Identities concerning k -balancing and k -Lucas-balancing numbers of arithmetic indexes

: In this article, we derive some identities involving k balancing and k -Lucas-balancing numbers of arithmetic indexes, say an + p , where a and p are some ﬁxed integers with 0 ≤ p ≤ a − 1 .

On the other hand, k-Lucas-balancing numbers that are the natural extension of Lucas-balancing numbers extensively studied in [9].The sequence of k-Lucas-balancing numbers {C k,n } = {1, 3k, 18k 2 − 1, 108k 3 − 9k, . ..} satisfies the same recurrence relation as that of k-balancing numbers with different initial conditions, i.e., Few properties that the k-balancing numbers satisfy are summarized below.
• Binet formula for k-balancing numbers: , where • Catalan identity for k-balancing numbers: • D' Ocagne identity for k-balancing numbers: • Generating function for k-balancing numbers: • First combinatorial formula for k-balancing numbers: • Second combinatorial formula for k-balancing numbers:

Identities concerning k-balancing numbers of arithmetic indexes
In this section, we study different sums of k-balancing numbers of arithmetic indexes, say an + p for fixed integers a and p with 0 ≤ p ≤ a − 1.Several identities concerning such numbers are established straightforwardly.
The following lemmas are useful while proving the subsequent results.
Lemma 2.1.For all integers n ≥ 1, Proof.The proof of this result can be easily shown by using Binet formula for k-balancing numbers and the fact Proof.Using Binet formula for k-balancing numbers and the result from Lemma 2.1, the first term of the right hand side expression reduces and the result follows.[9], the previous formula reduces to an identity This identity gives the general term of the sequence of k-balancing numbers {B k,an+p } as a linear combination of two preceding terms.Iterative application of this result gives the general term as a combination of first two terms as follows: In particular, for a = 1, then r = 0 and we have the corresponding identity for k-balancing numbers, Now we will find the generating function for the sequence {B k,an+p }.Let G(k, a, p, x) be the generating function for the sequence {B k,an+p }, where 0 ≤ p ≤ a − 1, then Multiplying 2C k,a x and x 2 in (2.1) by turns and subtracting the first one from (2.1) and then adding the second one, we obtain The expression within the summation vanishes in view of Lemma 2.2.On the other hand, using the convolution identity Therefore, (2.2) gives For a = 1, then p = 0 and G(k, 1, 0, x) = x 1−6kx+x 2 which is indeed the generating function for k-balancing numbers.While Choosing a = 2, p will be 0 and 1 and we have The following theorem establishes the sum for k-balancing numbers of the type an + p.
Theorem 2.3.Let a be any integer and 0 Proof.Using Binet formula, the formula for geometric series and the fact By virtue of Lemma 2.1 and by Binet formula, we get the desired result.
The following results are immediate consequence of Theorem 2.3 by setting a = 2t + 1 and a = 2t respectively.
Corollary 2.4.The sum of odd k-balancing numbers of the kind an + p is Consequently, which is the desired recurrence relation for the sequence {B k,an+p }.

Identities concerning k-Lucas-balancing numbers of arithmetic indexes
In this section, we study the k-Lucas-balancing numbers of arithmetic indexes of the form an + p.
A repeated application of the formula in Lemma 2.2 gives an identity that relates k-Lucas-balancing numbers with k-balancing numbers, that is, for natural numbers n and l, Proof.Clearly 2C k,a(n+1)+p = B k,a(n+1)+p+1 − B k,a(n+1)+p−1 .Therefore, using Lemma 2.2, we get and we obtain the desired result.
In particular, for p = 0 the above identity reduces to C k,an+a = 2C k,a C k,an − C k,an−a .Applying iteratively the identity in Lemma 3.1 gives rise to the following sum formula.
Further, for m = n, this formula reduces to In order to find the generating function of the sequence {C k,an+p }, we proceed in the following way.Let g(k, a, p, x) be the generating function for the sequence {C k,an+p }, where 0 ≤ p ≤ a − 1, then Multiply (3.1) by 2C k,a x and x 2 and proceed as in case of the sequence {B k,an+p }, we get the generating function for the sequence {C k,an+p } as It is observed that for a = 1, then r = 0 and we have the generating function for k-Lucas-balancing numbers, g(k, x) = 1−3kx 1−6kx+x 2 .Further, for k = 1, the generating function for Lucas-balancing numbers g(x) = 1−3x 1−6x+x 2 is obtained.Theorem 3.2.Let a be any integer and 0 Proof.The proof of this theorem is analogous to Theorem 2.3.
As an observation, one can see that, for a = 1 then r = 0 gives the identity , where c n is the n th Lucas-cobalancing number with We end this section by establishing an important relation between k-balancing and k-Lucas-balancing numbers.This completes the proof.
In particular, for t = 1, the above identity reduces to B k,2n = 2B k,n C k,n , a known identity for kbalancing numbers.