetric identities of Tribonacci numbers by a matrix-based approach and binomial inversion…

Let $(F_n)_{n\geq 0}$ be the Fibonacci sequence given by $F_0 = 0, F_1 = 1$ and $F_{n+2} = F_{n+1}+F_n$ for $n \geq 0$. In this paper, we have determined all the powers of 2 which are sums of five Fibonacci numbers with few exceptions that…

We study matrices which transform the sequence of Fibonacci or Lucas polynomials with even index to those with odd index and vice versa. They turn out to be intimately related to generalized Stirling numbers and to Bernoulli, Genocchi and…

For a pair of positive integers $n,k$ with $n\geq 2$, in this paper we prove that $$ \sum_{r=1}^k\sum_{|\bf\alpha|=k}{k\choose\bf\alpha} \zeta(n\bf\alpha)=\zeta(n)^k =\sum^k_{r=1}\sum_{|\bf\alpha|=k}…