On Recurrences in Generalized Arithmetic Triangle

Hacéne Belbachir; Abdelkader Bouyakoub; Fariza Krim

doi:10.1515/ms-2023-0025

Article Open Access

On Recurrences in Generalized Arithmetic Triangle

Hacéne Belbachir , Abdelkader Bouyakoub and Fariza Krim

Published/Copyright: March 31, 2023

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From the journal Mathematica Slovaca Volume 73 Issue 2

Abstract

In the present paper, we consider the generalized arithmetic triangle called GAT which is structurally identical to Pascal’s triangle for which we keep the Pascal’s rule of addition and we replace both legs by two sequences (a_n)_n≥1 and (b_n)_n≥1 with a₀ = b₀ = Ω. Our goal is to describe the recurrence relation associated to the sum of elements lying along a finite ray in this triangle. As consequences, we obtain some combinatorial properties and we establish that the sum of elements lying along a main rising diagonal is a convolution of generalized Fibonacci sequence and another sequence which one will determine. We also precise the corresponding generating function. Further, we establish some nice identities by using the Morgan-Voyce phenomenon. Finally, we generalize the Golden ratio.

2020 Mathematics Subject Classification: Primary 11B37; 05A10; Secondary 11B65; 11B39; 05A15

Keywords: Generalized Pascal triangle; arithmetic triangle; binomial coefficient; recurrence relation; generating function; generalized Golden ratio; Fibonacci sequence; convolution

1. Introduction

Pascal’s triangle appears in numerous situations and has many intriguing properties. Over the years, a number of Pascal-like triangular arrays have been developed. Koshy [20, 21] and Belbachir and Szalay [7] collected various generalizations. The generalization of Ensley [13] called Generalized Arithmetic Triangle, the Ensley’s GAT for short, changed the legs called generator sequences to an arbitrary sequences of real numbers (a_n)_n≥0 and (b_n)_n≥0. Then the GAT depends on the generator sequences (a_n)_n≥0 and (b_n)_n≥0. Ensley [13] studied the Fibonacci triangle, the case where a_n = b_n = F_n, the Fibonacci numbers. The cases, where (a_n, b_n) = (0, F_n), (an,bn)=(0,1n) and (a_n, b_n) = (F_2n−1, F_n−1), were treated by Dil and Mezo [12]. In [8], we can also find an interesting description of Fibonacci and Lucas triangles. As Fibonacci numbers can be generated from the rising diagonals of Pascal’s triangle, Feinberg, see [14], was motivated to develop a similar triangle that would generate Lucas numbers from its rising diagonal. This triangle is Lucas triangle which is the particular case of Ensley’s GAT when a_n = 1 and b_n = 2. For other recent works on the subject, we refer to references [2,3]

In the present paper, we consider similar construction of Ensley’s GAT and our aim is to determine the recurrence relation associated to the sum of elements lying along a finite ray. Then, we establish the corresponding generating function. Also, we study the Morgan-Voyce phenomenon that leads us to establish new identities. Finally, we provide some interesting applications, among which one will establish that the sum of elements lying along a main rising diagonal is a convolution of generalized Fibonacci sequence and another sequence which one will determine.

1.1. Ensley’s Generalized Arithmetic Triangle

The following definition is given in [13].

Definition 1.

Let (a_n)_n≥0 and (b_n)_n≥0 be real sequences such that a₀ = b₀ = Ω. The Ensley’s Generalized Arithmetic Triangle for (a_n)_n≥0 and (b_n)_n≥0 contains elements C(n, k) in the n^th row and k^th column defined as follows: C(n, 0) = a_n, C(n, n) = b_n and for n > k > 0,

C(n,k)=C(n−1,k)+C(n−1,k−1).

Belbachir and Szalay, see [7], considered the Ensley’s GAT with suitable change of the rule of addition and the generator sequences. It is defined as follows.

Definition 2.

Let (a_n)_n≥0 and (b_n)_n≥0 be two real sequences, A and B two real numbers. The GAT contains elements [ nk] in the n^th row and k^th column defined for n ≥ 2, as follows: [ n0]=Anan, [ nn]=Bnbn and for 1 ≤ k ≤ n − 1,

[nk]=A[n−1 k]+B[n−1k−1],

with the convention [ nk]=0 whenever k < 0 or k > n and [ 00] is denoted by Ω.

The convention certainly works also for the binomial coefficients: (nk)=0 whenever k < 0 or k > n.

In this paper, we keep the rule of addition in the triangle defined by Belbachir and Szalay [7] and the generator sequences of Ensley’s GAT. We call this triangle, generalized arithmetic triangle (GAT for short). It is defined as follows.

Definition 3.

Let (a_n)_n≥0 and (b_n)_n≥0 be two real sequences, A and B two numbers. The GAT contains elements 〈 nk〉 in the n^th row and k^th column defined for n ≥ 2, as follows: 〈 n0〉=an, 〈 nn〉=bn, and for 1 ≤ k ≤ n − 1,

〈nk〉=A〈n−1 k〉+B〈n−1k−1〉,

with the convention 〈 nk〉=0 whenever k < 0 or k > n and 〈 00〉 is denoted by Ω.

Both definitions given in [13] and [7] are special cases of the above definition.

Figure 1 shows the rows 0 to 4 of the GAT generated by the sequences (a_n)_n≥0 and (b_n)_n≥0 and the numbers A and B.

Clearly, a GAT depends only on the generator sequences (a_n)_n≥0, (b_n)_n≥0 and the numbers A and B. Therefore, when necessary, we will denote by 〈 nk〉(A,B,an,bn)(〈 nk〉 for short) to emphasize the parameters of the triangle.

Figure 1.

First rows of the GAT.

1.2. GAT’S elements and binomial coefficients

The aim of this section is to evaluate 〈 nk〉 the entry k of row n, of a GAT in terms of the sequences (a_n)_n≥0, (b_n)_n≥0, the values A and B and the binomial coefficients (nk) of Pascal’s triangle. It is established in [7] with a slight modification, that:

Theorem 1.1.

Let (a_n)_n∈ℕ and (b_n)_n∈ℕ be two sequences and A and B two numbers. The corresponding GAT has as entry the number 〈 nk〉 given for (n, k) ≠ (0, 0), by the expression

〈nk〉=An−kBk(∑i=1n−k(n−1−i k−1)aiAi+∑i=1k(n−1−i k−i)biBi)+⌊n−kn⌋an+⌊kn⌋bn.

As ∑i=1n−k(n−1−ik−1)=(n−1k) for k ≥ 1, n ≥ 1 and ∑i=1k(n−1−ik−i)=(n−1k−1) for k < n, n ≥ 1, we get the following.

Corollary 1.1.1.

For a_n = ΩAⁿ and a given sequence (b_n)_n, the corresponding GAT has as entry the number 〈 nk〉 given for (n, k) ≠ (0, 0), by the expression

〈nk〉=An−kBk(Ω(n−1 k)+∑i=1k(n−1−i k−i)biBi)+⌊kn⌋bn.

Corollary 1.1.2.

For b_n = ΩBⁿ and a given sequence (a_n)_n, the corresponding GAT has as entry the number 〈 nk〉 given for (n, k) ≠ (0, 0), by the expression

〈nk〉=An−kBk(∑i=1n−k(n−1−i k−1)aiAi+Ω(n−1k−1))+⌊n−kn⌋an.

Corollary 1.1.3.

For (a_n, b_n) = (ΩAⁿ, ΩBⁿ), the corresponding GAT has as entry the number 〈 nk〉 given for (n, k) ≠ (0, 0), by the expression

〈nk〉=ΩAn−kBk(nk).

2. Quasi-symmetry and duality in a GAT

Note that in a GAT, we generally lose the symmetry. In fact, one observes that in general

〈nk〉(A,B,an,bn)≠〈 nn−k〉(A,B,an,bn).

However one has the quasi-symmetry property.

Quasi-symmetry property

Let (a_n)_n≥0 and (b_n)_n≥0 be two sequences, A and B two numbers. Then

2.1 〈nk〉(A,B,an,bn)=〈 nn−k〉(B,A,bn,an).

Example 1.

Let A = B = 1. For a_n = b_n = F_n, where (F_n)_n is the Fibonacci sequence, we obtain the symmetrical GAT given in Table 1. Whereas, for a_n = F_n, b_n = J_n, where (J_n)_n is the Jacobsthal sequence, we obtain the asymmetrical GAT given in Table 2. The quasi-symmetry GAT are illustrated in Tables 3 and 4.

Table 1.

The (1,1,F_n, F_n) GAT.

Ω
1	1
1	2	1
2	3	3	2
3	5	6	5	3
5	8	11	11	8	5
⋮	⋮	⋮	⋮	⋮	⋮	⋱

Table 2.

The (1,1,F_n, J_n) GAT.

Ω
1	1
1	2	1
2	3	3	3
3	5	6	6	5
5	8	11	12	11	11
⋮	⋮	⋮	⋮	⋮	⋮	⋱

Table 3.

The (1, 1, F_{n+1, 1}) GAT.

Ω
1	1
2	2	1
3	4	3	1
5	7	7	4	1
8	12	14	11	5	1
⋮	⋮	⋮	⋮	⋮	⋮	⋱

Table 4.

The (1, 1, 1, F_n+1) GAT.

Ω
1	1
1	2	2
1	3	4	3
1	4	7	7	5
1	5	11	14	12	8
⋮	⋮	⋮	⋮	⋮	⋮	⋱

Duality

Let n ∈ ℕ, r ∈ ℕ, q ∈ ℤ, p = 0,…, r − 1 such that r + q > 0. Then the grid point (n, p) and the direction (r, q) define a transversal ray of the GAT which contains the elements

〈n−qkp+rk〉, k=0,1,…,⌊n−pr+q⌋.

For a fixed direction (r, q) and a fixed value of p, the sequence (Gn(r,q,p))n≥0 defined by

Gn+1(r,q,p):=∑k=0⌊(n−p)/(r+q)⌋〈n−qkp+rk〉,

and G0(r,q,p)=0, constitutes the sum of elements lying on the corresponding ray in the GAT. Note that (because a sum over empty set is zero)

2.2 G1(r,q,p)=⋯Gp(r,q,p)=0 and Gn+1(r,q,p)=〈np〉 for p≤n<p+r+q.

For more details concerning directions in Pascal’s triangle, see [6] and [7]. In the following, using the quasi-symmetry property, we prove that there exists a grid point (m, p′) and a direction (r′, −q), q > 0 which define a transversal ray of GAT containing the same elements in reverse order. More precisely, we have the result.

Proposition 2.1.

Let r ∈ ℕ, q < 0, p = 0,…, r − 1 such that r + q > 0. Then

∑k=0⌊(n−p)/(r+q)⌋〈n−qkp+rk〉(A,B,an,bn)=∑k=0⌊(n−p′)/(r′+q′)⌋〈m−q′kp′+r′k〉(B,A,bn,an),

with m = n − q [(n − p)/(r + q)], q′ = −q, r′ = r + q, p′ = n − p − (r + q) [(n − p)/(r + q)].

Proof. We have r′ + q′ > 0 and p′ < r′. It is easy to see that

⌊(m−p′)(r′+q′)⌋=⌊1r(p+r⌊(n−p)(r+q)⌋)⌋=⌊(p/r)+⌊(n−p)(r+q)⌋⌋.

Since p < r, ⌊(m − p′) / (r′ + q′)⌋ = ⌊(n − p) / (r + q)⌋ and

∑k=0⌊(n−p′)/(r′+q′)⌋〈m−q′kp′+r′k〉(B,A,bn,an)=∑k=0⌊(n−p)/(r+q)⌋〈 n−q(⌊(n−p)/(r+q)⌋−k)n−p−(r+q)(⌊(n−p)/(r+q)⌋−k)〉(B,A,bn,an)=∑k=0⌊(n−p)/(r+q)⌋〈 n−qln−p−(r+q)l〉B,A,bn,an,

with l = ⌊(n − p)/(r + q)⌋ − k.

We conclude by (2.1).

According to the quasi-symmetry property in the GAT, we will consider in the sequel only directions (r, q) with q ≥ 0.

3. Sum of elements lying along a finite ray in GAT

In what follows, we shall use the following notation given in [15]. Let S be a statement that can be true or false. The bracketed notation [S] stands for 1 if S is true, 0 otherwise.

Ensley [13] discussed some GAT’s properties generated by the Fibonacci sequence and A = B = 1. He showed that the sum s_n of the elements in the n^th row

sn=∑k=0n〈nk〉(1,1,Fn,Fn)

satisfy the recurrence relation

sn=sn−1+sn−2+2n−1.

Moreover (see for instance, [7]), the ascending diagonal sum

dn=∑k=0⌊n/2⌋〈n−k k〉(1,1,Fn,Fn)

satisfies the recursion formula

dn=dn−1+dn−2+Fn−2+Fn/2−2[n even].

In this section, we consider the GAT as defined in Definition 3. Our aim is to determine the sum of elements located in a finite ray of the GAT generated by any sequences (a_n)_n≥0 and (b_n)_n≥0.

The case a_n = Aⁿ and b_n = Bⁿ was treated in [7]. One can easily show the following.

Lemma 3.1.

For p ≥ 0 and α = (n − p)/(r + q), we have

∑k=0⌊α⌋⌊n−p−(r+q)kn−qk⌋an−qk=an[p=0], ∑k=0⌊α⌋⌊p+rkn−qk⌋bn−qk=bp+αr[α∈ℕ].

Recall that the bracketed notation [S] stands for 1 if S is true, 0 otherwise.

Theorem 1.1 and Lemma 3.1 allow us to state the following result which gives the expression of Gn+1(r,p,q) in terms of (a_n)_n≥0, (b_n)_n≥0, A, B and binomial coefficients.

Proposition 3.1.

For p ≥ 0 and α = (n − p)/(_r + q), we have

Gn+1(r,p,q)=∑k=0⌊α⌋An−p−(r+q)kBp+rk(∑i=1n−p−(r+q)k(n−qk−1−i p+rk−1)aiAi+∑i=1p+rk(n−qk−1−i p+rk−1)biBi) +an⌊n−Pn⌋+bp+αr[α∈ℕ].

Remark 1.

Note that for p = 0, the summation can starts at k = 1 as (n−1)=0.

Remark 2.

Formula (3) in [6] is a special case of Proposition 3.1. Indeed, when a_n = Aⁿ and b_n = Bⁿ for n ≥ 0, we get

3.1 Gn+1(r,p,q)=∑k=0⌊α⌋(n−qkp+rk)An−p−(r+q)kBp+rk.

To avoid some complication, we agree that a_−n = b_−n = 0 for n ≥ 1.

4. Recurrence relation associated to rays in the GAT

We deal now with the main result, establishing a recurrence relation associated to the sum of elements located in a finite ray of the GAT generated by any sequences (a_n)_n≥0 and (b_n)_n≥0.

According to Proposition 2.1, we have to consider only the case q > 0. First, we set, for m ≥ 1,

λ0(m):=(−1)m(r−1 m),

and for s ≥ 1,

λs(m):=(−1)m−1(r−s−1 m−1).

Note that the λs(m)'s, 1 ≤ s ≤ r − m, are the binomial coefficients given by the extension of Pascal’s triangle to negative rows. By induction, we have the following.

Lemma 4.1.

For r ≥ 2 and 1 ≤ m ≤ r − 1,

λ0(m)=−∑s=1r−mλs(m).

Let now αj:=n−1−j−pr+q, 0 ≤ j ≤ r − 1. For α_j ∈ ℕ and r ≥ 1 we define

4.1 M0:=−bp+rα0+Brbp+rα0−r,

and for j ≥ 1,

4.2 Mj:=−Aj(λ0(j)bp+rαj+λ1(j)bp+rαj−1B+⋯+λr−j(j)bp+rαj−r+jBr−j).

Let us consider the expression of Gn(r,q,p) as given in Proposition 3.1. We denote by (Gn(p))n the sequence (Gn(r,q,p))n. We are now able to give our main result.

Theorem 4.1.

The terms of the sequence (Gn(p))n satisfy the recurrence relation

∑j=0r(−A)j(rj)Gn−j(p)−Bp∑j=0r(−A)j(r−p j)an−p−j−1+∑j=0r−1Mj[αj∈ℕ]=BrGn−q−r(p).

Remark 3.

For the sake of clarity, the proof of Theorem 4.1 will be given at the end of this section in the case p = 0. A similar proof can easily be established in the case p ≥ 1.

From Lemma 4.1 and Theorem 4.1, we can deduce the following.

Corollary 4.1.1.

For b_n = Bⁿb, b ∈ ℝ, the terms of the sequence (Gn(p))n satisfy the recurrence relation

∑j=0r(−A)j(rj)Gn−j(p)−Bp∑j=0r−p(−A)j(r−p j)an−p−j−1=BrGn−q−r(p).

Corollary 4.1.2.

For a_n = Aⁿa, a ∈ ℝ, the terms of the sequence (Gn(p))n satisfy the recurrence relation

∑j=0r(−A)j(rj)Gn−j(p)+∑j=0r−1Mj[αj∈ℕ]=BrGn−q−r(p).

Remark 4.

[6: Theorem 1] is a special case of Theorem 4.1 with a_n = Aⁿ and b_n = Bⁿ, n ≥ 0.

As the last result of this section, we express the sequence (Gn(0))n as the convolution of two well determined sequences.

Let (X_n)_n be a sequence defined for n ≥ 0 by

Xn:=∑k=0⌊n/(r+q)⌋An−(r+q)kBrk(n+r−1−qk r−1+rk),

and (Y_n)_n a sequence defined by: Y₀ := 0 and

Yn:={∑j=0n−1(−A)j(rj)an−j−1,for 1≤n<r+q,∑j=0r(−A)j(rj)an−j−1−∑j=0r−1Mj[αj∈ℕ]for n≥r+q.

Theorem 4.2.

For n ≥ 1, we have

Gn(0)=∑k=0n−1XkYn−k.

Proof. We use the induction on the non negative integer n.

For n ≥ r + q, assume that Gm(0)=∑k=0m−1XkYm−k, m ≤ n − 1. We have from Theorem 4.1,

Gn(0)=−∑j=1r(−A)j(rj)Gn−j(0)+BrGn−q−r(0)+Yn=−∑j=1r(−A)j(rj)∑k=0n−j−1XkYn−j−k+Br∑k=0n−r−q−1XkYn−r−q−k+Yn.

Letting s = k + j and t = k + r + q, we get

Gn(0)=−∑j=1r(−A)j(rj)∑s=jn−1Xs−jYn−s+Br∑t=r+qn−1Xt−r−qYn−t+Yn.

Since X_n = 0 for n < 0, we can write

Gn(0)=−∑j=1r(−A)j(rj)∑s=1n−1Xs−jYn−s+Br∑t=1n−1Xt−r−qYn−t+Yn=∑s=1n−1Yn−s(−∑j=1r(−A)j(rj)Xs−j)+Br∑t=1n−1Xt−r−qYn−t+Yn=∑s=1n−1Yn−s(−∑j=1r(−A)j(rj)Xs−j+BrXs−r−q)+Yn.

We know that (X_n)_n satisfies the recurrence relation, (see [6])

∑j=0r(−A)j(rj)Xn−j=BrXn−q−r.

Therefore, Gn(0)=∑s=1n−1Yn−sXs+Yn=∑s=0n−1Yn−sXs as X₀ = 1.

We now assume that 1 ≤ n ≤ r + q − 1. So, we have Xn=(n+r−1r−1)An and from (2.2), Gn(0)=〈 n−10 〉=an−1.

Then, we have to prove that ∑k=0n−1XkYn−k=an−1.

We have

∑k=0n−1XkYn−k=X0Yn+∑k=1n−1XkYn−k.

Since X₀ = 1 and Yn=∑j=0n−1(−A)j(rj)an−j−1, we obtain that

∑k=0n−1XkYn−k=an−1+∑j=1n−1(−A)j(rj) an−j−1+∑k=1n−1XkYn−k.

Therefore, we aim to prove that for 1 ≤ n ≤ r + q − 1,

4.3 ∑k=1n−1XkYn−k=−∑j=1n−1(−A)j(rj) an−j−1,

by considering two cases.

1. If 1 ≤ n ≤ r + 1.

∑k=1n−1XkYn−k=∑k=1n−1Ak(k−1+r r−1)∑j=0n−k−1(−A)j(rj)an−k−j−1,∑k=1n−1XkYn−k=∑k=1n−1∑i=0n−k−1(−1)n−1−k−iAn−k(r+ir−1)( rn−1−k−i)ak−1.

For each 1 ≤ k ≤ n − 1, we set j = n − 1 − k − i . Then, we get

∑k=1n−1XkYn−k=∑k=1n−1An−k∑j=0n−k−1(−1)j(rj)( r+n−1−k−j r−1)ak−1.

By putting in (9.3), t = n − 1 − k for k ≥ 1, t ≤ r − 1 since n ≤ r + 1, one has

∑k=1n−1XkYn−k=∑k=1n−1(−1)n−1−kAn−k( rn−k)ak−1.

If we put j = n − k, we get (4.3).

2. If r + 2 ≤ n ≤ r + q − 1. In this case, Yn=∑j=0r(−1)jAj(rj)an−1−j since n − 1 > r. So,

∑k=1n−1XkYn−k=∑k=1n−1Ak(k−1+r r−1)∑j=0r(−A)j(rj)an−k−j−1.

Now, we separate the sum following the index k in three parts; 1 ≤ k ≤ r, k = r + 1 and r + 2 ≤ k ≤ n − 1 to obtain

∑k=1n−1XkYn−k=∑k=1r∑j=0k−1(−1)k−j−1Ak( rk−1−j)(r+jr−1)an−k−1+∑j=0r(−1)r−jAr+1( rr−j)(r+jr−1)an−r−2+∑k=r+2n−1∑j=0n−2(−1)k−j−2Ak( rr−j)(k−1+j r−1)an−k−1.

We set s = k − 1 − j, t = r − j and we get

∑k=1n−1XkYn−k=∑k=1rAk∑s=0k−1(−1)s(rs)(r+k−1−s r−1)an−k−1+Ar+1∑t=0r(−1)t(r t)(2r−tr−1)an−r−2+∑k=r+2n−1(−1)k−2−rAk∑t=0r(−1)t(rt)(k−1+r−t r−1)an−k−1.

By putting in (9.3), t = k − 1 ≤ r − 1 in the first sum of the right hand side, one has

∑k=1rAk∑s=0k−1(−1)s(rs)(r+k−1−s r−1)an−k−1=∑k=1r(−1)k−1Ak(rk)an−k−1.

From Lemma 9.3,

∑t=0r(−1)t(rt)(2r−tr−1)=0 and ∑t=0r(−1)t(rt)(k−1+r−t r−1)=0.

Therefore, we get (4.3). The proof is complete.

5. The generating function

In this section, we give the generating function associated to the sum of GAT’s elements lying along a finite ray. But first, we will need the following lemmas and theorem.

Lemma 5.1.

For a ≥ r ≥ 1,

∑j=0r(−A)j(rj)〈n−j a〉=Br〈n−r a−r〉.

Proof. Let Ir=∑j=0r(−A)j(rj)〈 n−ja 〉. We use induction on the positive integer r. From Definition 3,

I1〈na〉−A〈n−1 a〉=B〈n−1a−1〉.

Assume now that Ir=Br〈 n−ra−r 〉 We have

Ir+1=∑j=0r(−A)j(rj)〈n−j a〉+∑j=0r(−A)j(rj−1)〈n−j a〉 =∑j=0r(−A)j(rj)〈n−j a〉−A∑j=0r(−A)j(rj)〈n−j−1 a〉=Br〈n−ra−r〉−ABr〈n−r−1 a−r〉.

We conclude by Definition 3.

Lemma 5.2.

Let Zn:=∑k=0p(n+p−kp−k)bk. The following identity holds true

∑n≥0Zntn=(1−t)−p−1∑k=0p(1−t)kbk.

Proof.

∑n≥0(∑k=0p(n+p−k p−k)bk) tn=∑k=0p∑n≥0(n+p−k p−k)tnbk=∑k=0p(1−t)k−p−1bk

∑m≥0(m+s s)tm=(1−t)−s−1.

For fixed r and p, we now set

5.1 Ck(t):=∑n≥p+rk〈 np+rk〉tn and H(t):=∑n≥0antn.

Lemma 5.3.

For r ∈ ℕ, 0 ≤ p ≤ r, q ∈ ℤ and the sequences (a_n)_n and (b_n)_n, the generating function C_k(t) verifies the relation, for k ≥ 0,

5.2 Ck(t)=(Bt1−At)rkC0(t),

with

5.3 C0(t)=bptp+(Bt1−At)p(H(t)−Ω+At1−At∑k=1p(1−AtB)kbk).

Proof. For k ≥ 0,

(1−At)rCk+1(t)=∑j=0r(−A)j(rj)∑n≥p+rk+r〈np+rk+r〉tn+j =∑j=0r(−A)j(rj)∑n≥p+rk+r+j〈 n−jp+rk+r〉tn.

So,

(1−At)rCk+1(t)=∑j=0r(−A)j(rj)(∑n≥p+rk+r〈 n−jp+rk+r〉tn),

as 〈 n−jp+rk+r 〉=0 for p + rk + r ≤ n < p + rk + r + j.

From Lemma 5.1, we get

(1−At)rCk+1(t)=∑n≥p+rk+rBr〈 n−rp+rk〉tn=Brtr∑n≥p+rk〈 np+rk〉tn.

Thus from (5.1), (1 − At)^r C_k+1 (t) = B^rt^rC_k(t). It follows that Ck(t)=(Bt1−At)rkC0(t).

For p = 0, C0(t)=∑n≥0〈 n0〉tn=∑n≥0antn=H(t).

For p ≥ 1 and using Theorem 1.1,

C0(t)=〈pp〉tp+∑n≥p+1An−pBp(∑i=1n−p(n−1−i p−1)aiAi+∑i=1p(n−1−i p−i)biBi)tn,

as ⌊ n−pn ⌋=⌊ pn ⌋=0.

Let us set C₀ (t) = b_pt^p + T₁ + T₂ with

T1:=∑n≥p+1An−pBp∑i=1n−p(n−1−i p−1)aiAitn

and

T2:=∑n≥p+1An−pBp∑i=1p(n−1−i p−i)biBitn.

We have

T1=tpBp∑m≥1[∑i=1m(m−i+p−1 p−1)Am−iai]tm, with m=n−p,=tpBp(∑m≥1amtm)(∑m≥0(m+p−1 p−1)(At)m)=tpBp(H(t)−Ω)(1−At)−P.T2=tpBp∑m≥0[∑i=1p(m+p−1 p−i)biBi](At)m+1, with m=n−p−1,=Atp+1Bp(1−At)−p−1∑i=1p(1−AtB)ibi.

By replacing, the proof is complete.

Remark 5.

We can easily deduce the following special cases:

For b_n = ΩBⁿ, C0(t)=(Bt1−At)pH(t).
For a_n = ΩAⁿ, C0(t)=bptp+Atp+1∑k=0p(1−At)k−p−1Bp−kbk.
For a_n = ΩAⁿ and b_n = ΩBⁿ, C₀ (t) = Ω(Bt)^p (1 − At)^−p−1.

Now, we are able to give the generating function of (Gn(p))n.

Theorem 5.1.

For r ∈ ℕ, 0 ≤ p < r, q ∈ ℤ and B ≠ 0, the generating function G(t):=∑n≥0Gn+1(p)tn associated to (Gn(p))n is given by

G(t)=(1−At)r(1−At)r−Brtr+qC0(t),

with C₀ (t) as given by (5.3).

Proof. We have

G(t)=∑n≥0Gn+1(p)tn=∑k≥0∑n≥0〈n−qkp+rk〉tn,

as 〈 nk〉=0 for n < k. According to (5.1) and (5.2),

G(t)=∑k≥0tqkCk(t)=C0(t)∑k≥0(Brtr+q(1−At)r)k=(1−At)r(1−At)r−Brtr+qC0(t).

Corollary 5.1.1.

For r ∈ ℕ, 0 ≤ p < r, q ∈ ℤ and B ≠ 0, the generating function associated to (Gn(p))n is given for b_n = ΩBⁿ by

5.4 G(t)=(Bt)pH(t)(1−At)r−P(1−At)r−Brtr+q,

and for a_n = ΩAⁿ by

G(t)=(1−At)rbptp+Atp+1∑k=0p(1−At)r+k−p−1Bp−kbk(1−At)r−Brtr+q,

and for a_n = ΩAⁿ and b_n = ΩBⁿ by

G(t)=Ω(Bt)p(1−At)r−p−1(1−At)r−Brtr+q.

Remark 6.

For a_n = Aⁿ and b_n = Bⁿ, we get Theorem 2 given in [6] which holds not only for q > 0, but for q ≤ 0 as well.

6. Morgan-Voyce and Fibonacci numbers

In [6], the authors established that the sequence (υ_n) associated to different directions of the ray in Pascal’s triangle, defined for r ∈ ℕ, q ∈ ℤ, p ≤ r − 1 such that r + q > 0, by

υn+1=∑k=0⌊(n−p)/(r+q)⌋(n−qkp+rk)xn−p−(r+q)yp+rk,

satisfies the linear recurrence relation

υn−x(r1)υn−1+⋯+(−x)r(rr)υn−r=yrυn−r−q.

Thus, the sequence (υ_n) is of order r for q ≤ 0 and we can write

υn−x(r1)υn−1+⋯+((−x)r+q( rr+q)−yr)υn−r−q+⋯+(−x)r(rr)υn−r=0.

This is what we call the Morgan-Yoyce phenomenon. For more details, see [[22,[24,[25]. Using this phenomenon, we can infer remarkable identities for Fibonacci numbers.

Let δ₀ = Ω, δ₁ be two parameters and (δ_n)_n be a sequence defined by

6.1 δn=2Aδn−1−A2δn−2, n≥2.

In this section, we consider the GAT generated by the sequences (a_n)_n and (b_n)_n such that

6.2 a1=Aδ1,anδn,n≥2,bn=Bn,n≥1.

By setting in Corollary 4.1.1, r = 2, p = 0 and q = −1, we get for n ≥ 3,

Gn−(2A+B2)Gn−1+A2Gn−2=an−1−2Aan−2+A2an−3.

Assume that A = i, B² = 1 − 2i, with i² = −1. It follows from (6.1) and (6.2) that

Gn=Gn−1+Gn−2.

However, Gn+1=∑k=0n〈 n+k2k 〉, so G1=〈 00〉=Ω, G₂ = 1 + i(δ₁ − 2). If we write G_n = αF_n + βF_n+1, then from G₁ and G₂ it follows that

Gn=(2Ω−1)Fn+(1−Ω)Fn+1+i(δ1−2)Fn−1.

Thus

6.3 (2Ω−1)Fn+(1−Ω)Fn+1=Re(Gn) and (δ1−2)Fn−1=Im(Gn).

For δ₁ = 2,

Gn=(2Ω−1)Fn+(1−Ω)Fn+1,

and for δ₁ = 2, Ω = 0,

Gn=Fn−1.

For Ω = δ₁ = 1, using some combinatorial computations, Proposition 3.1 and (6.3) give for m ≥ 0,

F2m+1=∑k=02m∑s=0⌊k/2⌋(−1)m+k+s2k−2s(k2s)(2m+k 2k),F2m+2=∑k=02m+1∑s=0⌊(k−1)/2⌋(−1)1+m+k+s2k−2s−1( k2s+1)(2m+1+k 2k).

As a last result in this section, we quote an application of Theorem 5.1. Recall that the Morgan-Voyce polynomials B_n (x) are defined by

B0(x)=1,B1(x)=2+t,Bn(x)=(2+x)Bn−1(x)−Bn−2(x),n≥2.

They have been widely studied (see [[4,[22,[24,[25]). In [24], the author gives a closed form expression and a generating function

Bn(x)=∑k=0n(n+K+1 1+2k)xk, ∑n≥0Bn(x)tn=11−(x+2)t−t2.

Using (5.4), we are able to express B_n (x) in descending powers of x. Indeed, if we put A = x + 2, B = −1, r = q = 1, a_n = Aⁿ for n ≥ 0, we deduce

∑n≥0Gn+1(0)tn=∑n≥0Bn(x)tn.

So, from (3.1) we get for n ≥ 0,

Bn(x)=∑k=0n(n−k k)(−1)k(x+2)n−2k.

We get also, ∑n≥0Gn+1(0)tn=∑n≥0Bn(x)tn by setting A = 1, B=x(x≥0), r = 2, q = −1, a_n = n + 1. Therefore, for x ≥ 0 and n ≥ 0,

Bn(x)=∑k=0n〈n+k 2k〉(1,x,n+1,xn).

7. Some illustrations

We present some examples to illustrate Theorem 4.1 for p = 0. The sequence (Gn(0))n is denoted for short by (G_n)_n.

Example 2.

Consider (b_n)_n = (Bⁿ)_n and an=An(n+kk)Bk for a fixed k, k ≥ r. From Corollary 4.1.1, the terms of the sequence (G_n)_n satisfy the recurrence relation

7.1 ∑j=0r(−A)j(rj)Gn−j−∑j=0r(−A)j(rj)an−j−1=BrGn−q−r.

From Lemma 9.3,

∑j=0r(−A)j(rj)an−j−1=An−1∑j=0r(−1)j(rj)(n+k−1−j k)Bk=An−1(n+k−1−r k−r)Bk.

Then (7.1) becomes

7.2 ∑j=0r(−A)j(rj)Gn−j−BrGn−q−r=An−1(n+k−1−r k−r)Bk.

For (r, q) = (1,1), we obtain

Gn−AGn−1−BGn−2=An−1(n+k−2 k−1)Bk,

which is the nonhomogeneous recurrence relation of (Fn(k)), the bivariate hyper-Fibonacci polynomials given in [5]. More precisely,

Gn=Fn(k), where Fn+1(k)=∑j=0⌊n/2⌋An−2jBj+k(n+k−j j+k).

For (r, q) = (1, q), q > 0, we obtain

Gn−AGn−1−BGn−q−1=An−1(n+k−2 k−1)Bk,

which is the recurrence relation of (Un(k)), the hyper q-Fibonacci polynomials given in [1]:

Gn=Un+2(k), where Un+1(k)=∑j=0⌊n/(q+2)⌋An−(q+1)jBj+k(n+k−qj j+k).

Example 3.

The direction (r, q) = (1,1). It concerns the sequence Wn+1=Gn+1(1,1,0) given by

Wn+1=∑k=0⌊n/(r+q)⌋〈n−k k〉(A,B,an,bn).

From Theorem 5.1, (W_n) satisfies, for n ≥ 2,

7.3 Wn−AWn−1−BWn−2=(an−1−Aan−2)+(bn−12−Bbn−32)[n odd].

We give the following examples.

a_n	b_n	W_n
AⁿF_2n−1	BⁿF_n−1	Wn=AWn−1+BWn−2+An−1F2n−4+Bn−12Fn−72[ n odd ]
AⁿF_n	BⁿF_n	Wn=AWn−1+BWn−2+An−1Fn−3+Bn−12Fn−52[ n odd ]

Remark 7.

The recurrence relation associated to (W_n) in the last example of the above table, is given in [7] for A = B = 1.

From (7.3) and Theorem 4.2, we are able to write (W_n) as the convolution of two sequences. One has for n ≥ 2,

7.4 Wn=∑k=0n−1XkYn−k,

with X₀ = 1, X₁ = A, X_n = AX_n−1 + BX_n−2, n ≥ 2 and Y₀ = 0, Y₁ = Ω, Yn=(an−1−Aan−2)+(bn−12−Bbn−32)[ n odd ], n ≥ 2.

For A = B = 1 and (a_n)_n, (b_n)_n two given sequences, (W_n) is the convolution of the Fibonacci sequence and the sequence (Y_n)_n.

In the following, some examples are given, where (F_n) is the Fibonacci sequence, (l_n) is the Lucas sequence and (Hn(k)) the hyper harmonic sequence defined by Hn(0)=1n for k ≥ 1. For more details, see [9,11].

a_n	b_n	W_n	W_n =
F_{n + 1}	1	W_n = W_n−1 + W_n−2 + F_n−2.	∑k=0n−1Fk+1Fn−k−2
l_n+1	1	W_n = W_n−1 + W_n−2 + l_n−2.	∑k=0n−1Fk+1ln−k−2
−1n	1	Wn=Wn−1+Wn−2+1n(n+1).	∑k=0n−1Fk+1(n−k)(n−k+1)
Hn(k)	1	Wn=Wn−1+Wn−2+Hn−1(k−1).	∑k=0n−1Fk+1Hn−k−1(k−1)
1	F_n+1	Wn=Wn−1+Wn−2+Fn−32[ n odd ]	∑k=1⌊ (n−1)/2 ⌋Fk−1Fn−2k
1	l_n+1	Wn=Wn−1+Wn−2+ln−32[ n odd ]	ln+∑k=1⌊ (n−1)/2 ⌋lk−1ln−2k
1	1n+1	Wn=Wn−1+Wn−2+4n2−1[ n odd ]	Fn−∑k=1⌊ (n−1)/2 ⌋4Fn−2k(2k+1)2−1

Example 4.

As a consequence of (7.4), we infer an expression for the sum of hyper harmonic numbers in terms of Fibonacci numbers. Indeed, in the last case of the above table, it is easy to see that 〈 nk〉(1, 1, 1 1n+1)=Hk+1(n−k). So one has the following.

Proposition 7.1.

For any n ≥ 2, we have

∑k=1⌊n/2⌋Hk+1(n−2k)=Fn−∑k=2⌊(n−1)/2⌋4Fn−2k(2k+1)2−1.

Example 5.

The case A = 0, b_n = Bⁿ, p = 0. From Corollary 4.1.1, this case deals with the sequence (Vn(r,q))n satisfying for n ≥ r + q the recurrence

7.5 Vn(r,q)−BrVn−r−q(r,q)=an−1.

The arrays in Tables 1 and 2 were studied by Bruckman [10] and Hoggatt [16], respectively. The column 0 consists of the Fibonacci numbers F_2n−2 and F_2n+1, respectively. Each of the remaining columns j, j ≥ 1, is obtained by lowering the previous column j − 1 by one level. Note that these tables are the particular case of the GAT when A = 0, B = 1 and, respectively, (a_n)_n≥1 = (F_2n)_n≥0, (a_n)_n≥0 = (F_2n+1)_n≥0.

In the Table 5, the n-th row sum is Vn+1(1,0)=F2n+1 and the rising diagonal sum is Vn+1(1,1)=Fn+12. Indeed, from relation (7.5), the sequence (Vn(1,0))n satisfies the recurrence

Vn(1,0)−Vn−1(1,0)=F2n−2,

and the sequence (Vn(1,1))n satisfies the recurrence

Vn(1,1)−Vn−2(1,1)=F2n−2.

Also, in the Table 6, Vn+1(1,0)=F2n+2 and Vn+1(1,1)=Fn+1Fn.

Some examples, with t_n the triangular number and T_n the tetrahedral number, corresponding to this case are given as follows:

Table 5.

The (0,1,F_2n, 1) GAT.

1
1	1
3	1	1
8	3	1	1
21	8	3	1	1
55	21	8	3	1	1
⋮	⋮	⋮	⋮	⋮	⋮	⋱

Table 6.

The (0,1,F_2n+1, 1) GAT.

1
2	1
5	2	1
13	5	2	1
34	13	5	2	1
89	34	13	5	2	1
⋮	⋮	⋮	⋮	⋮	⋮	⋱

a_n	B	r	q	Vn(r,q)
n + 1	1	1	0	t_n
t_{n + 1}	1	1	0	t_n
F_2n+3	−1	1	0	Fn+12
l_n+1	−1	1	1	F_{n + 1}
F_{2n + 2}	1	1	1	Fn+12
l_n	−1	1	1	F_n
5F_{n + 1}	1	1	3	F_{n + 2}
l_n−1	1	1	3	F_n

8. The generalized golden ratio

Given two initial terms u₀ and u₁ not both zero, we consider the recurrence relation u_n = u_{n− 1} + u_n−2. It is a well-known property of such sequences that limn→∞un+1un=ϕ, where ϕ, called the golden ratio, is the greatest root of t² − t − 1 = 0. Raab [23] extended this result to the sequences defined by the recurrence

wn=xwn−1+ywn−m−1,

given m + 1 initial terms. He proves that limn→∞wn+1wn exists and is a root of t^m+1 − xt^m − y = 0.

In this section, we establish the existence of similar limits for more general sequences. For that, let us consider the sequence (T_n)_n defined for n ≥ r + q by

8.1 Tn−A(r1)Tn−1+⋯+(−A)r(r1)Tn−r=BrTn−q−r.

Note that, from Corollary 4.1.2 and Corollary 4.1.1, T_n is the sum of elements lying on the corresponding ray in the GAT generated by the sequences (a_n)_n = (Aⁿ)_n and (b_n)_n = (Bⁿ)_n. The characteristic polynomial of (T_n)_n is

8.2 tq(t−A)r−Br=0,

and, from (2.2) and Theorem 1.1, initial conditions are

Tn+1=(np)An−pBp, p≤n<p+r+q.

Let us set

Tn:=∑i=1r+qcitin,

where t_i, i = 1,…, r + q, are the roots of (8.2) and g(t) = t^q (t − A)^r − B^r.

We have g(t) = t^q−1 (t − A)^r−1 [(r + q)t − Aq]. or B ≠ 0, no root of g′(t) is a root of g(t). It follows that g(t) has no multiple root. Moreover, the determinant of the coefficients c_i of the linear system in c_i,i = 1,…, r + q,

8.3 ∑i=1r+qcitin+1=(np)An−PBp, p≤n≤p+r+q−1,

is not zero. Indeed, by writing (8.3) as

8.4 {∑i=1r+qcitip+1=(pp)A0Bp∑i=1r+qcitip+2=(p+1 p)A1Bp⋮∑i=1r+qcitip+r+q=(p+r+1−1 p)Ar+q−1Bp,

the determinant which is

D=|(t1p+1t1p+2⋯t1p+r+qt2p+1t2p+2⋯t2p+r+q⋮⋮⋮⋮tr+qp+1tr+qp+2⋯tr+qp+r+q)|=∏k=1r+qtkp+1∏1≤i<j≤r+qr+q(tj=ti).

Since all the roots are simple, then D ≠ 0 and the system (8.4) can be solved uniquely in c₁,…, c_r+1. Let now t_* be such that for each i, |t*| > |t_i|. We have the following.

Proposition 8.1.

If A > 0 and B > 0, then

limn→∞Tn+1Tn=t∗.

Proof. Let g(t) = t^q (t − A)^r − B^r. We have, g (A) = −B^r < 0 and limt→∞g(t)=+∞. It follows that g admits a real root t_* such that t_* > A > 0. Furthermore, we have for 1 ≤ i ≤ r + q, |t_*| > |t_*| with t_i a root of g.

Indeed, suppose that t_i is a root of g such that |t_i| > t_*. Thus,

|ti−A|≥||ti|−A|≥|ti|−A>t∗−A>0.

So,

|ti−A|r>|t∗−A|r

and it follows that

|tiq(ti−A)r|>|t∗|q|(t∗−A)|r=Br.

This is a contradiction.

Consider now the ratio

Tn+1Tn=∑i=1r+qcitin+1∑i=1r+qcitin=t∗n(c1t1(t1t∗)n+⋯+c∗t∗+⋯+cr+qtr+q(tr+qt∗)n)t∗n(c1(t1t∗)n+c∗+⋯+cr+q(tr+qt∗)n).

Since t_* > 0 and t_* > |t_i|, we get limn→∞Tn+1Tn=t*. (We suppose without loss of generality that c_* ≠ 0).

Example 6.

For (r, q) = (1,1), we have T_n = AT_n−1 + BT_n−2, T₀ = 0, T₁ = 1 (see relation (8.1)).

The characteristic polynomial of (T_n) is t² − At− B = 0, whose roots are t1=(A+A2+4B)/2 and t2=(A−A2+4B)/2. Then:

For (A, B) = (1,1), we obtain T_n = F_n, the Fibonacci number and
limn→∞Fn+1Fn=(1+5)/2.
For (A, B) = (2, 1), we obtain T_n = P_n, the Pell number and
limn→∞Pn+1Pn=1+2.
For (A, B) = (1, 2), we obtain T_n = J_n, the Jacobsthal number and
limn→∞Jn+1Jn=2.
For (A, B) = (3, 2), we obtain T_n = ϕ_n, the Fermat number and
limn→∞ϕn+1ϕn=(3+17)/2.

Remark 8.

For (r, q) = (1, q), the result is given in [23].

9. Lemmata

In this section, we establish some intermediate lemmata which will play a key role in the proof of the recurrence relation associated with the sum of elements along a transversal ray in a GAT.

Lemma 9.1.

Let (d_n)_n∈ℕ be a sequence. Then for r ≥ 1, 1 ≤ t ≤ r, c ≥ r − 1, all natural numbers,

∑j=0r(−1)j(rj)∑i=1c+2−j−t(c−j−i+1 t−1)di=∑j=0r−t(−1)j(r−t j)dc+2−t−j.

Proof. Let Sr=∑j=0r(−1)j(rj)∑i=1c+2−j−r(c−i−j+1r−1)di. Using induction on the nonnegative integer r, we will prove first that

9.1 Sr=dc−r+2.S1=∑j=01(−1)j(1j)∑i=1c−j+t(c−i−j+10)di=∑i=1c+1di−∑i=1cdi=dc+1.

Assume now that S_r = d_c−r+2 and prove that S_r+1 = d_c−r+1.

Sr+1=∑j=0r+1(−1)j(r+1 j)∑i=1c−j−r+1(c−i−j+1 r)di=∑j=0r+1(−1)j(rj)∑i=1c−j−r+1(c−i−j+1 r)di+∑j=1r+1(−1)j( rj−1)∑i=1c−j−r+1(c−i−j+1 r)di=∑j=0r(−1)j(rj)∑i=1c−j−r+1(c−i−j+1 r)di+∑j=0r(−1)j(rj)∑i=1c−j−r(c−i−j r )di=∑j=0r(−1)j(rj)∑i=1c−j−r+1(c−i−j r−1)di+dc−r+1.

Let now

Tr=∑j=0r(−1)j(rj)∑i=1c−j−r−t+1(c−i−j+1 t−1)di.Tr=∑j=0r(−1)j∑m=0t(r−tj−m)(tm)∑i=1c−j−r−t+1(c−i−j+1 t−1)di,Tr=∑m=0t∑j=0r(−1)j(r−tj−m)(tm)∑i=1c−j−r−t+1(c−i−j+1 t−1)di.

Let k = j − m. Since (nk)=0 if k < 0, we get,

Tr=∑m=0t∑k=0r−m(−1)k+m(r−t k)(tm)∑i=1c+2−k−m−t(c−i−k−m+1 t−1)di.

Note first, that the coefficient (r−tk) in the summation with index k is zero for all k > r − t and r − m ≥ r − t since m ≤ t. So, we can write the summation up to r − t instead of r − m without changing the total sum. Therefore

Tr=∑m=0t∑k=0r−t(−1)k+m(r−t k)(tm)∑i=1c+2−k−m−t(c−i−k−m+1 t−1)di,Tr=∑m=0r−t(−1)k(r−t k)∑m=0t(−1)m(tm)∑i=1c+2−k−m−t(c−k−m−i+1 t−1)di.

But according to (9.1),

∑m=0t(−1)m(tm)∑i=1c+2−k−m−t(c−k−m−i+1 t−1)di=dc−k−t+2,

then the proof is complete.

Lemma 9.2.

Let (d_n)_n∈ℕ be a sequence. Then for t ≥ r ≥ 1, c ≥ 0, all non negative integers

∑j=0r(−1)j(rj)∑i=1c−j−t+1(c−j−i+1 t)di=∑i=1c−t+1(c−i−r+1 t−r)di.

Proof. We use induction on the non negative integer r. For r =1,

∑j=01(−1)j(1j)∑i=1c−j−t+1(c−j−i+1 t)di=∑i=1c−t+1(c−i+1 t)di−∑i=1c−t(c−i t)di=(tt)dc−t+1+∑i=1c−t((c−i+1 t)−(c−i t))di=(tt)dc−t+1+∑i=0c−t(c−it−1)di=∑i=1c−t+1(c−it−1)di.

Let now, Xr=∑j=0r(−1)j(rj)∑i=1c−j−t+1(c−i−j+1t)di and assume that

Xr=∑i=1c−t+1(c−i−r+1t−r)di.

Xr+1=∑j=0r(−1)j(rj)∑i=1c−j−t+1(c−i−j+1t)di+∑j=0r+1(−1)j(rj−1)∑i=1c−j−t+1(c−i−j+1t)di=∑i=1c−t+1(c−i−r+1t−r)di−∑j=0r(−1)j(rj)∑i=1c−j−t(c−i−jt)di=∑i=1c−t+1(c−i−r+1t−r)di−∑i=1c−t(c−i−rt−r)di=(t−rt−r)dc−t+1−∑i=1c−t((c−i−r+1t−r)−(c−i−rt−r))di=(t−rt−r)dc−t+1−∑i=1c−t((c−i−rt−r−1))di=∑i=1c−t+1((c−i−rt−r−1))di.

Lemma 9.3.

Let a, b and r be non-negative integers. Then

∑j=0r(−1)j(rj)(a−jb)={ (a−rb−r)for r≤b≤a,0for r>b.

Proof. In case where r ≤ b ≤ a, the proof can be found in [6].

Let now r > b and

Xr=∑j=0r(−1)j(rj)(a−jr−1).

By using induction on r, it is easily seen that X_r = 0 for a ≥ 2r − 1. Moreover, one has (rj)=∑m=0b+1(r−b−1j−m)(b+1m), so

∑j=0r(−1)j(rj)(a−jb)=∑j=0r(−1)j∑m=0b+1(r−b−1j−m)(b+1m)(a−jb)=∑k=0r−b−1(−1)k(r−b−1k)∑m=0b+1(−1)m(b+1m)(a−k−mb).

According to the above, ∑m=0b+1(−1)m(b+1m)(a−k−mb)=0, then the proof is complete.

Lemma 9.4.

Let r ≥ 1. Then for 0 ≤ m ≤ r, 0 ≤ s ≤ r − 1,

∑j=0m(−1)j(rj)(s+m−js)=(−1)m(r−s−1m).

Proof. We again use induction on r. It is easy to verify the formula for r = 1. So assume it true for r.

∑j=0m(−1)j(r+1j)(s+m−js)=∑j=0m(−1)j(rj)(s+m−js)+∑j=1m(−1)j(rj−1)(s+m−js)=∑j=0m(−1)j(rj)(s+m−js)−∑j=0m−1(−1)j(rs)(s+m−j−1s)=(−1)m(r−s−1m)−(−1)m−1(r−s−1m−1)=(−1)m(r−sm).

Especially for s = 0, one has for r ≥ 1 and 0 ≤ m ≤ r,

9.2 ∑j=0m(−1)j(rj)=(−1)m(r−1m),

and for s = r − 1, one has for r ≥ 1 and 0 ≤ t ≤ r − 1,

9.3 ∑j=0t(−1)j(rj)(r+t−jr−1)=(−1)t(rt+1).

The proof of the principal theorem of this paper is essentially based on the following result.

Lemma 9.5.

Let r ∈ ℕ, q ∈ ℤ⁺ and n ≥ r + q + 1 and for j = 0,…, r, α_j = (n − 1 − j) / (r + q).

Then only two possibilities hold.

α_j ∈ ℕ for j = 0,…, r. In this case, ⌊α_j⌋ = ⌊α₀⌋, for j = 0, …, r.
There is a unique j₀ ∈ {0,…, r} such that α_j0 ∈ ℕ. In such case,
⌊ αj ⌋={ αj0for j≤j0,αj0−1for ≥j0+1.

Proof. Let n ≥ r + q + 1. Then, there exist k ∈ ℕ and β < r + q such that n − 1 = k(r + q) + β. Let us define, for j ∈ ℕ, f(j) = (β − j)/(r + q). Then for j₀ = β, f (j₀) = 0 and so α_j0 = k. But β < r + q and q > 0, so the two following cases arise.

β ≤ r. In this case, there exist a unique j₀ ≤ r such that α_j0 ∈ ℕ.
r < β < r + q. In this case, 0 < f (j) < 1 since 0 ≤ j ≤ r. It follows that 0 < α_j − k < 1 and so α_j ∈ ℕ for 0 ≤ j ≤ r.

Lemma 9.6.

Let r ∈ ℕ, q ∈ ℤ⁺, (d_n)_n a sequence and let

T1:=∑j=0r(−1)j(rj)∑k=1⌊ βj ⌋As+2−(r+q)kBrk∑i=1s+2−j−(r+q)k(s−j−qk−i+1rk−1)di,T2:=∑k=1⌊ β0−1 ⌋As+2−(r+q)(k+1)Br(k+1)∑i=1s+2−(r+q)(k+1)(s−q(k+1)−i−r+1rk−1)di,

with βj=s+2−jr+q. Then T₁ = T₂.

Proof. Let s ≥ r + q − 2. We will prove that T₁ − T₂ = 0 by using Lemma 9.5. There are two cases.

Case 1. β_j ∈ ℕ. So, ⌊β_j⌋ = ⌊β₀⌋ for 0 ≤ j ≤ r and

T1−T2=∑k=2⌊ β0 ⌋As+2−(r+q)kBrk∑j=0r(−1)j(rj)∑i=1s+2j−(r+q)k(s−j−qk−i+1rk−1)di−∑k=1⌊ β0 ⌋−1As+2−(r+q)(k+1)Br(k+1)∑i=1s+2−(r+q)(k+1)(s−q(k+1)−i−r+1rk−1)di.

By putting c = s − qk and t = rk − 1 in Lemma 9.2, we infer

T1−T2=∑k=2⌊ β0 ⌋As+2−(r+q)kBrk∑i=1s+2j−(r+q)k(s−qk−i−r+1r(k−1)−1)di−∑k=1⌊ β0 ⌋−1As+2−(r+q)(k+1)Br(k+1)∑i=1s+2−(r+q)(k+1)(s−q(k+1)−i−r+1rk−1)di.

So,

9.4 T1−T2=0.

Case 2. There is a unique j₀, 0 ≤ j₀ ≤ r such that β_j0 ∈ ℕ and β_j ∈ ℕ for j ≠ j₀ with ⌊ βj ⌋={ βj0for j≤j0,βj0−1for j≥j0+1. .

Let us write T1 as three sums by separating the sum indexed by j following 0 ≤ j ≤ j₀ − 1, j = j₀ and j₀ + 1 ≤ j ≤ r.

T1=∑k=2βj0As+2−(r+q)kBrk∑j=0j0−1(−1)j(rj)∑i=1s+2−j−(r+q)k(s−j−qk−i+1rk−1)di+(−1)j0(rj0)∑k=2βj0As+2−(r+q)kBrk∑i=1s+2−j0−(r+q)k(s−j−qk−i+1rk−1)di+∑k=2βj0−1As+2−(r+q)kBrk∑j=j0+1r(−1)j(rj)∑i=1s+2−j−(r+q)k(s−j−qk−i+1rk−1)di.

Separating now, by including β_j0 value, the sums indexed by k in two parts following 2 ≤ k ≤ β_j0 − 1 and k = β_j0. We get by noting that (s−j0−qβj0−i+1rβj0−1)=0 as s − j₀ − qβ_j0 − i + 1 < rβ_j0 − 1,

T1=∑k=2βj0−1As+2−(r+q)kBrk∑j=0j0(−1)j(rj)∑i=1s+2−j−(r+q)k(s−j−qk−i+1rk−1)di+Aj0Brβj0∑j=0j0−1(−1)j(rj)∑i=1s+2−j−(r+q)k(s−j−qβj0−i+1rβj0−1)di+(−1)j0(rj0)∑k=2βj0As+2−(r+q)kBrk∑i=1s+2−j−(r+q)k(s−j0−qk−i+1rk−1)di+∑k=2βj0−1As+2−(r+q)kBrk∑j=j0+1r(−1)j(rj)∑i=1s+2−j−(r+q)k(s−j−qk−i+1rk−1)di.

Then,

T1=Aj0Brβj0∑j=0j0(−1)j(rj)∑i=1s+2−j−(r+q)βj0(s−j−qβj0−i+1rβj0−1)di+∑k=2βj0−1As+2−(r+q)kBrk∑j=0r(−1)j(rj)∑i=1s+2−j−(r+q)k(s−j−qk−i+1rk−1)di.

If we put in Lemma 9.2, c = s − qk and t = rk − 1, we get

T1=Aj0Brβj0∑j0−1j=0(−1)j(rj)∑s+2−j−(r+q)βj0i=1(s−j−qβj0−i+1rβj0−1)di+∑βj0−1k=2As+2−(r+q)kBrk∑s+2−j−(r+q)ki=1(s−qk−i−r+1r(k−1)−1)di=Aj0Brβj0∑j0−1j=0(−1)j(rj)∑s+2−j−(r+q)βj0i=1(s−j−qβj0−i+1rβj0−1)di+∑βj0−1k=2As+2−(r+q)(k+1)Br(k+1)∑s+2−j−(r+q)(k+1)i=1(s−q(k+1)−i−r+1rk−1)di.

Similarly writing T₂ following 1 ≤ k ≤ β_j0 − 2 and k = β_j0 − 1. Noting that ⌊β₀ − 1⌋ = β_j0 − 1, we obtain

T2=∑k=1βj0−2As+2−(r+q)(k+1)Br(k+1)∑i=1s+2−(r+q)(k+1)(s−q(k+1)−i−r+1rk−1)di+Aj0Brαj0∑i=1s+2−(r+q)βj0(s−qβj0−i−r+1r(βj0−1)−1)di.

Therefore

9.5 T1−T2=Aj0Brαj0∑j=0j0(−1)j(rj)∑i=1s+2−j−(r+q)βj0(s−j−qβj0−i+1rβj0−1)di−∑i=1s+2−(r+q)βj0(s−qβj0−i−r+1r(βj0−1)−1)di.

For j ≥ j₀, (s−j−qβj0−i+1rβj0)=0 as s − j − qβ_j0 − i+1 < rβ_j0 −1 when 1 ≤ i ≤ s + 2−j − (r=q)β_j0. Then

T1−T2=Aj0Brαj0∑j=0r(−1)j(rj)∑i=1s+2−j−(r+q)βj0(s−j−qβj0−i+1rβj0−1)di−∑i=1s+2−(r+q)βj0(s−qβj0−i−r+1r(βj0−1)−1)di.

Applying Lemma 9.2 for c = s − qβ_j0 and t = rβ_j0 − 1, we get T₁ − T₂ = 0.

We set for r ≥ 1, q ∈ ℤ⁺, βj=s+2−jr+q for 0 ≤ j ≤ r. Let then T₃ and T₄ be

T3:=∑j=0r(−1)j(rj)∑k=2⌊ βj ⌋As+2−(r+q)kBrk∑i=1rk(s−j−qk−i+1rk−i)di,T4:=∑k=2⌊ β0−1 ⌋As+2−(r+q)k+1Br(k+1)∑i=1rk(s−j−q(k+1)−i−r+1rk−i)di,

Lemma 9.7.

Let r ≥ 1, q ∈ ℤ⁺, (d_n)_n a sequence. Then, only one of the two possibilities holds.

For any j, 0 ≤ j ≤ r, β_j ∈ ℕ and T₃ = T₄.
There is a unique j₀ ∈ {0,…, r} such that β_j0 ∈ ℕ. In which case,
T3−T4=Aj0Brαj0[ ∑j=0j0−1(−1)j(rj)∑i=1rβj0(s−j−qβj0−i+1rβj0−i)di−∑i=1rβj0(s−qβj0−i−r+1r(βj0−1)−i)di ].

Proof. We follow the same steps as in the proof of Lemma 9.6 by using Lemma 9.3 instead of Lemma 9.2.

Remark 9.

Note that in the second case of Lemma 9.7, as the sum over an empty set is zero and (nk)=0 when n < k, then

T3−T4=0for j0=0.

10. Proof of Theorem 4.1

Without loss of generality, we will prove Theorem 4.1 for p = 0. The sequence (Gn(0))n will be written (G_n)_n for short.

We set

Z:=∑j=0r(−A)j(rj)Gn−j−∑j=0r(−A)j(rj)an−j−1+∑j=0r−1Mj[ αj∈ℕ ],

and prove that for n ≥ r + q, Z = B^rG_n−q−r. So, replacing G_{n − j} by relation (4.3) and taking account of Remark 1, we obtain

Z=∑j=0r(−1)j(rj)∑k=1⌊ αj ⌋An−1−(r+q)kBrk( ∑i=1n−1−j−(r+q)k(n−j−2−qk−irk−1)aiAi+∑i=1rk(n−j−2−qk−irk−i)biBi)+∑j=0r(−A)j(rj)brαj[αj∈ℕ]+∑j=0r−1Mj[αj∈ℕ].

Let us set,

U=∑j=0r(−1)j(rj)∑k=1⌊ αj ⌋An−1−(r+q)kBrk∑i=1n−1−j−(r+q)k(n−j−2−qk−irk−1)aiAi

and

V=∑j=0r(−1)j(rj)∑k=1⌊ αj ⌋An−1−(r+q)kBrk∑i=1rk(n−j−2−qk−irk−1)biBi.

If we separate the sum in the right-hand side of U isolating the term with k = 1, we get

U=An−1−(r+q)Br∑j=0r(−1)j(rj)∑i=1n−1−j−(r+q)(n−j−2−q−ir−1)aiAi+∑j=0r(−1)j(rj)∑k=2⌊ αj ⌋An−1−(r+q)kBrk∑i=1n−1−j−(r+q)(n−j−2−qk−irk−1)aiAi.

Lemma 9.1 gives

U=Bran−1−q−r+∑j=0r(−1)j(rj)∑k=2⌊ αj ⌋An−1−(r+q)kBrk∑i=1n−1−j−(r+q)(n−j−2−qk−irk−1)aiAi.

We do the same for V,

V=An−1−(r+q)Br∑j=0r(−1)j(rj)∑i=1r(n−j−2−q−ir−i)biBi+∑j=0r(−1)j(rj)∑k=1⌊ αj ⌋An−1−(r+q)kBrk∑i=1rk(n−j−2−qk−irk−i)biBi.

Lemma 9.3 gives

V=∑j=0r(−1)j(rj)∑k=2⌊ αj ⌋An−1−(r+q)kBrk∑i=1rk(n−j−2−qk−irk−i)biBi.

Moreover, replacing G_n−q−r by relation (4.3), we get

BrGn−q−r=∑k=1⌊ α0−1 ⌋An−1−(r+q)(k+1)Br(k+1)( ∑i=1n−1−(r+q)(k+1)(n−2−q(k+1)−i−rrk−1)aiAi +∑i=1rk(n−2−q(k+1)−i−rrk−i)biBj )+Bran−q−r−1+Bran−q−r−1+Brb(α0−1)[ α0−1∈ℕ ].

Let us set now,

U*=∑k=1⌊ α0−1 ⌋An−1−(r+q)(k+1)Br(k+1)∑i=1n−1−(r+q)(k+1)(n−2−q(k+1)−i−rrk−1)aiAi+Bran−q−r−1,V*=∑k=1⌊ α0−1 ⌋An−1−(r+q)(k+1)Br(k+1)∑i=1rk(n−2−q(k+1)−i−rrk−1)biBi.

Then,

10.1 Z−BrGn−q−r=(U−U*)+(V−V*)+W,

with

U−U*=∑j=0r(−1)j(rj)∑k=2⌊ αj ⌋An−1−(r+q)kBrk∑i=1n−1−j−(r+q)k(n−j−2−qk−irk−1)aiAi−∑k=1⌊ α0−1 ⌋An−1−(r+q)(k+1)Br(k+1)∑i=1n−1−(r+q)(k+1)(n−2−q(k+1)−i−rrk−1)aiAi,V−V*=∑j=0r(−1)j(rj)∑k=2⌊ αj ⌋An−1−(r+q)kBrk∑i=1rk(n−j−2−qk−irk−i)biBi−∑k=1⌊ α0−1 ⌋An−1−(r+q)(k+1)Br(k+1)∑i=1rk(n−2−q(k+1)−i−rrk−i)biBi,

and

W=∑j=0r(−A)j(rj)brαj[ αj∈ℕ ]+∑j=0r−1Mj[ αj∈ℕ ]−Brbr(α0−1)[ α0−1∈ℕ ].

We aim to prove that Z − B^rG_n−q−r, given by formula (10.1), is zero by Lemma 9.5. So, we will consider the two cases for each of the terms U − U^*, V − V^* and W.

Case 1. If α_j ∈ ℕ, ⌊α_j⌋ = ⌊α₀⌋ for 0 ≤ j ≤ r, then W = 0 and by replacing s by n α 3 in Lemmata 9.6 and 9.7, we get

U−U*=T1−T2=0,V−V*=T3−T4=0.

Therefore Z − B^rG_n−q−r = 0.

Case 2. There exists j₀, 0 ≤ j₀ ≤ r such that α_j0 ∈ ℕ and for j = j₀, α_j ∈ ℕ. So, we are dealing with two alternatives: j₀ = 0 and j₀ = 0.

Suppose that j₀ = 0. So from Lemma 9.5, α₀ ∈ ℕ and α_j ∈ ℕ with ⌊α_j⌋ = α₀ − 1 for j ≠ 0.

From relation (4.1), W = 0 and by replacing s by n − 3 in Lemmata 9.6 and 9.7, we get
U−U*=T1−T2=0,V−V*=T3−T4.

From Remark 9, V − V^* = 0. Therefore Z − B^rG_n−q−r = 0.
Suppose now that j₀ ≠ 0. From Lemma 9.5, α_j0 ∈ ℕ and for j ≠ j₀, α_j ∈ ℕ with ⌊ αj ⌋={ αj0for j≤j0,αj0−1for j≥j0+1 . In this case, by relation (4.2),
W=(−A)j0(rj0)brαj0+Mj0.

Replacing s by n − 3 in Lemmata 9.6 and 9.7, we get, U − U^* = T₁ − T₂ = 0 and

V−V*=T3−T4=Aj0Brαj0[ ∑j=0j0−1(−1)j(rj)∑i=1rαj0(n−2−j−qαj0−irαj0−i)biBi−∑i=1rαj0−r(n−2−qαj0−i−rr(αj0−1)−i)biBi ].

Introducing the value of α_j0, we rewrite V − V^* as

V−V*=Aj0Brαj0(∑j=0j0−1(−1)j(rj)∑i=1rαj0(rαj0−1+j0−i−jj0−j−1)biBi−∑i=1rαj0−r(rαj0−1+j0−i−rj0−1)biBi).

By replacing U − U^*, V − V^*, W in (10.1), we get

Z−BrGn−q−r=Aj0Brαj0∑j=0j0−1(−1)j(rj)∑i=1rαj0(rαj0−1+j0−i−jj0−j−1)biBi−Aj0Brαj0∑i=1rαj0−r(rαj0−1+j0−i−rj0−1)biBi+(−A)j0(rj0)brαj0+Mj0.

Now, we write the first sum in the right-hand side of Z − B^rG_n−q−r by isolating i = rα_j0 to obtain

Z−BrGn−q−r=Aj0Brαj0∑j=0j0−1(−1)j(rj)∑i=1rαj0−1(rαj0−1+j0−i−jj0−j−1)biBi+Aj0∑j=0j0−1(−1)j(rj)(j0−j−1j0−j−1)brαj0−Aj0Brαj0∑i=1rαj0−r(rαj0−1+j0−i−rj0−1)biBi+(−A)j0(rj0)brαj0+Mj0.

From relation (4.2) and (9.2)

Mj0=Aj0(∑s=0j0(−1)s(rs)brαj0+∑t=1r−j0Btλt(j0)brαj0−t).

Then

Z−BrGn−q−r=Aj0Brαj0(∑j=0j0−1(−1)j(rj))∑i=1rαj0−1(rαj0−1+j0−i−jj0−j−1)biBi −∑i=1rαj0−r (rαj0−1+j0−i−rj0−1)biBi )−Aj0∑t=1r−j0Btλt(j0)brαj0−t.

Now, we separate the first sum in Z−BrGn−q−r0 following the index i in three parts; 1 ≤ i ≤ rα_j0 − r, rα_j0 − r + 1 ≤ i ≤ rα_j0 − r 1 + j₀ and rα_j0 − r + j₀ ≤ i ≤ rα_j0 − 1 to obtain

Z−BrGn−q−r=Aj0Brαj0(Z1+Z2+Z3−Z4−Aj0∑t=1r−j0Btλt(j0)brαj0−t),

with

Z1=∑j=0j0−1(−1)(rt)∑i=1rαj0−r(rαj0−1+j0−i−jj0−j−1)biBi,Z2=∑j=0j0−1(−1)j(rj)∑i=rαj0−r+1rαj0−r−1+j0(rαj0−1+j0−i−jj0−j−1)biBi,Z3=∑j=0j0−1(−1)j(rj)∑i=rαj0−r+j0rαj0−1(rαj0−1+j0−i−jj0−j−1)biBiZ4=∑i=1rαj0−r(rαj0−1+j0−i−rj0−1)biBi.

Since (rαj0−1+j0−i−jj0−j−1)=0 for j ≥ j₀, Z₁ becomes

Z1=∑j=0r(−1)j(rj)∑i=1rαj0−r(rαj0−1+j0−i−jj0−j−1)biBi=∑j=0r(−1)j(rj)∑i=1rαj0−r(rαj0−1+j0−i−jeαj0−1)biBi as (nk)=(nn−k).

Then Lemma 9.3 gives

Z1=∑i=1rαj0−r(rαj0−1+j0−i−rj0−1)biBi.

So,

Z1−Z4=0.

Otherwise, by setting k = i − rα_j0 + r, it is easy to see that

Z2=∑j=0r(−1)j(rj)∑k=1j0−1(r−1+j0−k−jr−k)bk+rαj0−rBk+rαj0−r=∑k=1j0−1jbk+rαj0−rBk+rαj0−r∑j=0r(−1)j(rj)=(r−1+j0−k−jr−k).

Then according to Lemma 9.3, Z₂ = 0. So,

Z−BrGn−q−r=Aj0Brαj0Z3−Aj0∑t=1r−j0Btλt(j0)brαj0−t,

with

Z3=∑j=0j−0−1∑j=1r−j0(−1)j0−1(rj)(s+j0−j−1s)brαj0−sBrαj0−s

by setting s = rα_j0 − i, and from Lemma 9.4,

Z3=∑s=1r−j0(−1)j0−1(r−s−1j0−1)brαj0−sBrαj0−s=∑s=1r−j0λs(j0)brαj0−sBrαj0−s.

Therefore, Z − B^rG_n−q−r = 0 and the proof is complete.

(Communicated by István Gaál)

Funding statement: For Hacéne Belbachir and Fariza Krim, the paper was partially supported by the DGRSDT grant C0656701.

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Received: 2021-07-04

Accepted: 2022-05-17

Published Online: 2023-03-31

Published in Print: 2023-04-01

This work is licensed under the Creative Commons Attribution 4.0 International License.

Articles in the same Issue

https://doi.org/10.1515/ms-2023-0025

Keywords for this article

Generalized Pascal triangle; arithmetic triangle; binomial coefficient; recurrence relation; generating function; generalized Golden ratio; Fibonacci sequence; convolution

Creative Commons

BY 4.0

On Recurrences in Generalized Arithmetic Triangle

Article

Abstract

1. Introduction

1.1. Ensley’s Generalized Arithmetic Triangle

Definition 1.

Definition 2.

Definition 3.

First rows of the GAT.

1.2. GAT’S elements and binomial coefficients

Theorem 1.1.

Corollary 1.1.1.

Corollary 1.1.2.

Corollary 1.1.3.

2. Quasi-symmetry and duality in a GAT

Quasi-symmetry property

Example 1.

The (1,1,Fn, Fn) GAT.

The (1,1,Fn, Jn) GAT.

The (1, 1, Fn+1, 1) GAT.

The (1, 1, 1, Fn+1) GAT.

Duality

Proposition 2.1.

3. Sum of elements lying along a finite ray in GAT

Lemma 3.1.

Proposition 3.1.

Remark 1.

Remark 2.

4. Recurrence relation associated to rays in the GAT

Lemma 4.1.

Theorem 4.1.

Remark 3.

Corollary 4.1.1.

Corollary 4.1.2.

Remark 4.

Theorem 4.2.

5. The generating function

Lemma 5.1.

Lemma 5.2.

Lemma 5.3.

Remark 5.

Theorem 5.1.

Corollary 5.1.1.

Remark 6.

6. Morgan-Voyce and Fibonacci numbers

7. Some illustrations

Example 2.

Example 3.

Remark 7.

Example 4.

Proposition 7.1.

Example 5.

The (0,1,F2n, 1) GAT.

The (0,1,F2n+1, 1) GAT.

8. The generalized golden ratio

Proposition 8.1.

Example 6.

Remark 8.

9. Lemmata

Lemma 9.1.

Lemma 9.2.

Lemma 9.3.

Lemma 9.4.

Lemma 9.5.

Lemma 9.6.

Lemma 9.7.

Remark 9.

10. Proof of Theorem 4.1

References

Articles in the same Issue

Articles in the same Issue

Articles in the same Issue

The (1,1,F_n, F_n) GAT.

The (1,1,F_n, J_n) GAT.

The (1, 1, F_{n+1, 1}) GAT.

The (1, 1, 1, F_n+1) GAT.

The (0,1,F_2n, 1) GAT.

The (0,1,F_2n+1, 1) GAT.