In this paper, we study the k–Lucas numbers of arithmetic indexes of the form an+r , where n is a natural number and r is less than r. We prove a formula for the sum of these numbers and particularly the sums of the first k-Lucas numbers, and then for the even and the odd k-Lucas numbers. Later, we find the generating function of these numbers. Below we prove these same formulas for the alternated k-Lucas numbers. Then, we prove a relation between the k–Fibonacci numbers of indexes of the form 2rn and the k–Lucas numbers of indexes multiple of 4. Finally, we find a formula for the sum of the square of the k-Fibonacci even numbers by mean of the k–Lucas numbers.
Let us remember the k-Lucas numbers Lk,n are defined [
Among other properties, the Binnet Identity establishes being and the characteristic roots of the recurrence equation.
Evidently,.
Moreover, it is verified [1, Theorem 2.4] that.
If we apply iteratively the equation then we will find a formula that relates the k–Lucas numbers to the k–Fibonacci numbers:
This formula is similar to the Convolution formula for the k–Fibonacci numbers [2,3].
Moreover, we define. Then, if we do p = −n in Formula (1.1) obtain
.
We begin this section with a formula that relates each other some k-Lucas numbers.
If a is a nonnull natural number and r = 0, 1, 2, ... a − 1, then
Proof. In [
.
Then
If r = 0, then
In this case, if a = 2p + 1, then an odd k-Lucas number can be expressed in the form
Applying iteratively Formula (2.1), the general term, for, can be written like a non-linear combination of the form
In particular, if m = n, then
Let be the generating function of the sequence. That is,
Then,
and
from where no more to take into account Formula (2.1). So, the generating function of the sequence is
.
As particular case, if a = 1, then r = 0 and the generating function of the k-Lucas sequence is , that, for the classical Lucas sequence is
If we want to take out the two bisection sequences of the classical Lucas sequence (k = 1), the respective generating functions are a = 2 and r = 0: that generates the sequence a = 2 and r = 1: that generates the sequence.
If a is a nonnull natural number and r = 0, 1, 2, ... a − 1, then
Proof.
because
and after applying the formula for the sum of a geometric progression.
If r = 0 and a = 2p + 1, Equation (2.2) is
In this case, the sum of the first k-Lucas numbers is (for p = 0),
that for the classical Lucas numbers is
If r = 0 and a = 2p, then Equation (2.2) is
In this case, if p = 1 we obtain the formula for the sum of the first even k-Lucas numbers, and for the classical Lucas numbers is
For a > 0 and r = 0, 1, 2, ...a − 1, the sum of alternated k-Lucas numbers is
Proof. As in the previous theorem,
As particular case, if a = 2p + 1 and r = 0,
Then, for p = 0 we obtain the sum of the first alternated k-Lucas numbers
, that for the classical Lucas numbers is
.
If r = 0 and a = 2p + 1, then
And for the first consecutive alternated even k-Lucas numbers
that for the classical Lucas numbers is
.
In this section we will study a relation between the numbers and.
For r ≥ 1, it is
Proof.
In particular, if r = 1, it is
Taking into account, if we expand Formula (3.1), we find that this formula can be expressed as or, that is the same,
Then, applying Formula (2.2) to the second hand right of this equation with, a = 4n, and r = 3n for the first term and r = n for the second,
We tray to simplify the second hand right of this equation. For that, we will prove the following Lemma.
Proof. We will apply the following formulas:
(relation)
(negative)
(convolution)
(definition)
Then:
(by relation)
(by convolution)
(by negative)
(by definition)
And applying this Lemma to Equation (3.2), we will have:
that is
from where
If in Equation (3.3) it is a = 0, then it is , and applying the Formulas (2.5) and (2.4),
That is
In particular, for the classical Lucas numbers (k = 1), it is.
This work has been supported in part by CICYT Project number MTM200805866-C03-02 from Ministerio de Educación y Ciencia of Spain.