On s-Fibonomials

Claudio de Jesús Pita Ruiz Velasco Universidad Panamericana Mexico City, Mexico

                        cpita@up.edu.mx

Abstract

For a given natural number s, we study s-Fibonacci sequences F_{sn} and the corresponding s-Fibonomial coefficients \binom{n}{p}_{F_{s}} = \frac{F_{sn}F_{s(n−1)}···F_{s(n−p+1)}}{F_{s}F_{2s}···F_{ps}}. We obtain the Z transform of products of powers of s-Fibonacci sequences. Since the s-Fibonomials are involved in this Z transform, we obtain from it some new results involving products of sequences of the type F_{sn+m}^{k} together with s-Fibonomials.

1 Introduction

We use ℕ for the natural numbers and ℕ′ for ℕ∪ {0}. We will be using without further comments the basic formulas for the Fibonacci sequence F_{n} and Lucas sequence L_{n} (\underline{A000045} and \underline{A000032} of Sloane’s Encyclopedia, respectively), such as F_{n} = \frac{1}{\sqrt{5}} (α^{n} − β^{n}) and L_{n} = α^{n} + β^{n}, where α = \frac{1}{2} (1 + \sqrt{5}) and β = \frac{1}{2} (1 − \sqrt{5}), together with the well-known relations involving α and β. What we use about Fibonacci identities is contained in the references [7] and [14].

Throughout this work, s will denote a natural number.

For a given Fibonacci number F_{n}, n ∈ ℕ, the s-Fibonacci factorial of F_{n}, denoted by (F_{n}!)_{s}, is defined as (F_{n}!)_{s} = F_{sn}F_{s(n−1)} ⋯ F_{s}. Given n ∈ ℕ′ and k ∈ {0, 1, …, n}, the s-Fibonomial coefficient \binom{n}{k}_{F_{s}} is defined by \binom{n}{0}_{F_{s}} = \binom{n}{n}_{F_{s}} = 1, and

\binom{n}{k}_{F_{s}} = \begin{matrix}(F_{n}!) & s \\ (F_{k}!) & s & (F_{n−k}!) & s \end{matrix} , k = 1, 2, …, n − 1,

that is

\binom{n}{k}_{F_{s}} = \frac{F_{sn}F_{s(n−1)} ⋯ F_{s(n−k+1)}}{F_{s}F_{2s} ⋯ F_{ks}} .  (1)

(The 1-Fibonomials are called simply Fibonomials.) It is clear that \binom{n}{k}_{F_{s}} = \binom{n}{n−k}_{F_{s}}. From the identity F_{s(n−k)+1}F_{sk} + F_{sk−1}F_{s(n−k)} = F_{sn} one can see at once that

\binom{n}{k}_{F_{s}} = F_{s(n−k)+1}\binom{n − 1}{k − 1}_{F_{s}} + F_{sk−1}\binom{n − 1}{k}_{F_{s}},  (2)

which shows (with a simple induction argument) that s-Fibonomials are integers. (See [5].)

Some examples are the following (as triangular arrays, where the lines correspond to n ∈ ℕ′, the columns to k, 0 ≤ k ≤ n, and the array is filled-out with formula (1) and/or rule (2):

(a) s = 1 (the Fibonomials: \underline{A010048} of Sloane’s Encyclopedia):

1

1  1
1  1_{⋆}  1_{♭}
1  2_{△}  2_{♭}  1
1  3_{✠}  6_{♭}  3  1
1  5  15_{♭}  15  5  1
1  8  40_{♭}  60  40  8  1
·  ·  ·

(b) s = 2 (\underline{A034801} of Sloane’s Encyclopedia):

1

1  1
1  3_{⋆}  1_{♭}
1  8_{△}  8_{♭}  1
1  21_{✠}  56_{♭}  21  1
1  55  385_{♭}  385  55  1
1  144  2640_{♭}  6930  2640  144  1
·  ·  ·

(c) s = 3 (\underline{A034802} of Sloane’s Encyclopedia):

1

1  1
1  4_{⋆}  1_{♭}
1  17_{△}  17_{♭}  1
1  72_{✠}  306_{♭}  72  1
1  305  5490_{♭}  5490  305  1
1  1292  98515_{♭}  417240  98515  1292  1
·  ·  ·

We have marked some patterns that can be recognized as general relations among the s-Fibonomials \binom{n}{1}_{F_{s}} (that is, among the quotients \frac{F_{sn}}{F_{s}}, as s-sequences —with fixed n) as follows:

• (a)

\binom{2^{n}}{1}_{F_{s}} = \frac{F_{2(2^{n−1}s)}}{F_{s}} = L_{2^{n−1}s} \frac{F_{2^{n−1}s}}{F_{s}} = L_{2^{n−1}s}\binom{2^{n−1}}{1}_{F_{s}},

and then

\binom{2^{n}}{1}_{F_{s}} = L_{s}L_{2s} ⋯ L_{2^{n−1}s}.  (3)

Thus, for n = 1 we have the sequence of Lucas numbers \binom{2}{1}_{F_{s}} = \frac{F_{2s}}{F_{s}} = L_{s} (corresponding to the mark ⋆ in previous tables), and for n = 2 we have the sequence \binom{4}{1}_{F_{s}} = \frac{F_{4s}}{F_{s}} = L_{s}L_{2s} = (3, 21, 72, …) (\underline{A083564} of Sloane’s Encyclopedia; corresponding to the mark ✠ in previous tables).

• (b)

\binom{3^{n}}{1}_{F_{s}} = \frac{F_{3^{n}s}}{F_{s}} = \frac{(α^{3^{n−1}s})^{3} − (β^{3^{n−1}s})^{3}}{F_{s}} = (α^{3^{n−1}2s} + β^{3^{n−1}2s} + (−1)^{3^{n−1}s}) \frac{F_{3^{n−1}s}}{F_{s}}
= (L_{3^{n−1}2s} + (−1)^{s}) \binom{3^{n−1}}{1}_{F_{s}},

and then

\binom{3^{n}}{1}_{F_{s}} = (L_{3^{n−1}2s} + (−1)^{s}) (L_{3^{n−2}2s} + (−1)^{s}) ⋯ (L_{2s} + (−1)^{s}) .  (4)

Thus for n = 1 we have the sequence \binom{3}{1}_{F_{s}} = \frac{F_{3s}}{F_{s}} = L_{2s} + (−1)^{s} = (2, 8, 17, …) (\underline{A047946} of Sloane’s Encyclopedia; corresponding to the mark △ in previous tables).

• (c)

\binom{2^{m}3^{n}}{1}_{F_{s}} = \frac{F_{2(2^{m−1}3^{n}s)}}{F_{s}} = L_{2^{m−1}3^{n}s} \frac{F_{2^{m−1}3^{n}s}}{F_{s}} = L_{2^{m−1}3^{n}s}\binom{2^{m−1}3^{n}}{1}_{F_{s}},

and then

\binom{2^{m}3^{n}}{1}_{F_{s}} = L_{2^{m−1}3^{n}s}L_{2^{m−2}3^{n}s} ⋯ L_{3^{n}s}\binom{3^{n}}{1}_{F_{s}}.  (5)

Finally, note that for fixed s, the sequence (1, ⋆, ▲, ✠, …) corresponds to \binom{n}{1}_{F_{s}} = \frac{F_{ns}}{F_{s}}, n ≥ 1. Thus, for s = 1 we have the sequence \frac{F_{n}}{F_{1}} = F_{n}. For s = 2 we have the sequence \frac{F_{2n}}{F_{2}} = F_{2n} = (1, 3, 8, 21, …) (\underline{A001906} of Sloane’s Encyclopedia), and for s = 3 we have the sequence \frac{F_{3n}}{F_{3}} = \frac{1}{2}F_{3n} = (1, 4, 17, 72, …) (\underline{A001076} of Sloane’s Encyclopedia).

Remark 1. For fixed s, the sequence \binom{n}{2}_{F_{s}} = \frac{F_{sn}F_{s(n−1)}}{F_{s}F_{2s}}, n ≥ 2, (marked with ♭ for each s = 1, 2, 3 in previous tables) is, in the case of Fibonomials (s = 1), the famous golden rectangle sequence \binom{n}{2}_{F} = F_{n}F_{n−1} = (1, 2, 6, 15, …) (\underline{A001654} of Sloane’s Encyclopedia). The corresponding sequences for the cases s = 2 and s = 3 are also included in the mentioned Encyclopedia (\underline{A092521} and \underline{A156085}, respectively), but there are not references to the formulas in which they appear in this work, namely \binom{n}{2}_{F_{2}} = \frac{1}{3}F_{2n}F_{2(n−1)} = (1, 8, 56, 385, …) and \binom{n}{2}_{F_{3}} = \frac{1}{16}F_{3n}F_{3(n−1)} = (1, 17, 306, 5490, …), respectively.

We have found just a few references on s-Fibonomials, even though Hoggatt [5] considered them in 1967. Besides the sequences of s-Fibonomials for s = 1, s = 2 and s = 3 included in Sloane’s Encyclopedia mentioned above, Gould [4] studied divisibility properties of s-Fibonomials.

The context in which s-Fibonomials are studied in this article is related to Z transforms (and then to generating functions) of certain Fibonacci sequences. This story begins with the works of Riordan [9], Carlitz [2] and Horadam [6] first, and Shannon [12] later. From all these works we know that the Z transform of the sequence F_{n}^{k} is

Z (F_{n}^{k}) = z \frac{∑_{i=0}^{k} ∑_{j=0}^{i} (−1)^{\frac{j(j+1)}{2}} \binom{k + 1}{j}_{F} F_{i−j}^{k}z^{k−i}}{∑_{i=0}^{k+1} (−1)^{\frac{i(i+1)}{2}} \binom{k + 1}{i}_{F} z^{k+1−i}} ,  (6)

(See also [10] and [11].) In a recent work [8], we proved the following formula for the Z transform of the sequence F_{n+m_{1}}^{k_{1}}F_{n+m_{2}}^{k_{2}}:

Z (F_{n+m_{1}}^{k_{1}}F_{n+m_{2}}^{k_{2}}) = z \frac{∑_{i=0}^{k_{1}+k_{2}} ∑_{j=0}^{i} (−1)^{\frac{j(j+1)}{2}} \binom{k_{1} + k_{2} + 1}{j}_{F} F_{m_{1}+i−j}^{k_{1}}F_{m_{2}+i−j}^{k_{2}}z^{k_{1}+k_{2}−i}}{∑_{i=0}^{k_{1}+k_{2}+1} (−1)^{\frac{i(i+1)}{2}} \binom{k_{1} + k_{2} + 1}{i}_{F} z^{k_{1}+k_{2}+1−i}} , (7)

where m_{1}, m_{2} ∈ ℤ and k_{1}, k_{2} ∈ ℕ′ are given. So (6) became a particular case of (7). After our work [8], we were able to find that (7) is in fact a particular case of a more general result: it turns out that for given m_{1}, m_{2} ∈ ℤ, s ∈ ℕ and k_{1}, k_{2}, t_{1}, t_{2} ∈ ℕ′, the Z transform of the sequence F_{t_{1}sn+m_{1}}^{k_{1}}F_{t_{2}sn+m_{2}}^{k_{2}} is

Z (F_{t_{1}sn+m_{1}}^{k_{1}}F_{t_{2}sn+m_{2}}^{k_{2}})  (8)

= z \frac{∑_{i=0}^{t_{1}k_{1}+t_{2}k_{2}} ∑_{j=0}^{i} (−1)^{\frac{(sj+2(s+1))(j+1)}{2}} \binom{t_{1}k_{1} + t_{2}k_{2} + 1}{j}_{F_{s}}F_{m_{1}+t_{1}s(i−j)}^{k_{1}}F_{m_{2}+t_{2}s(i−j)}^{k_{2}}z^{t_{1}k_{1}+t_{2}k_{2}−i}}{∑_{i=0}^{t_{1}k_{1}+t_{2}k_{2}+1} (−1)^{\frac{(si+2(s+1))(i+1)}{2}} \binom{t_{1}k_{1} + t_{2}k_{2} + 1}{i}_{F_{s}}z^{t_{1}k_{1}+t_{2}k_{2}+1−i}} ,

which means that for z ∈ ℂ outside the disk \overline{D} = \{z ∈ ℂ : |z| ≤ α^{s(t_{1}k_{1}+t_{2}k_{2})}\}, the right-hand side of this formula equals to

∑_{n=0}^{∞} \frac{F_{t_{1}sn+m_{1}}^{k_{1}}F_{t_{2}sn+m_{2}}^{k_{2}}}{z^{n}} .

We have to mention that Fibonomials can be seen as combinatorial objects (see [1]), and it is natural to expect that s-Fibonomials could enclose more general combinatorial interpretations. Formula (8) is the main result in this work. But we will see that some interesting results are obtained as consequences of (8), so its proof is not our final goal. In section 2 we introduce some basic facts about Z transform that will be used in the rest of the work, together with some preliminary results to be used mainly in section 3. Is in section 3 where we prove (8). In section 4 we establish some corollaries of (8), and finally in section 5 we do some comments on the representation of s-Fibonomials as linear combinations of certain “homogeneous terms”.

2 Preliminaries

We will be working with the so called “Z transform”, of which we recall some basic facts in this section. (For more details see [3] and [15].) The Z transform maps complex sequences (a_{0}, a_{1}, a_{2}, …) into complex (holomorphic) functions A : U ⊂ ℂ → ℂ given by the Laurent series A (z) = ∑_{n=0}^{∞} a_{n}z^{−n}. This function A is defined outside the closure \overline{D} of the disk D of convergence of the Taylor series ∑_{n=0}^{∞} a_{n}z^{n}. We will also denote the Z transform of the sequence (a_{n})_{n=0}^{∞} by Z (a_{n}). Some properties of the Z transform which we will be using throughout this work are the following: (avoiding the details of regions of convergence)

(a) Z is linear and injective.

(b) Advance-shifting property. For k ∈ ℕ we have

Z (a_{n+k}) = z^{k} (Z (a_{n}) − a_{0} − \frac{a_{1}}{z} − ⋯ − \frac{a_{k−1}}{z^{k−1}} ) .  (9)

Here a_{n+k} is the sequence a_{n+k} = (a_{k}, a_{k+1}, …).

(c) Multiplication by the sequence λ^{n}. If Z (a_{n}) = A (z), then

Z (λ^{n}a_{n}) = A (\frac{z}{λ}) .  (10)

(d) Convolution theorem. If a_{n} and b_{n} are two given sequences, then

Z (a_{n} ∗ b_{n}) = Z (a_{n}) Z (b_{n}) ,  (11)

where a_{n} ∗ b_{n} = ∑_{t=0}^{n} a_{t}b_{n−t} is the convolution of the sequences a_{n} and b_{n}.

If A (z) = Z (a_{n}), we also write a_{n} = Z^{−1} (A (z)), and we say that the sequence a_{n} is the inverse Z transform of A (z). Clearly Z^{−1} is also linear and injective.

Observe that according to (10), if Z (a_{n}) = A (z) then

Z ((−1)^{n} a_{n}) = A (−z) ,  (12)

and

Z (L_{sn+m}a_{n}) = α^{m}A (\frac{z}{α^{s}} ) + β^{m}A (\frac{z}{β^{s}} ) .  (13)

For given λ ∈ ℂ, λ ≠ 0, the Z transform of the sequence λ^{n} is plainly

Z (λ^{n}) = ∑_{n=0}^{∞} \frac{λ^{n}}{z^{n}} = \frac{z}{z − λ},  (14)

(defined for |z| > |λ|). This is an important formula in this article since Fibonacci sequences are combinations of sequences of the form λ^{n}. In particular we have that the Z transform of the constant sequence 1 is

Z (1) = \frac{z}{z − 1}.  (15)

(Observe that if t_{1} = t_{2} = 0, formula (8) says that Z (F_{m_{1}}^{k_{1}}F_{m_{2}}^{k_{2}}) = z \frac{(−1)^{s+1}F_{m_{1}}^{k_{1}}F_{m_{2}}^{k_{2}}}{(−1)^{s+1}z+(−1)^{s}}, which is essentially (15).)

For given m ∈ ℤ one has

Z (F_{sn+m}) = \frac{z (F_{m}z + (−1)^{m} F_{s−m})}{z^{2} − L_{s}z + (−1)^{s}},  (16)

and from this expression one can see that

F_{s}F_{sn+m} = F_{m}F_{s(n+1)} + (−1)^{m} F_{s−m}F_{sn}.  (17)

(See [8].)

Now we begin with a list of preliminary results that will be used in sections 3 and 4.

Proposition 2. For given k ∈ ℕ′ we have

(−1)^{s+1} ∏_{j=0}^{k} (z − α^{sj}β^{s(k−j)}) = ∑_{i=0}^{k+1} (−1)^{\frac{(si+2(s+1))(i+1)}{2}} \binom{k + 1}{i}_{F_{s}}z^{k+1−i}.  (18)

Proof. We proceed by induction on k. For k = 0 the result is clearly true (both sides are equal to (−1)^{s+1} (z − 1)). Let us suppose the formula is true for a given k. Then

(−1)^{s+1} ∏_{j=0}^{k+1} (z − α^{sj}β^{s(k+1−j)})

= (−1)^{s+1} (z − α^{s(k+1)}) β^{s(k+1)} ∏_{j=0}^{k} (\frac{z}{β^{s}} − α^{sj}β^{s(k−j)})

= (−1)^{s+1} (z − α^{s(k+1)}) β^{s(k+1)} (−1)^{s+1} ∑_{i=0}^{k+1} (−1)^{\frac{(si+2(s+1))(i+1)}{2}} \binom{k + 1}{i}_{F_{s}} (\frac{z}{β^{s}} )^{k+1−i}

= (z − α^{s(k+1)}) ∑_{i=0}^{k+1} (−1)^{\frac{(si+2(s+1))(i+1)}{2}} \binom{k + 1}{i}_{F_{s}}β^{si}z^{k+1−i}

= ∑_{i=0}^{k+1} (−1)^{\frac{(si+2(s+1))(i+1)}{2}} \binom{k + 1}{i}_{F_{s}}β^{si}z^{k+2−i} − α^{s(k+1)} ∑_{i=1}^{k+2} (−1)^{\frac{(s(i−1)+2(s+1))i}{2}} \binom{k + 1}{i − 1}_{F_{s}}β^{s(i−1)}z^{k+2−i}

= ∑_{i=0}^{k+2} (−1)^{\frac{(si+2(s+1))(i+1)}{2}} \binom{k + 2}{i}_{F_{s}} \frac{1}{F_{s(k+2)}} (β^{si}F_{s(k+2−i)} + (−1)^{−s(i+1)} α^{s(k+1)}β^{s(i−1)}F_{si}) z^{k+2−i}

= ∑_{i=0}^{k+2} (−1)^{\frac{(si+2(s+1))(i+1)}{2}} \binom{k + 2}{i}_{F_{s}}z^{k+2−i}

as wanted. Here we used that

β^{si}F_{s(k+2−i)} + (−1)^{−s(i+1)} α^{s(k+1)}β^{s(i−1)}F_{si} = F_{s(k+2)},

which can be proved easily by writing the F ’s in terms of α and β.

We will denote the (k + 1)-th degree polynomial (−1)^{s+1} ∏_{j=0}^{k} (z − α^{sj}β^{s(k−j)}) as D_{s,k+1} (z). By combining pairs of adequate linear factors, it is possible to express D_{s,k+1} (z) as product of “Lucas factors” (quadratic factors in which the coefficient of the linear term is a Lucas number, like in the denominator of (16); see also [13]). We have two cases:

(a) If k is even, k = 2p say, then

D_{s,2p+1} (z) = (−1)^{s+1} (z − (−1)^{sp}) ∏_{j=0}^{p−1} (z^{2} − (−1)^{sj} L_{2s(p−j)}z + 1) .  (19)

(b) If k is odd, k = 2p − 1 say, then

D_{s,2p} (z) = (−1)^{s+1} ∏_{j=0}^{p−1} (z^{2} − (−1)^{sj} L_{s(2p−1−2j)}z + (−1)^{s}) .  (20)

Indeed, we have

D_{s,2p+1} (z) = (−1)^{s+1} ∏_{j=0}^{2p} (z − α^{sj}β^{s(2p−j)})

= (−1)^{s+1} (z − (−1)^{sp}) ∏_{j=0}^{p−1} (z − α^{sj}β^{s(2p−j)}) ∏_{j=p+1}^{2p} (z − α^{sj}β^{s(2p−j)})

= (−1)^{s+1} (z − (−1)^{sp}) ∏_{j=0}^{p−1} (z − α^{sj}β^{s(2p−j)}) ∏_{j=0}^{p−1} (z − α^{s(2p−j)}β^{sj})

= (−1)^{s+1} (z − (−1)^{sp}) ∏_{j=0}^{p−1} (z^{2} − (−1)^{sj} L_{2s(p−j)}z + 1) ,

which proves (19). Similarly we have

D_{s,2p} (z) = (−1)^{s+1} ∏_{j=0}^{2p−1} (z − α^{sj}β^{s(2p−1−j)})

= (−1)^{s+1} ∏_{j=0}^{p−1} (z − α^{sj}β^{s(2p−1−j)}) ∏_{j=p}^{2p−1} (z − α^{sj}β^{s(2p−1−j)})

= (−1)^{s+1} ∏_{j=0}^{p−1} (z − α^{sj}β^{s(2p−1−j)}) ∏_{j=0}^{p−1} (z − α^{s(2p−1−j)}β^{sj})

= (−1)^{s+1} ∏_{j=0}^{p−1} (z^{2} − (−1)^{sj} L_{s(2p−1−2j)}z + (−1)^{s}) ,

which proves (20).

Before we establish the next proposition, we would like to see how (18) and (19) have some interesting identities (involving s-Fibonomials) hidden. First observe that we can write (18) as follows:

∏_{j=0}^{k} (z − α^{sj}β^{s(k−j)}) = ∑_{i=0}^{k+1} (−1)^{\frac{i(s(i+1)+2(s+1))}{2}} \binom{k + 1}{i}_{F_{s}}z^{k+1−i}.  (21)

Since

α^{sj}β^{s(k−j)} = (−1)^{s(k+j)} α^{s(2j−k)} = \frac{(−1)^{s(k+j)}}{2} (L_{s(2j−k)} + \sqrt{5}F_{s(2j−k)}) ,

we see that the numbers a_{s,j,k} = \frac{(−1)^{s(k+j)}}{2} (L_{s(2j−k)} + \sqrt{5}F_{s(2j−k)}), j, k ∈ ℤ, form an abelian multiplicative group a_{s,j_{1},k_{1}}a_{s,j_{2},k_{2}} = a_{s,j_{1}+j_{2},k_{1}+k_{2}}, (with identity a_{s,0,0} and inverses a_{s,j,k}^{−1} = a_{s,−j,−k}). Thus, (21) together with Viète’s formulas tell us that

(−1)^{\frac{r(s(r+1)+2(s+1))}{2} −ksr+r} 2\binom{k + 1}{r}_{F_{s}}  (22)
= ∑_{0≤j_{1}<j_{2}<···<j_{r}≤k} (−1)^{s(j_{1}+j_{2}+···+j_{r})} (L_{2s(j_{1}+j_{2}+···+j_{r})−ksr} + \sqrt{5}F_{2s(j_{1}+j_{2}+···+j_{r})−ksr}) ,

where r ∈ {1, 2, …, k + 1} is given. Moreover, since the left-hand side of (22) is an integer, we have

∑_{0≤j_{1}<j_{2}<···<j_{r}≤k} (−1)^{s(j_{1}+j_{2}+···+j_{r})} L_{2s(j_{1}+j_{2}+···+j_{r})−ksr} = (−1)^{\frac{r(s(r+1)+2(s+1))}{2} −ksr+r} 2\binom{k + 1}{r}_{F_{s}},

(23)

and

∑_{0≤j_{1}<j_{2}<···<j_{r}≤k} (−1)^{s(j_{1}+j_{2}+···+j_{r})} F_{2s(j_{1}+j_{2}+···+j_{r})−ksr} = 0.  (24)

In particular we have the following identities, corresponding to (23) and (24) with r = 1 :

∑_{j=0}^{k} (−1)^{sj} L_{s(2j−k)} = \frac{(−1)^{ks}}{F_{s}} 2F_{s(k+1)} , ∑_{j=0}^{k} (−1)^{sj} F_{s(2j−k)} = 0,  (25)

and with r = 2 :

∑_{j=1}^{k} ∑_{i=0}^{j−1} (−1)^{s(i+j)} L_{2s(i+j−k)} = \frac{2 (−1)^{s}}{F_{s}F_{2s}} F_{sk}F_{s(k+1)} , ∑_{j=1}^{k} ∑_{i=0}^{j−1} (−1)^{s(i+j)} F_{2s(i+j−k)} = 0.

(26)

On the other hand, observe that according to (18) and (19) we have

∑_{i=0}^{2p+1} (−1)^{\frac{(si+2(s+1))(i+1)}{2}} \binom{2p + 1}{i}_{F_{s}}z^{2p+1−i} = (−1)^{s+1} (z − (−1)^{sp}) ∏_{j=0}^{p−1} (z^{2} − (−1)^{sj} L_{2s(p−j)}z + 1) ,

which can be written as follows:

∑_{i=0}^{p} ((−1)^{\frac{(si+2(s+1))(i+1)}{2}} z^{2p+1−i} + (−1)^{\frac{(s(2p+1−i)+2(s+1))(2p+2−i)}{2}} z^{i}) \binom{2p + 1}{i}_{F_{s}} (27)
= (−1)^{s+1} (z − (−1)^{sp}) ∏_{j=0}^{p−1} (z^{2} − (−1)^{sj} L_{2s(p−j)}z + 1) .

If s is even, s = 2σ say (σ ∈ ℕ), and we set z = −1 we get from (27) that

∑_{i=0}^{p} ((−1)^{σi(i+1)} + (−1)^{σi(i−1)}) \binom{2p + 1}{i}_{F_{2σ}} = 2 ∏_{j=0}^{p−1} (2 + L_{4σ(p−j)}) .

That is, we have the identity

∑_{i=0}^{p} \binom{2p + 1}{i}_{F_{2σ}} = ∏_{j=1}^{p} L_{2σj}^{2}.  (28)

Similarly, if s is odd, s = 2σ − 1 say (σ ∈ ℕ), formula (27) can be written as follows:

∑_{i=0}^{p} ((−1)^{\frac{i(i+1)}{2}} z^{2p+1−i} + (−1)^{\frac{(2p+1−i)(2p+2−i)}{2}} z^{i}) \binom{2p + 1}{i}_{F_{2σ−1}}  (29)
= (z − (−1)^{p}) ∏_{j=0}^{p−1} (z^{2} − (−1)^{j} L_{2(2σ−1)(p−j)}z + 1) .

If in (29) p is replaced by 2p − 1 and we set z = 1, then we get

∑_{i=0}^{2p−1} ((−1)^{\frac{i(i+1)}{2}} + (−1)^{\frac{i(i+1)}{2}} ) \binom{2 (2p − 1) + 1}{i}_{F_{2σ−1}} = 2 ∏_{j=0}^{2p−2} (2 − (−1)^{j} L_{2(2σ−1)(2p−1−j)}) ,

from where we obtain the identity

∑_{i=0}^{2p−1} (−1)^{\frac{i(i+1)}{2}} \binom{4p − 1}{i}_{F_{2σ−1}} = (−1)^{p} ∏_{j=1}^{2p−1} L_{(2σ−1)j}^{2}.  (30)

If now in (29) p is replaced by 2p and we set z = −1, then we get

∑_{i=0}^{2p} 2 (−1)^{\frac{i(i−1)}{2} +1} \binom{4p + 1}{i}_{F_{2σ−1}} = −2 ∏_{j=0}^{2p−1} (2 + (−1)^{j} L_{2(2σ−1)(2p−j)}) ,

from where we obtain the identity

∑_{i=0}^{2p} (−1)^{\frac{i(i−1)}{2}} \binom{4p + 1}{i}_{F_{2σ−1}} = (−1)^{p} ∏_{j=1}^{2p} L_{(2σ−1)j}^{2}.  (31)

Proposition 3. Let t, k ∈ ℕ′, m ∈ ℤ be given. Then

(a)

\frac{α^{sk}}{∑_{i=0}^{t+1} (−1)^{\frac{(si+2(s+1))(i+1)}{2}} \binom{t + 1}{i}_{F_{s}}α^{si}z^{t+1−i}} + \frac{β^{sk}}{∑_{i=0}^{t+1} (−1)^{\frac{(si+2(s+1))(i+1)}{2}} \binom{t + 1}{i}_{F_{s}}β^{si}z^{t+1−i}}

= \frac{L_{sk}z + (−1)^{sk+1} L_{s(t−k+1)}}{∑_{i=0}^{t+2} (−1)^{\frac{(si+2(s+1))(i+1)}{2}} \binom{t + 2}{i}_{F_{s}}z^{t+2−i}} .  (32)

(b)

\frac{α^{m+sk}}{∑_{i=0}^{t+1} (−1)^{\frac{(si+2(s+1))(i+1)}{2}} \binom{t + 1}{i}_{F_{s}}α^{si}z^{t+1−i}} − \frac{β^{m+sk}}{∑_{i=0}^{t+1} (−1)^{\frac{(si+2(s+1))(i+1)}{2}} \binom{t + 1}{i}_{F_{s}}β^{si}z^{t+1−i}}

= \frac{\sqrt{5} (F_{sk+m}z + (−1)^{sk+m} F_{s(t−k+1)−m})}{∑_{i=0}^{t+2} (−1)^{\frac{(si+2(s+1))(i+1)}{2}} \binom{t + 2}{i}_{F_{s}}z^{t+2−i}} .  (33)

Proof. Observe that, according to proposition 2, we have that

∑_{i=0}^{t+1} (−1)^{\frac{(si+2(s+1))(i+1)}{2}} \binom{t + 1}{i}_{F_{s}}α^{si}z^{t+1−i} = α^{s(t+1)} ∑_{i=0}^{t+1} (−1)^{\frac{(si+2(s+1))(i+1)}{2}} \binom{t + 1}{i}_{F_{s}} (\frac{z}{α^{s}} )^{t+1−i}
= (−1)^{s+1} α^{s(t+1)} ∏_{j=0}^{t} (\frac{z}{α^{s}} − α^{sj}β^{s(t−j)})
= (−1)^{s+1} ∏_{j=0}^{t} (z − α^{s(j+1)}β^{s(t−j)}) ,

and similarly

∑_{i=0}^{t+1} (−1)^{\frac{(si+2(s+1))(i+1)}{2}} \binom{t + 1}{i}_{F_{s}}β^{si}z^{t+1−i} = (−1)^{s+1} β^{s(t+1)} ∏_{j=0}^{t} (\frac{z}{β^{s}} − α^{sj}β^{s(t−j)})
= (−1)^{s+1} ∏_{j=0}^{t} (z − α^{sj}β^{s(t−j+1)}) .

Then

\frac{α^{sk}}{∑_{i=0}^{t+1} (−1)^{\frac{(si+2(s+1))(i+1)}{2}} \binom{t + 1}{i}_{F_{s}}α^{si}z^{t+1−i}} + \frac{β^{sk}}{∑_{i=0}^{t+1} (−1)^{\frac{(si+2(s+1))(i+1)}{2}} \binom{t + 1}{i}_{F_{s}}β^{si}z^{t+1−i}}

= \begin{matrix}α & sk \\ (−1) & s+1 & ∏ & {t \atop j=0} & (z − α^{s(j+1)}β^{s(t−j)}) \end{matrix} + \begin{matrix}β & sk \\ (−1) & s+1 & ∏ & {t \atop j=0} & (z − α^{sj}β^{s(t−(j−1))}) \end{matrix}
= \begin{matrix}α & sk \\ (−1) & s+1 & ∏ & {t+1 \atop j=1} & (z − α^{sj}β^{s(t−(j−1))}) \end{matrix} + \begin{matrix}β & sk \\ (−1) & s+1 & ∏ & {t \atop j=0} & (z − α^{sj}β^{s(t−(j−1))}) \end{matrix}
= \frac{α^{sk} (z − β^{s(t+1)}) + β^{sk} (z − α^{s(t+1)})}{(−1)^{s+1} ∏_{j=0}^{t+1} (z − α^{sj}β^{s(t+1−j)})}
= \frac{L_{sk}z + (−1)^{sk+1} L_{s(t−k+1)}}{∑_{i=0}^{t+2} (−1)^{\frac{(si+2(s+1))(i+1)}{2}} \binom{t + 2}{i}_{F_{s}}z^{t+2−i}} ,

which proves (a). The proof of (b) is similar. Lemma 4. For given i, t ∈ ℕ′, the following identity holds

(−1)^{si} F_{s(t+2−i)}F_{s(t+1−i)} + L_{s(t+1)}F_{s(t+2−i)}F_{si} + (−1)^{s(t+i)} F_{si}F_{s(i−1)} = F_{s(t+2)}F_{s(t+1)}. (34)

Proof. Leaving the details for the reader, the proof is as follows: first, use α’s and β’s to prove that F_{s(t+1−i)} (−1)^{si} + L_{s(t+1)}F_{si} = F_{s(t+1+i)}. Then write the left-hand side of (34) as F_{s(t+2−i)}F_{s(t+1+i)}+(−1)^{s(t+i)} F_{si}F_{s(i−1)}. Finally use the standard identity F_{m}F_{n}−F_{m+k}F_{n−k} = (−1)^{n−k} F_{m+k−n}F_{k} to obtain (34). Proposition 5. Let t ∈ ℕ′ be given. Then

∑_{i=0}^{t+2} (−1)^{\frac{(si+2(s+1))(i+1)}{2}} \binom{t + 2}{i}_{F_{s}}z^{t+2−i}  (35)
= (z^{2} − L_{s(t+1)}z + (−1)^{s(t+1)}) ∑_{i=0}^{t} (−1)^{\frac{(si+2(s+1))(i+1)}{2}} \binom{t}{i}_{F_{s}} (−1)^{si} z^{t−i}.

Proof. We have

(z^{2} − L_{s(t+1)}z + (−1)^{s(t+1)}) ∑_{i=0}^{t} (−1)^{\frac{(si+2(s+1))(i+1)}{2}} \binom{t}{i}_{F_{s}} (−1)^{si} z^{t−i}

= ∑_{i=0}^{t} (−1)^{\frac{(si+2(s+1))(i+1)}{2}} \binom{t}{i}_{F_{s}} (−1)^{si} z^{t+2−i} − L_{s(t+1)} ∑_{i=1}^{t+1} (−1)^{\frac{(si−s+2(s+1))i}{2}} \binom{t}{i − 1}_{F_{s}} (−1)^{s(i−1)} z^{t+2−i}

+ (−1)^{s(t+1)} ∑_{i=2}^{t+2} (−1)^{\frac{(si−2s+2(s+1))(i−1)}{2}} \binom{t}{i − 2}_{F_{s}} (−1)^{si} z^{t+2−i}

= ∑_{i=0}^{t+2} (−1)^{\frac{(si+2(s+1))(i+1)}{2}} \binom{t + 2}{i}_{F_{s}} \frac{1}{F_{s(t+2)}F_{s(t+1)}} ( \begin{matrix}(−1)^{si} F_{s(t+2−i)}F_{s(t+1−i)} \\ +L_{s(t+1)}F_{s(t+2−i)}F_{si} \\ + (−1)^{s(t+i)} F_{si}F_{s(i−1)} \end{matrix} ) z^{t+2−i}
= ∑_{i=0}^{t+2} (−1)^{\frac{(si+2(s+1))(i+1)}{2}} \binom{t + 2}{i}_{F_{s}}z^{t+2−i},

as wanted. In the last step we used lemma 4. Proposition 6. For given i, t ∈ ℕ′ the following identity holds

∑_{j=0}^{i} (−1)^{\frac{(s(i−j)+2(s+1))(i−j+1)}{2}} \binom{t + 1}{i − j}_{F_{s}}F_{stj+m} = (−1)^{\frac{is}{2} (i−1)+i+s+m} \binom{t}{i}_{F_{s}}F_{is−m}.  (36)

Proof. We proceed by induction on t. For t = 0 we need to check that

∑_{j=0}^{i} (−1)^{\frac{(s(i−j)+2(s+1))(i−j+1)}{2}} \binom{1}{i − j}_{F_{s}}F_{m} = (−1)^{\frac{is}{2} (i−1)+i+s+m} \binom{0}{i}_{F_{s}}F_{is−m}.

If i = 0 we have a trivial equality (both sides are equal to (−1)^{s+1} F_{m}). If i ≥ 1 we have

∑_{j=0}^{i} (−1)^{\frac{(s(i−j)+2(s+1))(i−j+1)}{2}} \binom{1}{i − j}_{F_{s}}F_{m} = (−1)^{s+1} F_{m} + (−1)^{s} F_{m} = 0,

as expected. Suppose now that (36) is valid for a given t. We have (by using (2))

∑_{j=0}^{i} (−1)^{\frac{(s(i−j)+2(s+1))(i−j+1)}{2}} \binom{t + 2}{i − j}_{F_{s}}F_{s(t+1)j+m}

= ∑_{j=0}^{i} (−1)^{\frac{(s(i−j)+2(s+1))(i−j+1)}{2}} F_{s(t+1)j+m} (F_{s(t+2−i+j)+1}\binom{t + 1}{i − j − 1}_{F_{s}} + F_{s(i−j)−1}\binom{t + 1}{i − j}_{F_{s}})
= ∑_{j=1}^{i+1} (−1)^{\frac{(s(i−j+1)+2(s+1))(i−j+2)}{2}} F_{s(t+1)(j−1)+m}F_{s(t+1−i+j)+1}\binom{t + 1}{i − j}_{F_{s}}

+ ∑_{j=0}^{i} (−1)^{\frac{(s(i−j)+2(s+1))(i−j+1)}{2}} F_{s(t+1)j+m}F_{s(i−j)−1}\binom{t + 1}{i − j}_{F_{s}}

= ∑_{j=0}^{i} (−1)^{\frac{(s(i−j)+2(s+1))(i−j+1)}{2}} \binom{t + 1}{i − j}_{F_{s}} ( \begin{matrix}(−1)^{s(i−j)+1} F_{s(t+1)(j−1)+m}F_{s(t+1−i+j)+1} \\ +F_{s(t+1)j+m}F_{s(i−j)−1} \end{matrix} )

− (−1)^{\frac{(s(i+1)+2(s+1))(i+2)}{2}} F_{s(t+1)(−1)+m}F_{s(t+1−i)+1}\binom{t + 1}{i}_{F_{s}}.

Next we will use the identity

(−1)^{s(i−j)+1} F_{s(t+1)(j−1)+m}F_{s(t+1−i+j)+1} + F_{s(t+1)j+m}F_{s(i−j)−1}  (37)
= \frac{F_{s(t+1)}}{F_{st}} (F_{s(i−1)−1}F_{stj+m} + (−1)^{s(i−1)−1} F_{s(t−i+1)+1}F_{st(j−1)+m}) ,

which can be proved in two steps (we leave the details for the reader): first use α’s and β’s to prove that

(−1)^{s(i−j)+1} F_{s(t+1)(j−1)+m}F_{s(t+1−i+j)+1} + F_{s(t+1)j+m}F_{s(i−j)−1} = F_{s(t+1)}F_{st(j−1)+m+s(i−1)−1}, and then use (17) to write

F_{st(j−1)+m+s(i−1)−1} = \frac{1}{F_{st}} (F_{s(i−1)−1}F_{stj+m} + (−1)^{s(i−1)−1} F_{s(t−i+1)+1}F_{st(j−1)+m}) ,

obtaining in this way (37).

Thus, by using (37) we can write

∑_{j=0}^{i} (−1)^{\frac{(sj+2(s+1))(j+1)}{2}} \binom{t + 2}{j}_{F_{s}}F_{s(t+1)(i−j)+m}

= \frac{F_{s(t+1)}}{F_{st}} ∑_{j=0}^{i} (−1)^{\frac{(s(i−j)+2(s+1))(i−j+1)}{2}} \binom{t + 1}{i − j}_{F_{s}} ( \begin{matrix}F_{s(i−1)−1}F_{stj+m} \\ + (−1)^{s(i−1)−1} F_{s(t−i+1)+1}F_{st(j−1)+m} \end{matrix} )

− (−1)^{\frac{(s(i+1)+2(s+1))(i+2)}{2}} F_{s(t+1)(−1)+m}F_{s(t+1−i)+1}\binom{t + 1}{i},

and by using the induction hypothesis (together with some simplifications), we get

∑_{j=0}^{i} (−1)^{\frac{(sj+2(s+1))(j+1)}{2}} \binom{t + 2}{j}_{F_{s}}F_{s(t+1)(i−j)+m}

= \frac{F_{s(t+1)}}{F_{st}} F_{s(i−1)−1} ∑_{j=0}^{i} (−1)^{\frac{(s(i−j)+2(s+1))(i−j+1)}{2}} F_{stj+m}\binom{t + 1}{i − j}_{F_{s}}

+\frac{F_{s(t+1)}}{F_{st}} F_{s(t−i+1)+1} (−1)^{s(i−1)−1} ∑_{j=0}^{i−1} (−1)^{\frac{(s(i−j−1)+2(s+1))(i−j)}{2}} F_{stj+m}\binom{t + 1}{i − 1 − j}_{F_{s}}

− (−1)^{\frac{(s(i+1)+2(s+1))(i+2)}{2}} F_{s(t+1)(−1)+m}F_{s(t+1−i)+1}\binom{t + 1}{i}

+\frac{F_{s(t+1)}}{F_{st}} F_{s(t−i+1)+1} (−1)^{s(i−1)−1} (−1)^{\frac{(si+2(s+1))(i+1)}{2}} F_{−st+m}\binom{t + 1}{i}_{F_{s}}

= \frac{F_{s(t+1)}}{F_{st}} F_{s(i−1)−1} (−1)^{\frac{is}{2} (i−1)+i+s+m} \binom{t}{i}_{F_{s}}F_{is−m}

+\frac{F_{s(t+1)}}{F_{st}} F_{s(t−i+1)+1} (−1)^{s(i−1)−1} (−1)^{\frac{(i−1)s}{2} i+i−1+(2−i)s+m} \binom{t}{i − 1}_{F_{s}}F_{(i−1)s−m}

+\binom{t + 1}{i} ( \begin{matrix}\frac{F_{s(t+1)}}{F_{st}} F_{s(t−i+1)+1} (−1)^{s(i−1)−1} (−1)^{\frac{(si+2(s+1))(i+1)}{2}} F_{−st+m} \\ − (−1)^{\frac{(s(i+1)+2(s+1))(i+2)}{2}} F_{s(t+1)(−1)+m}F_{s(t+1−i)+1} \end{matrix} )

= (−1)^{\frac{is}{2} (i−1)+i+s+m} \binom{t + 1}{i}_{F_{s}} \frac{1}{F_{st}} (F_{is−m}F_{s(i−1)−1}F_{s(t−i+1)} + F_{is}F_{(i−1)s−m}F_{s(t−i+1)+1})

+\binom{t + 1}{i} (−1)^{\frac{is}{2} (i−1)+i+s+m} \frac{(−1)^{si+m}}{F_{st}} F_{s(t−i+1)+1} ( \begin{matrix}(−1)^{s} F_{s(t+1)}F_{−st+m} \\ − (−1)^{si+m} F_{st}F_{s(t+1)(−1)+m} \end{matrix} )

= (−1)^{\frac{is}{2} (i−1)+i+s+m} \binom{t + 1}{i}_{F_{s}}F_{is−m},

as wanted. In the last step we used the identity F_{is−m}F_{s(i−1)−1}F_{s(t−i+1)} + F_{is}F_{(i−1)s−m}F_{s(t−i+1)+1}

+ (−1)^{si+m} F_{s(t−i+1)+1} ((−1)^{s} F_{s(t+1)}F_{−st+m} − F_{st}F_{−s(t+1)+m})
= F_{st}F_{is−m},

which proof is an easy exercise left to the reader.

The proof of the following proposition is similar to the proof of proposition 6, with some changes in signs and, of course, some F ’s substituted by L’s. We leave it for the reader. Proposition 7. For given i, t ∈ ℕ′ the following identity holds

∑_{j=0}^{i} (−1)^{\frac{(s(i−j)+2(s+1))(i−j+1)}{2}} \binom{t + 1}{i − j}_{F_{s}}L_{stj+m} = (−1)^{(i+1)(\frac{is}{2} +1)−is+m+s} \binom{t}{i}_{F_{s}}L_{is−m}. (38)

3 The main results

We begin this section by noting that we can write explicitly the sequence F_{sn+m_{1}}^{k_{1}}F_{sn+m_{2}}^{k_{2}} (where m_{1}, m_{2} ∈ ℤ and k_{1}, k_{2} ∈ ℕ′ are given) as follows:

F_{sn+m_{1}}^{k_{1}}F_{sn+m_{2}}^{k_{2}}

= (\frac{α^{sn+m_{1}} − β^{sn+m_{1}}}{\sqrt{5}} )^{k_{1}} (\frac{α^{sn+m_{2}} − β^{sn+m_{2}}}{\sqrt{5}} )^{k_{2}}

= 5^{− \frac{k_{1}+k_{2}}{2}} ∑_{i=0}^{k_{1}} \binom{k_{1}}{i} (α^{sn+m_{1}})^{i} (−β^{sn+m_{1}})^{k_{1}−i} ∑_{j=0}^{k_{2}} \binom{k_{2}}{j} (α^{sn+m_{2}})^{j} (−β^{sn+m_{2}})^{k_{2}−j}

= 5^{− \frac{k_{1}+k_{2}}{2}} β^{m_{1}k_{1}+m_{2}k_{2}} ∑_{j=0}^{k_{1}+k_{2}} ∑_{i=0}^{k_{1}} (−1)^{k_{1}+k_{2}−j} \binom{k_{1}}{i}\binom{k_{2}}{j − i} (\frac{α}{β})^{(m_{1}−m_{2})i+m_{2}j} (α^{sj}β^{s(k_{1}+k_{2}−j)})^{n} .

Then the Z transform of F_{sn+m_{1}}^{k_{1}}F_{sn+m_{2}}^{k_{2}} is

Z (F_{sn+m_{1}}^{k_{1}}F_{sn+m_{2}}^{k_{2}})  (39)
= 5^{− \frac{k_{1}+k_{2}}{2}} β^{m_{1}k_{1}+m_{2}k_{2}} ∑_{j=0}^{k_{1}+k_{2}} ∑_{i=0}^{k_{1}} (−1)^{k_{1}+k_{2}−j} \binom{k_{1}}{i}\binom{k_{2}}{j − i} (\frac{α}{β})^{(m_{1}−m_{2})i+m_{2}j} \frac{z}{z − α^{sj}β^{s(k_{1}+k_{2}−j)}} .

Our first main result says that this expression can be written in a special form.

Theorem 8. Let m_{1}, m_{2} ∈ ℤ and k_{1}, k_{2} ∈ ℕ′ be given. The Z transform of the sequence F_{sn+m_{1}}^{k_{1}}F_{sn+m_{2}}^{k_{2}} is

Z (F_{sn+m_{1}}^{k_{1}}F_{sn+m_{2}}^{k_{2}}) = z \frac{∑_{i=0}^{k_{1}+k_{2}} ∑_{j=0}^{i} (−1)^{\frac{(si+2(s+1))(i+1)}{2}} \binom{k_{1} + k_{2} + 1}{j}_{F_{s}}F_{m_{1}+s(i−j)}^{k_{1}}F_{m_{2}+s(i−j)}^{k_{2}}z^{k_{1}+k_{2}−i}}{∑_{i=0}^{k_{1}+k_{2}+1} (−1)^{\frac{(si+2(s+1))(i+1)}{2}} \binom{k_{1} + k_{2} + 1}{i}_{F_{s}}z^{k_{1}+k_{2}+1−i}} .

(40)

Proof. We have to show that

5^{− \frac{k_{1}+k_{2}}{2}} β^{m_{1}k_{1}+m_{2}k_{2}} ∑_{j=0}^{k_{1}+k_{2}} ∑_{i=0}^{k_{1}} \binom{k_{1}}{i}\binom{k_{2}}{j − i} (−1)^{k_{1}+k_{2}−j} (\frac{α}{β})^{(m_{1}−m_{2})i+m_{2}j} \frac{z}{z − α^{sj}β^{s(k_{1}+k_{2}−j)}}

= z \frac{∑_{i=0}^{k_{1}+k_{2}} ∑_{j=0}^{i} (−1)^{\frac{(sj+2(s+1))(j+1)}{2}} \binom{k_{1} + k_{2} + 1}{j}_{F_{s}}F_{m_{1}+s(i−j)}^{k_{1}}F_{m_{2}+s(i−j)}^{k_{2}}z^{k_{1}+k_{2}−i}}{∑_{i=0}^{k_{1}+k_{2}+1} (−1)^{\frac{(si+2(s+1))(i+1)}{2}} \binom{k_{1} + k_{2} + 1}{i}_{F_{s}}z^{k_{1}+k_{2}+1−i}} .  (41)

We will proceed by induction on k_{1} and/or k_{2}. Observe that the case k_{1} = k_{2} = 0 is (15). Let us consider the case k_{1} = k_{2} = 1 and then we construct the induction argument on k_{1} (note that (40) is symmetric with respect to k_{1} and k_{2}). For the case k_{1} = k_{2} = 1 we use that L_{2s} + (−1)^{s} = \frac{F_{3s}}{F_{s}} to write

5^{− \frac{1+1}{2}} β^{m_{1}+m_{2}} ∑_{i=0}^{1} ∑_{j=0}^{1} \binom{1}{i}\binom{1}{j} (−1)^{i+j} (\frac{α}{β})^{m_{1}i+m_{2}j} \frac{z}{z − α^{s(i+j)}β^{s(2−i−j)}}

= 5^{−1}z (\frac{(α^{m_{1}+m_{2}} + β^{m_{1}+m_{2}}) z − β^{m_{1}+m_{2}}α^{2s} − α^{m_{1}+m_{2}}β^{2s}}{z^{2} − L_{2s}z + 1} − \frac{α^{m_{1}}β^{m_{2}} + α^{m_{2}}β^{m_{1}}}{z − (−1)^{s}} )

= 5^{−1} \frac{z}{z^{3} − (L_{2s} + (−1)^{s}) z^{2} + (1 + (−1)^{s} L_{2s}) z − (−1)^{s}} ×

× (\begin{matrix}(α^{m_{1}+m_{2}} + β^{m_{1}+m_{2}} − α^{m_{1}}β^{m_{2}} − α^{m_{2}}β^{m_{1}}) z^{2} \\ − (β^{m_{1}+m_{2}}α^{2s}+α^{m_{1}+m_{2}}β^{2s}−(α^{m_{1}}β^{m_{2}} +α^{m_{2}}β^{m_{1}}) (α^{2s}+β^{2s})+(−1)^{s} (α^{m_{1}+m_{2}} +β^{m_{1}+m_{2}})) z \\ + (−1)^{s} β^{m_{1}+m_{2}}α^{2s} + (−1)^{s} α^{m_{1}+m_{2}}β^{2s} − α^{m_{1}}β^{m_{2}} − α^{m_{2}}β^{m_{1}} \end{matrix})

= \frac{z}{(−1)^{s+1} z^{3} + (−1)^{s} \frac{F_{3s}}{F_{s}} z^{2} − \frac{F_{3s}}{F_{s}} z + 1} ×

× ( \begin{matrix}(−1)^{s+1} F_{m_{1}}F_{m_{2}}z^{2} + (−1)^{s+1} (F_{m_{1}+s}F_{m_{2}+s} − \frac{F_{3s}}{F_{s}} F_{m_{1}}F_{m_{2}}) z \\ + (−1)^{s+1} F_{m_{1}+2s}F_{m_{2}+2s} + (−1)^{s} \frac{F_{3s}}{F_{s}} F_{m_{1}+s}F_{m_{2}+s} − \frac{F_{3s}}{F_{s}} F_{m_{1}}F_{m_{2}} \end{matrix} )

= z \frac{∑_{i=0}^{2} ∑_{j=0}^{i} (−1)^{\frac{(sj+2(s+1))(j+1)}{2}} \binom{3}{j}_{F_{s}}F_{m_{1}+s(i−j)}F_{m_{2}+s(i−j)}z^{2−i}}{∑_{i=0}^{3} (−1)^{\frac{(si+2(s+1))(i+1)}{2}} \binom{3}{i}_{F_{s}}z^{3−i}} ,

which is (41) with k_{1} = k_{2} = 1. Suppose now that (41) is true for a given k_{1}. We will show that it is also true for k_{1} + 1. We have

Z (F_{sn+m_{1}}^{k_{1}+1}F_{sn+m_{2}}^{k_{2}}) = 5^{− \frac{k_{1}+k_{2}+1}{2}} β^{m_{1}(k_{1}+1)+m_{2}k_{2}} ∑_{j=0}^{k_{1}+k_{2}+1} ∑_{i=0}^{k_{1}+1} (−1)^{k_{1}+k_{2}+1−j} \binom{k_{1} + 1}{i}\binom{k_{2}}{j − i} ×

× (\frac{α}{β})^{(m_{1}−m_{2})i+m_{2}j} \frac{z}{z − α^{sj}β^{s(k_{1}+k_{2}+1−j)}} .

If we use that \binom{k_{1}+1}{i} = \binom{k_{1}}{i} + \binom{k_{1}}{i−1}, separate in the corresponding two terms, and shift the indices of the second term, we get

Z (F_{sn+m_{1}}^{k_{1}+1}F_{sn+m_{2}}^{k_{2}})  (42)
= −5^{− \frac{k_{1}+k_{2}+1}{2}} β^{m_{1}k_{1}+m_{2}k_{2}}β^{m_{1}} ∑_{j=0}^{k_{1}+k_{2}}∑_{i=0}^{k_{1}}(−1)^{k_{1}+1+k_{2}−j}\binom{k_{1}}{i}\binom{k_{2}}{j − i}(\frac{α}{β})^{(m_{1}−m_{2})i+m_{2}j} \frac{\frac{z}{β^{s}}}{\frac{z}{β^{s}} − α^{sj}β^{s(k_{1}+k_{2}−j)}}

+5^{− \frac{k_{1}+k_{2}+1}{2}} β^{m_{1}k_{1}+m_{2}k_{2}}α^{m_{1}} ∑_{j=0}^{k_{1}+k_{2}}∑_{i=0}^{k_{1}}(−1)^{k_{1}+k_{2}−j}\binom{k_{1}}{i}\binom{k_{2}}{j − i}(\frac{α}{β})^{(m_{1}−m_{2})i+m_{2}j} \frac{\frac{z}{α^{s}}}{\frac{z}{α^{s}} − α^{sj}β^{s(k_{1}+k_{2}−j)}} .

By using the induction hypothesis and proposition 3 (b), we can write (42) as

Z(F_{sn+m_{1}}^{k_{1}+1}F_{sn+m_{2}}^{k_{2}})
= \frac{α^{m_{1}}}{5^{\frac{1}{2}}} \frac{\frac{z}{α^{s}} ∑_{i=0}^{k_{1}+k_{2}} ∑_{j=0}^{i} (−1)^{\frac{(sj+2(s+1))(j+1)}{2}} \binom{k_{1} + k_{2} + 1}{j}_{F_{s}}F_{m_{1}+s(i−j)}^{k_{1}}F_{m_{2}+s(i−j)}^{k_{2}} (\frac{z}{α^{s}} )^{k_{1}+k_{2}−i}}{∑_{i=0}^{k_{1}+k_{2}+1} (−1)^{\frac{(si+2(s+1))(i+1)}{2}} \binom{k_{1} + k_{2} + 1}{i}_{F_{s}} (\frac{z}{α^{s}} )^{k_{1}+k_{2}+1−i}}

−\frac{β^{m_{1}}}{5^{\frac{1}{2}}} \frac{\frac{z}{β^{s}} ∑_{i=0}^{k_{1}+k_{2}} ∑_{j=0}^{i} (−1)^{\frac{(sj+2(s+1))(j+1)}{2}} \binom{k_{1} + k_{2} + 1}{j}_{F_{s}}F_{m_{1}+s(i−j)}^{k_{1}}F_{m_{2}+s(i−j)}^{k_{2}} (\frac{z}{β^{s}} )^{k_{1}+k_{2}−i}}{∑_{i=0}^{k_{1}+k_{2}+1} (−1)^{\frac{(si+2(s+1))(i+1)}{2}} \binom{k_{1} + k_{2} + 1}{i}_{F_{s}} (\frac{z}{β^{s}} )^{k_{1}+k_{2}+1−i}}

= \frac{z}{5^{\frac{1}{2}}} ∑_{i=0}^{k_{1}+k_{2}} ∑_{j=0}^{i} (−1)^{\frac{(sj+2(s+1))(j+1)}{2}} \binom{k_{1} + k_{2} + 1}{j}_{F_{s}}F_{m_{1}+s(i−j)}^{k_{1}}F_{m_{2}+s(i−j)}^{k_{2}} ×

× ( \frac{α^{m_{1}+si}}{∑_{i=0}^{k_{1}+k_{2}+1} (−1)^{\frac{(si+2(s+1))(i+1)}{2}} \binom{k_{1} + k_{2} + 1}{i}_{F_{s}}α^{si}z^{k_{1}+k_{2}+1−i}} ; − \frac{β^{m_{1}+si}}{∑_{i=0}^{k_{1}+k_{2}+1} (−1)^{\frac{(si+2(s+1))(i+1)}{2}} \binom{k_{1} + k_{2} + 1}{i}_{F_{s}}β^{si}z^{k_{1}+k_{2}+1−i}} ) z^{k_{1}+k_{2}−i}

= \frac{z ∑_{i=0}^{k_{1}+k_{2}} ∑_{j=0}^{i} (−1)^{\frac{(sj+2(s+1))(j+1)}{2}} \binom{k_{1} + k_{2} + 1}{j}_{F_{s}}F_{m_{1}+s(i−j)}^{k_{1}}F_{m_{2}+s(i−j)}^{k_{2}}}{∑_{i=0}^{k_{1}+k_{2}+2} (−1)^{\frac{(si+2(s+1))(i+1)}{2}} \binom{k_{1} + k_{2} + 2}{i}_{F_{s}}z^{k_{1}+k_{2}+2−i}} ×

× (F_{m_{1}+si}z + (−1)^{si+m_{1}} F_{s(k_{1}+k_{2}−i+1)−m_{1}}) z^{k_{1}+k_{2}−i}

= \frac{z}{∑_{i=0}^{k_{1}+k_{2}+2} (−1)^{\frac{(si+2(s+1))(i+1)}{2}} \binom{k_{1} + k_{2} + 2}{i}_{F_{s}}z^{k_{1}+k_{2}+2−i}} ×

× ( \begin{matrix}∑_{i=0}^{k_{1}+k_{2}} ∑_{j=0}^{i} (−1)^{\frac{(sj+2(s+1))(j+1)}{2}} \binom{k_{1} + k_{2} + 1}{j}_{F_{s}}× \\ ×F_{m_{1}+s(i−j)}^{k_{1}}F_{m_{2}+s(i−j)}^{k_{2}}F_{m_{1}+si}z^{k_{1}+k_{2}+1−i} \\ + \\ ∑_{i=0}^{k_{1}+k_{2}} ∑_{j=0}^{i} (−1)^{\frac{(sj+2(s+1))(j+1)}{2}} \binom{k_{1} + k_{2} + 1}{j}_{F_{s}}× \\ ×F_{m_{1}+s(i−j)}^{k_{1}}F_{m_{2}+s(i−j)}^{k_{2}} (−1)^{si+m_{1}} F_{s(k_{1}+k_{2}−i+1)−m_{1}}z^{k_{1}+k_{2}−i} \end{matrix} )

Some further simplifications give us

Z(F_{sn+m_{1}}^{k_{1}+1}F_{sn+m_{2}}^{k_{2}})
= \frac{z}{∑_{i=0}^{k_{1}+k_{2}+2} (−1)^{\frac{(si+2(s+1))(i+1)}{2}} \binom{k_{1} + k_{2} + 2}{i}_{F_{s}}z^{k_{1}+k_{2}+2−i}} ×

× ( \begin{matrix}∑_{i=0}^{k_{1}+k_{2}} ∑_{j=0}^{i} (−1)^{\frac{(sj+2(s+1))(j+1)}{2}} \binom{k_{1} + k_{2} + 1}{j}_{F_{s}}× \\ ×F_{m_{1}+s(i−j)}^{k_{1}}F_{m_{2}+s(i−j)}^{k_{2}}F_{m_{1}+si}z^{k_{1}+k_{2}+1−i} \\ + \\ ∑_{i=1}^{k_{1}+k_{2}+1} ∑_{j=1}^{i} (−1)^{\frac{(sj−s+2(s+1))j}{2}} \binom{k_{1} + k_{2} + 1}{j − 1}_{F_{s}}× \\ ×F_{m_{1}+s(i−j)}^{k_{1}}F_{m_{2}+s(i−j)}^{k_{2}} (−1)^{si−s+m_{1}} F_{s(k_{1}+k_{2}−i+2)−m_{1}}z^{k_{1}+k_{2}+1−i} \end{matrix} )

= \frac{z}{∑_{i=0}^{k_{1}+k_{2}+2} (−1)^{\frac{(si+2(s+1))(i+1)}{2}} \binom{k_{1} + k_{2} + 2}{i}_{F_{s}}z^{k_{1}+k_{2}+2−i}} ×

× ∑_{i=0}^{k_{1}+k_{2}+1} ∑_{j=0}^{i} ( \begin{matrix}(−1)^{\frac{(sj+2(s+1))(j+1)}{2}} F_{m_{1}+s(i−j)}^{k_{1}}F_{m_{2}+s(i−j)}^{k_{2}}\binom{k_{1} + k_{2} + 2}{j}_{F_{s}} \frac{1}{F_{s(k_{1}+k_{2}+2)}} × \\ × (F_{s(k_{1}+k_{2}+2−j)}F_{m_{1}+si} + (−1)^{m_{1}+1+s(i−j)} F_{sj}F_{s(k_{1}+k_{2}−i+2)−m_{1}}) \end{matrix} ) z^{k_{1}+k_{2}−i+1}

= z \frac{∑_{i=0}^{k_{1}+k_{2}+1} ∑_{j=0}^{i} (−1)^{\frac{(sj+2(s+1))(j+1)}{2}} F_{m_{1}+s(i−j)}^{k_{1}+1}F_{m_{2}+s(i−j)}^{k_{2}}\binom{k_{1} + k_{2} + 2}{j}_{F_{s}}z^{k_{1}+k_{2}−i+1}}{∑_{i=0}^{k_{1}+k_{2}+2} (−1)^{\frac{(si+2(s+1))(i+1)}{2}} \binom{k_{1} + k_{2} + 2}{i}_{F_{s}}z^{k_{1}+k_{2}+2−i}} ,

as wanted. We used that

∑_{j=0}^{k_{1}+k_{2}+1} (−1)^{\frac{(sj+2(s+1))(j+1)}{2}} \binom{k_{1} + k_{2} + 1}{j}_{F_{s}}F_{m_{1}+s(k_{1}+k_{2}+1−j)}^{k_{1}}F_{m_{2}+s(k_{1}+k_{2}+1−j)}^{k_{2}} = 0,

(which comes from the induction hypothesis). We also used in the last step that

F_{s(k_{1}+k_{2}+2−j)}F_{m_{1}+si} + (−1)^{m_{1}+1+s(i−j)} F_{sj}F_{s(k_{1}+k_{2}−i+2)−m_{1}} = F_{s(k_{1}+k_{2}+2)}F_{m_{1}+s(i−j)}, which is a direct consequence of the known identity F_{m}F_{n}−F_{m+k}F_{n−k} = (−1)^{n−k} F_{m+k−n}F_{k}.

Our second important result of this section is the following theorem. Theorem 9. Let m_{1}, m_{2} ∈ ℤ and t_{1}, t_{2} ∈ ℕ′ be given. The Z transform of the sequence F_{t_{1}sn+m_{1}}^{k_{1}}F_{t_{2}sn+m_{2}}^{k_{2}} is given by

Z (F_{t_{1}sn+m_{1}}F_{t_{2}sn+m_{2}})  (43)
= z \frac{∑_{i=0}^{t_{1}+t_{2}} ∑_{j=0}^{i} (−1)^{\frac{(sj+2(s+1))(j+1)}{2}} \binom{t_{1} + t_{2} + 1}{j}_{F_{s}}F_{m_{1}+t_{1}s(i−j)}F_{m_{2}+t_{2}s(i−j)}z^{t_{1}+t_{2}−i}}{∑_{i=0}^{t_{1}+t_{2}+1} (−1)^{\frac{(si+2(s+1))(i+1)}{2}} \binom{t_{1} + t_{2} + 1}{i}_{F_{s}}z^{t_{1}+t_{2}+1−i}} .

Proof. We will proceed by induction on the parameters t_{1} and/or t_{2}. Observe that case t_{1} = t_{2} = 0 is trivial (it is essentially (15)) and in the case t_{1} = t_{2} = 1 the result is true by theorem 8. Suppose now the result is true for a given t_{1} together with all its values ≤ t_{1}, and let us prove it is also true for t_{1} + 1. (As in the proof of 8, we have a symmetry property that allows us to proceed in this way.) We will use that

F_{(t_{1}+1)sn+m_{1}} = F_{t_{1}sn+m_{1}}L_{sn} − (−1)^{sn} F_{(t_{1}−1)sn+m_{1}},

(easy to prove). Then we have that

Z (F_{(t_{1}+1)sn+m_{1}}F_{t_{2}sn+m_{2}}) = Z ((F_{t_{1}sn+m_{1}}L_{sn} − (−1)^{sn} F_{(t_{1}−1)sn+m_{1}}) F_{t_{2}sn+m_{2}})  (44)
= Z (L_{sn}F_{t_{1}sn+m_{1}}F_{t_{2}sn+m_{2}}) − Z ((−1)^{sn} F_{(t_{1}−1)sn+m_{1}}F_{t_{2}sn+m_{2}}) .

Now we use (10) and (13), together with the induction hypothesis to write

Z (F_{(t_{1}+1)sn+m_{1}}F_{t_{2}sn+m_{2}})

= \frac{z}{α^{s}} \frac{∑_{i=0}^{t_{1}+t_{2}} ∑_{j=0}^{i} (−1)^{\frac{(sj+2(s+1))(j+1)}{2}} \binom{t_{1} + t_{2} + 1}{j}_{F_{s}}F_{m_{1}+t_{1}s(i−j)}F_{m_{2}+t_{2}s(i−j)} (\frac{z}{α^{s}} )^{t_{1}+t_{2}−i}}{∑_{i=0}^{t_{1}+t_{2}+1} (−1)^{\frac{(si+2(s+1))(i+1)}{2}} \binom{t_{1} + t_{2} + 1}{i}_{F_{s}} (\frac{z}{α^{s}} )^{t_{1}+t_{2}+1−i}}

+\frac{z}{β^{s}} \frac{∑_{i=0}^{t_{1}+t_{2}} ∑_{j=0}^{i} (−1)^{\frac{(sj+2(s+1))(j+1)}{2}} \binom{t_{1} + t_{2} + 1}{j}_{F_{s}}F_{m_{1}+t_{1}s(i−j)}F_{m_{2}+t_{2}s(i−j)} (\frac{z}{β^{s}} )^{t_{1}+t_{2}−i}}{∑_{i=0}^{t_{1}+t_{2}+1} (−1)^{\frac{(si+2(s+1))(i+1)}{2}} \binom{t_{1} + t_{2} + 1}{i}_{F_{s}} (\frac{z}{β^{s}} )^{t_{1}+t_{2}+1−i}}

−\frac{z}{(−1)^{s}} \frac{∑_{i=0}^{t_{1}+t_{2}−1} ∑_{j=0}^{i} (−1)^{\frac{(sj+2(s+1))(j+1)}{2}} \binom{t_{1} + t_{2}}{j}_{F_{s}}F_{m_{1}+(t_{1}−1)s(i−j)}F_{m_{2}+t_{2}s(i−j)} (\frac{z}{(−1)^{s}} )^{t_{1}−1+t_{2}−i}}{∑_{i=0}^{t_{1}+t_{2}} (−1)^{\frac{(si+2(s+1))(i+1)}{2}} \binom{t_{1} + t_{2}}{i}_{F_{s}} (\frac{z}{(−1)^{s}} )^{t_{1}+t_{2}−i}}

= z ∑_{i=0}^{t_{1}+t_{2}} ∑_{j=0}^{i} (−1)^{\frac{(sj+2(s+1))(j+1)}{2}} \binom{t_{1} + t_{2} + 1}{j}_{F_{s}}F_{m_{1}+t_{1}s(i−j)}F_{m_{2}+t_{2}s(i−j)} ×

× ( \frac{α^{si}}{∑_{i=0}^{t_{1}+t_{2}+1} (−1)^{\frac{(si+2(s+1))(i+1)}{2}} \binom{t_{1} + t_{2} + 1}{i}_{F_{s}}α^{si}z^{t_{1}+t_{2}+1−i}} ; + \frac{β^{si}}{∑_{i=0}^{t_{1}+t_{2}+1} (−1)^{\frac{(si+2(s+1))(i+1)}{2}} \binom{t_{1} + t_{2} + 1}{i}_{F_{s}}β^{si}z^{t_{1}+t_{2}+1−i}} ) z^{t_{1}+t_{2}−i}

−z \frac{∑_{i=0}^{t_{1}+t_{2}−1} ∑_{j=0}^{i} (−1)^{\frac{(sj+2(s+1))(j+1)}{2}} \binom{t_{1} + t_{2}}{j}_{F_{s}}F_{m_{1}+(t_{1}−1)s(i−j)}F_{m_{2}+t_{2}s(i−j)} (−1)^{si} z^{t_{1}−1+t_{2}−i}}{∑_{i=0}^{t_{1}+t_{2}} (−1)^{\frac{(si+2(s+1))(i+1)}{2}} \binom{t_{1} + t_{2}}{i}_{F_{s}} (−1)^{si} z^{t_{1}+t_{2}−i}} .

Next we use proposition 3 (a) to obtain that

Z (F_{(t_{1}+1)sn+m_{1}}F_{t_{2}sn+m_{2}})

= \frac{z}{∑_{i=0}^{t_{1}+t_{2}+2} (−1)^{\frac{(si+2(s+1))(i+1)}{2}} \binom{t_{1} + t_{2} + 2}{i}_{F_{s}}z^{t_{1}+t_{2}+2−i}} ×

× ∑_{i=0}^{t_{1}+t_{2}} ∑_{j=0}^{i} (−1)^{\frac{(sj+2(s+1))(j+1)}{2}} \binom{t_{1} + t_{2} + 1}{j}_{F_{s}}F_{m_{1}+t_{1}s(i−j)}F_{m_{2}+t_{2}s(i−j)} ×

× (L_{si}z + (−1)^{si+1} L_{s(t_{1}+t_{2}−i+1)}) z^{t_{1}+t_{2}−i}

−z \frac{∑_{i=0}^{t_{1}+t_{2}−1} ∑_{j=0}^{i} (−1)^{\frac{(sj+2(s+1))(j+1)}{2}} \binom{t_{1} + t_{2}}{j}_{F_{s}}F_{m_{1}+(t_{1}−1)s(i−j)}F_{m_{2}+t_{2}s(i−j)} (−1)^{si} z^{t_{1}−1+t_{2}−i}}{∑_{i=0}^{t_{1}+t_{2}} (−1)^{\frac{(si+2(s+1))(i+1)}{2}} \binom{t_{1} + t_{2}}{i}_{F_{s}} (−1)^{si} z^{t_{1}+t_{2}−i}} ,

and then, by proposition 5 we have

Z (F_{(t_{1}+1)sn+m_{1}}F_{t_{2}sn+m_{2}})  (45)
= \frac{z}{∑_{i=0}^{t_{1}+t_{2}+2} (−1)^{\frac{(si+2(s+1))(i+1)}{2}} \binom{t_{1} + t_{2} + 2}{i}_{F_{s}}z^{t_{1}+t_{2}+2−i}} ×

× ( \begin{matrix}∑_{i=0}^{t_{1}+t_{2}} ∑_{j=0}^{i} (−1)^{\frac{(sj+2(s+1))(j+1)}{2}} \binom{t_{1} + t_{2} + 1}{j}_{F_{s}}F_{m_{1}+t_{1}s(i−j)}F_{m_{2}+t_{2}s(i−j)}× \\ × (L_{si}z + (−1)^{si+1} L_{s(t_{1}+t_{2}−i+1)}) z^{t_{1}+t_{2}−i} \\ − ∑_{i=0}^{t_{1}+t_{2}−1} ∑_{j=0}^{i} (−1)^{\frac{(sj+2(s+1))(j+1)}{2}} \binom{t_{1} + t_{2}}{j}_{F_{s}}F_{m_{1}+(t_{1}−1)s(i−j)}F_{m_{2}+t_{2}s(i−j)}× \\ × (−1)^{si} z^{t_{1}−1+t_{2}−i} (z^{2} − L_{s(t_{1}+t_{2}+1)}z + (−1)^{s(t_{1}+t_{2}+1)}) \end{matrix} ) .

We have now the expected denominator (of (43) with t_{1} replaced by t_{1} + 1). Let us work with the numerator of (45), z (A − B) say, where

A = ∑_{i=0}^{t_{1}+t_{2}} ∑_{j=0}^{i} (−1)^{\frac{(sj+2(s+1))(j+1)}{2}} \binom{t_{1} + t_{2} + 1}{j}_{F_{s}}F_{m_{1}+t_{1}s(i−j)}F_{m_{2}+t_{2}s(i−j)} ×
× (L_{si}z + (−1)^{si+1} L_{s(t_{1}+t_{2}−i+1)}) z^{t_{1}+t_{2}−i},

and

B = ∑_{i=0}^{t_{1}+t_{2}−1} ∑_{j=0}^{i} (−1)^{\frac{(sj+2(s+1))(j+1)}{2}} \binom{t_{1} + t_{2}}{j}_{F_{s}}F_{m_{1}+(t_{1}−1)s(i−j)}F_{m_{2}+t_{2}s(i−j)} ×
× (−1)^{si} z^{t_{1}−1+t_{2}−i} (z^{2} − L_{s(t_{1}+t_{2}+1)}z + (−1)^{s(t_{1}+t_{2}+1)}) .

We have that

A = ∑_{i=0}^{t_{1}+t_{2}} ∑_{j=0}^{i} (−1)^{\frac{(sj+2(s+1))(j+1)}{2}} \binom{t_{1} + t_{2} + 1}{j}_{F_{s}}F_{m_{1}+t_{1}s(i−j)}F_{m_{2}+t_{2}s(i−j)}L_{si}z^{t_{1}+t_{2}+1−i}

+ ∑_{i=1}^{t_{1}+t_{2}+1} ∑_{j=1}^{i} (−1)^{\frac{(sj−s+2(s+1))j}{2}} \binom{t_{1} + t_{2} + 1}{j − 1}_{F_{s}} ×

×F_{m_{1}+t_{1}s(i−j)}F_{m_{2}+t_{2}s(i−j)} (−1)^{s(i−1)+1} L_{s(t_{1}+t_{2}−i+2)}z^{t_{1}+t_{2}+1−i}

= ∑_{i=0}^{t_{1}+t_{2}+1} ∑_{j=0}^{i} (−1)^{\frac{(sj+2(s+1))(j+1)}{2}} \binom{t_{1} + t_{2} + 2}{j}_{F_{s}} \frac{F_{m_{2}+t_{2}s(i−j)}F_{m_{1}+t_{1}s(i−j)}}{F_{s(t_{1}+t_{2}+2)}} ×

× (F_{s(t_{1}+t_{2}+2−j)}L_{si} + (−1)^{s(i−j)} F_{sj}L_{s(t_{1}+t_{2}−i+2)}) z^{t_{1}+1+t_{2}−i}. But we have the identity

F_{s(t_{1}+t_{2}+2−j)}L_{si} + (−1)^{s(i−j)} F_{sj}L_{s(t_{1}+t_{2}−i+2)} = L_{s(i−j)}F_{s(t_{1}+t_{2}+2)},

thus

A = ∑_{i=0}^{t_{1}+t_{2}+1} ∑_{j=0}^{i} (−1)^{\frac{(sj+2(s+1))(j+1)}{2}} \binom{t_{1} + t_{2} + 2}{j}_{F_{s}}F_{m_{2}+t_{2}s(i−j)}F_{m_{1}+t_{1}s(i−j)}L_{s(i−j)}z^{t_{1}+1+t_{2}−i}.

Now let us work with B :

B = ∑_{i=0}^{t_{1}+t_{2}−1} ∑_{j=0}^{i} (−1)^{\frac{(sj+2(s+1))(j+1)}{2}} \binom{t_{1} + t_{2}}{j}_{F_{s}}F_{m_{1}+(t_{1}−1)s(i−j)}F_{m_{2}+t_{2}s(i−j)} (−1)^{si} z^{t_{1}+1+t_{2}−i}

− ∑_{i=1}^{t_{1}+t_{2}} ∑_{j=1}^{i} (−1)^{\frac{(sj−s+2(s+1))j}{2}} \binom{t_{1} + t_{2}}{j − 1}_{F_{s}} ( \begin{matrix}F_{m_{1}+(t_{1}−1)s(i−j)}F_{m_{2}+t_{2}s(i−j)}× \\ ×L_{s(t_{1}+t_{2}+1)} (−1)^{s(i−1)} z^{t_{1}+t_{2}+1−i} \end{matrix} )

+ ∑_{i=2}^{t_{1}+t_{2}+1} ∑_{j=2}^{i} (−1)^{\frac{(sj−2s+2(s+1))(j−1)}{2}} \binom{t_{1} + t_{2}}{j − 2}_{F_{s}} ( \begin{matrix}F_{m_{1}+(t_{1}−1)s(i−j)}F_{m_{2}+t_{2}s(i−j)}× \\ × (−1)^{si} z^{t_{1}+1+t_{2}−i} (−1)^{s(t_{1}+t_{2}+1)} \end{matrix} )

= ∑_{i=0}^{t_{1}+t_{2}+1} ∑_{j=0}^{i} (−1)^{\frac{(sj+2(s+1))(j+1)}{2}} \binom{t_{1} + t_{2} + 2}{j}_{F_{s}} \frac{F_{m_{2}+t_{2}s(i−j)}F_{m_{1}+(t_{1}−1)s(i−j)}}{F_{s(t_{1}+t_{2}+2)}F_{s(t_{1}+t_{2}+1)}} ×

× (−1)^{s(i−j)} ( \begin{matrix}(−1)^{sj} F_{s(t_{1}+t_{2}+1−j)}F_{s(t_{1}+t_{2}+2−j)} \\ +F_{sj}F_{s(t_{1}+t_{2}+2−j)}L_{s(t_{1}+t_{2}+1)} \\ + (−1)^{s(t_{1}+t_{2}+j)} F_{sj}F_{s(j−1)} \end{matrix} ) z^{t_{1}+1+t_{2}−i}

= ∑_{i=0}^{t_{1}+t_{2}+1} ∑_{j=0}^{i} (−1)^{\frac{(sj+2(s+1))(j+1)}{2}} \binom{t_{1} + t_{2} + 2}{j}_{F_{s}} (−1)^{s(i−j)} F_{m_{2}+t_{2}s(i−j)}F_{m_{1}+(t_{1}−1)s(i−j)}z^{t_{1}+1+t_{2}−i}.

(We used lemma 4 in the last step.)

Thus, the numerator of (45) is

z (A − B) = z ∑_{i=0}^{t_{1}+t_{2}+1} ∑_{j=0}^{i} (−1)^{\frac{(sj+2(s+1))(j+1)}{2}} \binom{t_{1} + t_{2} + 2}{j}_{F_{s}}F_{m_{2}+t_{2}s(i−j)} ×  (46)
× (F_{m_{1}+t_{1}s(i−j)}L_{s(i−j)} − (−1)^{s(i−j)} F_{m_{1}+(t_{1}−1)s(i−j)}) z^{t_{1}+1+t_{2}−i}.

Finally, with a simple calculation with α’s and β’s (left to the reader), we see that

F_{m_{1}+t_{1}s(i−j)}L_{s(i−j)} − (−1)^{s(i−j)} F_{m_{1}+(t_{1}−1)s(i−j)} = F_{m_{1}+(t_{1}+1)s(i−j)},

so (46) is

z (A − B) = z ∑_{i=0}^{t_{1}+t_{2}+1} ∑_{j=0}^{i} (−1)^{\frac{(sj+2(s+1))(j+1)}{2}} \binom{t_{1} + t_{2} + 2}{j}_{F_{s}}F_{m_{1}+(t_{1}+1)s(i−j)}F_{m_{2}+t_{2}s(i−j)}z^{t_{1}+1+t_{2}−i},

which is the expected numerator (of (43) with t_{1} replaced by t_{1} +1). This ends our induction argument.

The natural generalization of (40) (theorem 8) is clearly as follows:

Z (F_{sn+m_{1}}^{k_{1}} ⋯ F_{sn+m_{l}}^{k_{l}})  (47)

= z \frac{∑_{i=0}^{k_{1}+···+k_{l}} ∑_{j=0}^{i} (−1)^{\frac{(sj+2(s+1))(j+1)}{2}} \binom{k_{1} + ⋯ + k_{l} + 1}{j}_{F_{s}}F_{m_{1}+s(i−j)}^{k_{1}} ⋯ F_{m_{l}+s(i−j)}^{k_{l}}z^{k_{1}+···+k_{l}−i}}{∑_{i=0}^{k_{1}+···+k_{l}+1} (−1)^{\frac{(si+2(s+1))(i+1)}{2}} \binom{k_{1} + ⋯ + k_{l} + 1}{i}_{F_{s}}z^{k_{1}+···+k_{l}+1−i}} ,

(which can be proved with the same sort of arguments used in the proof of (40)), and the natural generalization of (43) (theorem 9) is

Z (F_{t_{1}sn+m_{1}} ⋯ F_{t_{l}sn+m_{l}})  (48)

= z \frac{∑_{i=0}^{t_{1}+···+t_{l}} ∑_{j=0}^{i} (−1)^{\frac{(sj+2(s+1))(j+1)}{2}} \binom{t_{1} + ⋯ + t_{l} + 1}{j}_{F_{s}}F_{m_{1}+t_{1}s(i−j)} ⋯ F_{m_{l}+t_{l}s(i−j)}z^{t_{1}+···+t_{l}−i}}{∑_{i=0}^{t_{1}+···+t_{l}+1} (−1)^{\frac{(si+2(s+1))(i+1)}{2}} \binom{t_{1} + ⋯ + t_{l} + 1}{i}_{F_{s}}z^{t_{1}+···+t_{l}+1−i}} .

But then, from (48) we see that

Z (F_{t_{1}sn+m_{1}}^{k_{1}} ⋯ F_{t_{l}sn+m_{l}}^{k_{l}})  (49)

= z \frac{∑_{i=0}^{k_{1}t_{1}+···+k_{l}t_{l}}∑_{j=0}^{i} (−1)^{\frac{(sj+2(s+1))(j+1)}{2}} \binom{k_{1}t_{1} + ⋯ + k_{l}t_{l} + 1}{j}_{F_{s}}F_{m_{1}+t_{1}s(i−j)}^{k_{1}}⋯F_{m_{l}+t_{l}s(i−j)}^{k_{l}}z^{k_{1}t_{1}+···+k_{l}t_{l}−i}}{∑_{i=0}^{k_{1}t_{1}+···+k_{l}t_{l}+1} (−1)^{\frac{(si+2(s+1))(i+1)}{2}} \binom{k_{1}t_{1} + ⋯ + k_{l}t_{l} + 1}{i}_{F_{s}}z^{k_{1}t_{1}+···+k_{l}t_{l}+1−i}} .

That is, the generalization (47) is indeed a particular case of (49). This is the result we wanted to prove in this section.

4 Some corollaries

Our starting point in this section is formula (49). We want to present here some consequences of it.

Corollary 10. For p ∈ ℕ′ given, the Z transform of the sequence \binom{n}{p}_{F_{s}} is

Z (\binom{n}{p}_{F_{s}}) = \frac{(−1)^{s+1} z}{D_{s,p+1} (z)}.  (50)

Proof. We use (49) to write

Z (\binom{n}{p}_{F_{s}}) = \frac{1}{(F_{p}!)_{s}} Z (F_{s(n−p+1)}F_{s(n−p+2)} ⋯ F_{sn})

= \frac{1}{(F_{p}!)_{s}} z \frac{∑_{i=0}^{p} ∑_{j=0}^{i} (−1)^{\frac{(sj+2(s+1))(j+1)}{2}} \binom{p + 1}{j}_{F_{s}}F_{s(1−p+i−j)}F_{s(2−p+i−j)} ⋯ F_{s(i−j)}z^{p−i}}{∑_{i=0}^{p+1} (−1)^{\frac{(si+2(s+1))(i+1)}{2}} \binom{p + 1}{i}_{F_{s}}z^{p+1−i}} .

But the product F_{s(1−p+i−j)}F_{s(2−p+i−j)} ⋯ F_{s(i−j)} is different from zero if and only if i = p and j = 0. In such a case that product is (F_{p}!)_{s} and the numerator reduces to (−1)^{s+1} (F_{p}!)_{s}. Thus (50) follows.

(Formula (50) corresponding to s = 1 was proved by a different method in [13].)

Observe that according to the advance-shifting property (9) and (50), if 0 ≤ p_{0} ≤ p, then

Z (\binom{n + p_{0}}{p}_{F_{s}}) = z^{p_{0}} \frac{(−1)^{s+1} z}{∑_{i=0}^{p+1} (−1)^{\frac{(si+2(s+1))(i+1)}{2}} \binom{p + 1}{i}_{F_{s}}z^{p+1−i}} .  (51)

By using that G_{sn+m} = G_{m}F_{sn−1} + G_{m+1}F_{sn}, we can see that formula (49) is valid for Gibonacci sequences G_{n} replacing the Fibonacci ones. In fact (it suffices to check the case G_{st_{1}n+m_{1}}^{k_{1}}G_{st_{2}n+m_{2}}^{k_{2}}), we have

Z (G_{st_{1}n+m_{1}}^{k_{1}}G_{st_{2}n+m_{2}}^{k_{2}})

= ∑_{l_{1}=0}^{k_{1}} ∑_{l_{2}=0}^{k_{2}} \binom{k_{1}}{l_{1}}\binom{k_{2}}{l_{2}}G_{m_{1}}^{l_{1}}G_{m_{1}+1}^{k_{1}−l_{1}}G_{m_{2}}^{l_{2}}G_{m_{2}+1}^{k_{2}−l_{2}}Z (F_{st_{1}n−1}^{l_{1}}F_{st_{1}n}^{k_{1}−l_{1}}F_{st_{2}n−1}^{l_{2}}F_{st_{2}n}^{k_{2}−l_{2}})
= ∑_{l_{1}=0}^{k_{1}} ∑_{l_{2}=0}^{k_{2}} \binom{k_{1}}{l_{1}}\binom{k_{2}}{l_{2}}G_{m_{1}}^{l_{1}}G_{m_{1}+1}^{k_{1}−l_{1}}G_{m_{2}}^{l_{2}}G_{m_{2}+1}^{k_{2}−l_{2}}z ×

×\frac{∑_{i=0}^{t_{1}k_{1}+t_{2}k_{2}}∑_{j=0}^{i}(−1)^{\frac{(sj+2(s+1))(j+1)}{2}}\binom{t_{1}k_{1}+t_{2}k_{2}+1}{j}_{F_{s}}F_{st_{1}(i−j)−1}^{l_{1}}F_{st_{1}(i−j)}^{k_{1}−l_{1}}F_{st_{2}(i−j)−1}^{l_{2}}F_{st_{2}(i−j)}^{k_{2}−l_{2}}z^{t_{1}k_{1}+t_{2}k_{2}−i}}{∑_{i=0}^{t_{1}k_{1}+t_{2}k_{2}+1} (−1)^{\frac{(si+2(s+1))(i+1)}{2}} \binom{t_{1}k_{1} + t_{2}k_{2} + 1}{i}_{F_{s}}z^{t_{1}k_{1}+t_{2}k_{2}+1−i}}

= \frac{z}{∑_{i=0}^{t_{1}k_{1}+t_{2}k_{2}+1} (−1)^{\frac{(si+2(s+1))(i+1)}{2}} \binom{t_{1}k_{1} + t_{2}k_{2} + 1}{i}_{F_{s}}z^{t_{1}k_{1}+t_{2}k_{2}+1−i}} ×

× ( \begin{matrix}∑_{i=0}^{t_{1}k_{1}+t_{2}k_{2}} ∑_{j=0}^{i} (−1)^{\frac{(sj+2(s+1))(j+1)}{2}} \binom{t_{1}k_{1} + t_{2}k_{2} + 1}{j}_{F_{s}}× \\ ×∑_{l_{1}=0}^{k_{1}} ∑_{l_{2}=0}^{k_{2}}\binom{k_{1}}{l_{1}}\binom{k_{2}}{l_{2}}G_{m_{1}}^{l_{1}}G_{m_{1}+1}^{k_{1}−l_{1}}G_{m_{2}}^{l_{2}}G_{m_{2}+1}^{k_{2}−l_{2}}F_{st_{1}(i−j)−1}^{l_{1}}F_{st_{1}(i−j)}^{k_{1}−l_{1}}F_{st_{2}(i−j)−1}^{l_{2}}F_{st_{2}(i−j)}^{k_{2}−l_{2}}z^{t_{1}k_{1}+t_{2}k_{2}−i} \end{matrix} )

= z \frac{∑_{i=0}^{t_{1}k_{1}+t_{2}k_{2}} ∑_{j=0}^{i} (−1)^{\frac{(sj+2(s+1))(j+1)}{2}} \binom{t_{1}k_{1} + t_{2}k_{2} + 1}{j}_{F_{s}}G_{st_{1}(i−j)+m_{1}}^{k_{1}}G_{st_{2}(i−j)+m_{2}}^{k_{2}}z^{t_{1}k_{1}+t_{2}k_{2}−i}}{∑_{i=0}^{t_{1}k_{1}+t_{2}k_{2}+1} (−1)^{\frac{(si+2(s+1))(i+1)}{2}} \binom{t_{1}k_{1} + t_{2}k_{2} + 1}{j}_{F_{s}}z^{t_{1}k_{1}+t_{2}k_{2}+1−i}} ,

as expected. In general we have

Z (G_{st_{1}n+m_{1}}^{k_{1}} ⋯ G_{st_{l}n+m_{l}}^{k_{l}})  (52)
= z \frac{∑_{i=0}^{t_{1}k_{1}+···+t_{l}k_{l}}∑_{j=0}^{i}(−1)^{\frac{(sj+2(s+1))(j+1)}{2}}\binom{t_{1}k_{1} + ⋯ + t_{l}k_{l} + 1}{j}_{F_{s}}G_{m_{1}+st_{1}(i−j)}^{k_{1}} ⋯ G_{m_{l}+st_{l}(i−j)}^{k_{l}}z^{t_{1}k_{1}+···+t_{l}k_{l}−i}}{∑_{i=0}^{t_{1}k_{1}+···+t_{l}k_{l}+1} (−1)^{\frac{(si+2(s+1))(i+1)}{2}} \binom{t_{1}k_{1} + ⋯ + t_{l}k_{l} + 1}{i}_{F_{s}}z^{t_{1}k_{1}+···+t_{l}k_{l}+1−i}} .

Corollary 11. Let m_{1}, …, m_{l} ∈ ℤ and t_{1}, …, t_{l}, k_{1}, …, k_{l} ∈ ℕ′ be given. The sequence ∏_{i=1}^{l} G_{t_{i}sn+m_{i}}^{k_{i}} can be expressed as a linear combination of s-Fibonomials \binom{n+t_{1}k_{1}+···+t_{l}k_{l}−i}{t_{1}k_{1}+···+t_{l}k_{l}}_{F_{s}}, i = 0, 1, …, t_{1}k_{1} + ⋯ + t_{l}k_{l}, according to

G_{t_{1}sn+m_{1}}^{k_{1}} ⋯ G_{t_{l}sn+m_{l}}^{k_{l}} = (−1)^{s+1} ∑_{i=0}^{t_{1}k_{1}+···+t_{l}k_{l}} ∑_{j=0}^{i} (−1)^{\frac{(sj+2(s+1))(j+1)}{2}} \binom{t_{1}k_{1} + ⋯ + t_{l}k_{l} + 1}{j}_{F_{s}} ×
×G_{m_{1}+t_{1}s(i−j)}^{k_{1}} ⋯ G_{m_{l}+t_{l}s(i−j)}^{k_{l}}\binom{n + t_{1}k_{1} + ⋯ + t_{l}k_{l} − i}{t_{1}k_{1} + ⋯ + t_{l}k_{l}}_{F_{s}}.  (53)

Proof. This follows directly from (52) and (51).

Some examples of (53) are the following (after simplifications on the coefficients of the s-Fibonomials sequences of the right-hand side):

F_{2sn+m} = F_{m}\binom{n + 2}{2}_{F_{s}} + (−1)^{m} F_{s−m}L_{s}\binom{n + 1}{2}_{F_{s}} + (−1)^{s+m+1} F_{2s−m}\binom{n}{2}_{F_{s}}. (54)

F_{sn}^{3} = F_{s}^{3} (\binom{n + 2}{3}_{F_{s}} + 2 (−1)^{s} L_{s}\binom{n + 1}{3}_{F_{s}} + (−1)^{s} \binom{n}{3}_{F_{s}}) .  (55)

F_{2sn}^{2} = F_{2s}^{2} (\binom{n + 3}{4}_{F_{s}} + \binom{n}{4}_{F_{s}} + (−1)^{s+1} \frac{L_{3s}}{L_{s}} (\binom{n + 2}{4}_{F_{s}} + \binom{n + 1}{4}_{F_{s}})) . (56)

L_{sn}^{2} = 4\binom{n + 2}{2}_{F_{s}} − (3L_{s}^{2} + 4 (−1)^{s+1}) \binom{n + 1}{2}_{F_{s}} + (−1)^{s} L_{s}^{2}\binom{n}{2}_{F_{s}}.  (57)

F_{sn}^{4} = F_{s}^{4} (\binom{n + 3}{4}_{F_{s}} + \binom{n}{4}_{F_{s}} + (3 (−1)^{s} \frac{F_{3s}}{F_{s}} + 2) (\binom{n + 2}{4}_{F_{s}} + \binom{n + 1}{4}_{F_{s}})) . (58)

L_{2sn}F_{sn} = F_{s} (L_{2s}\binom{n + 2}{3}_{F_{s}} − 2L_{s}\binom{n + 1}{3}_{F_{s}} + (−1)^{s} L_{2s}\binom{n}{3}_{F_{s}}) .  (59)

L_{2sn}F_{sn}^{2} = F_{s}^{2} (L_{2s} (\binom{n + 3}{4}_{F_{s}} + \binom{n}{4}_{F_{s}}) + (−1)^{s} (5F_{s}F_{3s} − 2) (\binom{n + 2}{4}_{F_{s}} + \binom{n + 1}{4}_{F_{s}})) .

(60)

From (53) we see that

G_{tsn+m} = (−1)^{s+1} ∑_{i=0}^{t} ∑_{j=0}^{i} (−1)^{\frac{(sj+2(s+1))(j+1)}{2}} \binom{t + 1}{j}_{F_{s}}G_{ts(i−j)+m}\binom{n + t − i}{t}_{F_{s}}.  (61)

But when G = F or G = L, this formula can be written in a simpler form.

Corollary 12. Let t ∈ ℕ′ be given. The following identities hold (a)

F_{tsn+m} = ∑_{i=0}^{t} (−1)^{\frac{is(i−1)}{2} +i+m+1} \binom{t}{i}_{F_{s}}F_{is−m}\binom{n + t − i}{t}_{F_{s}}.  (62)

(b)

L_{tsn+m} = ∑_{i=0}^{t} (−1)^{\frac{is(i−1)}{2} +i+m} \binom{t}{i}_{F_{s}}L_{is−m}\binom{n + t − i}{t}_{F_{s}}.  (63)

Proof. We have that

F_{tsn+m} = (−1)^{s+1} ∑_{i=0}^{t} ∑_{j=0}^{i} (−1)^{\frac{(sj+2(s+1))(j+1)}{2}} \binom{t + 1}{j}_{F_{s}}F_{ts(i−j)+m}\binom{n + t − i}{t}_{F_{s}}

= (−1)^{s+1} ∑_{i=0}^{t} ∑_{j=0}^{i} (−1)^{\frac{(s(i−j)+2(s+1))(i−j+1)}{2}} \binom{t + 1}{i − j}_{F_{s}}F_{tsj+m}\binom{n + t − i}{t}_{F_{s}}.

According to proposition 6 we can write

F_{tsn+m} = (−1)^{s+1} ∑_{i=0}^{t} (−1)^{\frac{is}{2} (i−1)+i+s+m} \binom{t}{i}_{F_{s}}F_{is−m}\binom{n + t − i}{t}_{F_{s}}

= ∑_{i=0}^{t} (−1)^{\frac{is}{2} (i−1)+i+m+1} \binom{t}{i}_{F_{s}}F_{is−m}\binom{n + t − i}{t}_{F_{s}},

which proves (62).

The proof of (63) is similar (using proposition 7).

In the following corollary we consider sequences involving “s-Gibonomials” \binom{n}{p}_{G_{s}} =

\frac{G_{sn}···G_{s(n−p+1)}}{G_{s}···G_{sp}}.

Corollary 13. Let t_{1}, …, t_{l} ∈ ℕ and r_{1} …, r_{l}, p_{1} …, p_{l} ∈ ℕ′ be given. Then the Z transform of the sequence \binom{n}{p_{1}}_{G_{st_{1}}}^{r_{1}} ⋯ \binom{n}{p_{k}}_{G_{st_{k}}}^{r_{k}} is given by

Z (\binom{n}{p_{1}}_{G_{st_{1}}}^{r_{1}} ⋯ \binom{n}{p_{k}}_{G_{st_{k}}}^{r_{k}} )  (64)
= \frac{z}{∑_{i=0}^{t_{1}r_{1}p_{1}+···+t_{k}r_{k}p_{k}+1} (−1)^{\frac{(si+2(s+1))(i+1)}{2}} \binom{t_{1}r_{1}p_{1} + ⋯ + t_{k}r_{k}p_{k} + 1}{i}_{F_{s}}z^{t_{1}r_{1}p_{1}+···+t_{k}r_{k}p_{k}+1−i}} ×

× ∑_{i=0}^{t_{1}r_{1}p_{1}+···+t_{k}r_{k}p_{k}} ∑_{j=0}^{i} (−1)^{\frac{(sj+2(s+1))(j+1)}{2}} \binom{t_{1}r_{1}p_{1} + ⋯ + t_{k}r_{k}p_{k} + 1}{j}_{F_{s}} ×

×\binom{i − j}{p_{1}}_{G_{st_{1}}}^{r_{1}} ⋯ \binom{i − j}{p_{k}}_{G_{st_{k}}}^{r_{k}} z^{t_{1}r_{1}p_{1}+···+t_{k}r_{k}p_{k}−i}.

Proof. First we write

Z (\binom{n}{p_{1}}_{G_{st_{1}}}^{r_{1}} ⋯ \binom{n}{p_{k}}_{G_{st_{k}}}^{r_{k}} )

= \frac{1}{G_{st_{1}}^{r_{1}} ⋯ G_{st_{1}p_{1}}^{r_{1}} ⋯ G_{st_{k}}^{r_{k}} ⋯ G_{st_{k}p_{k}}^{r_{k}}} Z (G_{st_{1}n}^{r_{1}} ⋯ G_{st_{1}(n−p_{1}+1)}^{r_{1}} ⋯ G_{st_{k}n}^{r_{k}} ⋯ G_{st_{k}(n−p_{k}+1)}^{r_{k}}) ,

and then we use (52) to get

Z (\binom{n}{p_{1}}_{G_{st_{1}}}^{r_{1}} ⋯ \binom{n}{p_{k}}_{G_{st_{k}}}^{r_{k}} )

= \frac{1}{G_{st_{1}}^{r_{1}} ⋯ G_{st_{1}p_{1}}^{r_{1}} ⋯ G_{st_{k}}^{r_{k}} ⋯ G_{st_{k}p_{k}}^{r_{k}}} ×

×\frac{z}{∑_{i=0}^{t_{1}r_{1}p_{1}+···+t_{k}r_{k}p_{k}+1} (−1)^{\frac{(si+2(s+1))(i+1)}{2}} \binom{t_{1}r_{1}p_{1} + ⋯ + t_{k}r_{k}p_{k} + 1}{i}_{F_{s}}z^{t_{1}r_{1}p_{1}+···+t_{k}r_{k}p_{k}+1−i}} ×

× ∑_{i=0}^{t_{1}r_{1}p_{1}+···+t_{k}r_{k}p_{k}} ∑_{j=0}^{i} (−1)^{\frac{(sj+2(s+1))(j+1)}{2}} \binom{t_{1}r_{1}p_{1} + ⋯ + t_{k}r_{k}p_{k} + 1}{j}_{F_{s}} ×

×G_{st_{1}(i−j)}^{r_{1}} ⋯ G_{st_{1}(i−j−p_{1}+1)}^{r_{1}} ⋯ G_{st_{k}(i−j)}^{r_{k}} ⋯ G_{st_{k}(i−j−p_{k}+1)}^{r_{k}}z^{t_{1}r_{1}p_{1}+···+t_{k}r_{k}p_{k}−i}, which implies the desired formula (64). Corollary 14. Let t_{1}, …, t_{l} ∈ ℕ and r_{1} …, r_{l}, p_{1} …, p_{l} ∈ ℕ′ be given. The sequence

∏_{i=1}^{k} \binom{n}{p_{i}}_{G_{st_{i}}}^{r_{i}} can be expressed as a linear combination of s-Fibonomials \binom{n+t_{1}r_{1}p_{1}+···+t_{k}r_{k}p_{k}−i}{t_{1}r_{1}p_{1}+···+t_{k}r_{k}p_{k}}_{F_{s}},

i = 0, 1, …, t_{1}r_{1}p_{1} + ⋯ + t_{k}r_{k}p_{k}, according to

∏_{i=1}^{k} \binom{n}{p_{i}}_{G_{st_{i}}}^{r_{i}} = (−1)^{s+1} ∑_{i=0}^{t_{1}r_{1}p_{1}+···+t_{k}r_{k}p_{k}} ∑_{j=0}^{i} (−1)^{\frac{(sj+2(s+1))(j+1)}{2}} \binom{t_{1}r_{1}p_{1} + ⋯ + t_{k}r_{k}p_{k} + 1}{j}_{F_{s}} ×
×\binom{i − j}{p_{1}}_{G_{st_{1}}}^{r_{1}} ⋯ \binom{i − j}{p_{k}}_{G_{st_{k}}}^{r_{k}} \binom{n + t_{1}r_{1}p_{1} + ⋯ + t_{k}r_{k}p_{k} − i}{t_{1}r_{1}p_{1} + ⋯ + t_{k}r_{k}p_{k}}_{F_{s}}.  (65)

Proof. This comes directly from (64) and (51).

Some examples of (65) are the following (after simplifications on the coefficients of the s-Fibonomials sequences of the right-hand side):

\binom{n}{2}_{F_{2s}} = \binom{n + 2}{4}_{F_{s}} + (−1)^{s+1} L_{2s}\binom{n + 1}{4}_{F_{s}} + \binom{n}{4}_{F_{s}}.  (66)
\binom{n}{2}_{F_{s}} \binom{n}{3}_{F_{s}} = \frac{F_{3s}}{F_{s}} \binom{n + 2}{5}_{F_{s}} + \frac{L_{s}F_{3s}}{F_{s}} \binom{n + 1}{5}_{F_{s}} + \binom{n}{5}_{F_{s}}.  (67)
\binom{n}{2}_{F_{s}}^{2} = \binom{n + 2}{4}_{F_{s}} + (−1)^{s} L_{s}^{2}\binom{n + 1}{4}_{F_{s}} + \binom{n}{4}_{F_{s}}.  (68)
\binom{n}{2}_{L_{s}} = \frac{2 (−1)^{s}}{L_{2s}} \binom{n + 2}{2}_{F_{s}} + 2 (−1)^{s+1} \binom{n + 1}{2}_{F_{s}} + \binom{n}{2}_{F_{s}}.  (69)
\binom{n}{2}_{L_{s}} \binom{n}{2}_{F_{s}} = \binom{n + 2}{4}_{F_{s}} + (−1)^{s+1} L_{2s}\binom{n + 1}{4}_{F_{s}} + \binom{n}{4}_{F_{s}}.  (70)

\binom{n}{2}_{L_{s}} \binom{n}{3}_{F_{s}} = \frac{L_{3s}}{L_{s}} \binom{n + 2}{5}_{F_{s}} − L_{3s}\binom{n + 1}{5}_{F_{s}} + \binom{n}{5}_{F_{s}}.  (71)

\binom{n}{2}_{L_{s}} \binom{n}{2}_{F_{s}}^{2} = \binom{n + 4}{6}_{F_{s}} + L_{2s}\binom{n + 3}{6}_{F_{s}} + (−1)^{s} L_{2s} (L_{2s}^{2} − 1) \binom{n + 2}{6}_{F_{s}}
+L_{2s}\binom{n + 1}{6}_{F_{s}} + \binom{n}{6}_{F_{s}}.  (72)

Corollary 15. (a) Let m_{1}, …, m_{l} ∈ ℤ and t_{1}, …, t_{l}, k_{1} …, k_{l} ∈ ℕ′ be given. For n ≥ k_{1} + ⋯ + k_{l} + 1 we have that

∑_{j=0}^{t_{1}k_{1}+···+t_{l}k_{l}+1} (−1)^{\frac{(sj+2(s+1))(j+1)}{2}} \binom{t_{1}k_{1} + ⋯ + t_{l}k_{l} + 1}{j}_{F_{s}}G_{m_{1}+st_{1}(n−j)}^{k_{1}} ⋯ G_{m_{l}+t_{l}s(n−j)}^{k_{l}} = 0.

(73)

(b) Let t_{1}, …, t_{l} ∈ ℕ and r_{1} …, r_{l}, p_{1} …, p_{l} ∈ ℕ′ be given. For n ≥ t_{1}r_{1}p_{1}+⋯+t_{k}r_{k}p_{k}+1 we have that

∑_{j=0}^{t_{1}r_{1}p_{1}+···+t_{k}r_{k}p_{k}+1} (−1)^{\frac{(sj+2(s+1))(j+1)}{2}} \binom{t_{1}r_{1}p_{1} + ⋯ + t_{k}r_{k}p_{k} + 1}{j}_{F_{s}} \binom{n − j}{p_{1}}_{G_{st_{1}}}^{r_{1}} ⋯ \binom{n − j}{p_{k}}_{G_{st_{k}}}^{r_{k}} = 0.

(74)

Proof. These results are direct consequences of (the numerators in) formulas (52) and (64).

Corollary 16. Let p ∈ ℕ′ be given. The following identities hold (a)

\binom{n + 1}{p + 2}_{F_{s}} = \frac{1}{F_{s(p+2)}} F_{s(p+2)n} ∗ (−1)^{s(n+p)} \binom{n}{p}_{F_{s}}.  (75)

(b)

\binom{n + 2}{p + 4}_{F_{s}} = \frac{1}{F_{s(p+4)}F_{s(p+2)}} F_{s(p+4)n} ∗ (−1)^{s(n+1)} F_{s(p+2)n} ∗ \binom{n}{p}_{F_{s}}.  (76)

Proof. (a) First observe that

D_{s,p+3} (z)
= ∏_{j=0}^{p+2} (z − α^{sj}β^{s(p+2−j)}) = ∏_{j=−1}^{p+1} (z − α^{s(j+1)}β^{s(p+1−j)}) = ∏_{j=−1}^{p+1} (z − (−1)^{s} α^{sj}β^{s(p−j)})

= (z − (−1)^{s} α^{−s}β^{s(p+1)}) (z − (−1)^{s} α^{s(p+1)}β^{−s}) ∏_{j=0}^{p} (−1)^{s} ((−1)^{s} z − α^{sj}β^{s(p−j)})
= (−1)^{s(p+1)} (z^{2} − L_{s(p+2)}z + (−1)^{sp}) D_{s,p+1} ((−1)^{s} z) .

Then

Z (\binom{n + 1}{p + 2}_{F_{s}}) = \frac{(−1)^{s+1} z^{2}}{D_{s,p+3} (z)}
= \frac{(−1)^{s+1} z^{2}}{(−1)^{s(p+1)} (z^{2} − L_{s(p+2)}z + (−1)^{sp}) D_{s,p+1} ((−1)^{s} z)}

= (−1)^{s(p+1)} (−1)^{s} \frac{1}{F_{s(p+2)}} \frac{F_{s(p+2)}z}{z^{2} − L_{s(p+2)}z + (−1)^{sp}} \frac{(−1)^{s+1} (−1)^{s} z}{D_{s,p+1} ((−1)^{s} z)},

from where (according to (16) and the convolution theorem)

\binom{n + 1}{p + 2}_{F_{s}} = (−1)^{sp} \frac{1}{F_{s(p+2)}} F_{s(p+2)n} ∗ (−1)^{sn} \binom{n}{p}_{F_{s}}
= \frac{1}{F_{s(p+2)}} F_{s(p+2)n} ∗ (−1)^{s(n+p)} \binom{n}{p}_{F_{s}},

as wanted.

(b) Let us consider the polynomial D_{s,p+5} (z) and observe that

D_{s,p+5} (z)

= ∏_{j=0}^{p+4} (z − α^{sj}β^{s(p+4−j)}) = ∏_{j=−2}^{p+2} (z − α^{s(j+2)}β^{s(p+2−j)}) = ∏_{j=−2}^{p+2} (z − α^{sj}β^{s(p−j)})

= (z − α^{−2s}β^{s(p+2)}) (z − α^{s(p+2)}β^{−2s}) (z − α^{−s}β^{s(p+1)}) (z − α^{s(p+1)}β^{−s}) ∏_{j=0}^{p} (z − α^{sj}β^{s(p−j)})
= (z^{2} − L_{s(p+4)}z + (−1)^{sp}) (z^{2} − (−1)^{s} L_{s(p+2)}z + (−1)^{sp}) D_{s,p+1} (z) .

Then

Z (\binom{n + 2}{p + 4}_{F_{s}})

= \frac{(−1)^{s+1} z^{3}}{D_{s,p+5} (z)}
= \frac{(−1)^{s+1} z^{3}}{(z^{2} − L_{s(p+4)}z + (−1)^{sp}) (z^{2} − (−1)^{s} L_{s(p+2)}z + (−1)^{sp}) D_{s,p+1} (z)}
= \frac{1}{F_{s(p+4)}} \frac{F_{s(p+4)}z}{z^{2} − L_{s(p+4)}z + (−1)^{sp}} \frac{1}{F_{s(p+2)}} (−1)^{s} \frac{F_{s(p+2)} (−1)^{s} z}{(z^{2} − (−1)^{s} L_{s(p+2)}z + (−1)^{sp})} \frac{(−1)^{s+1} z}{D_{s,p+1} (z)},

from where (according to (16) and the convolution theorem)

\binom{n + 2}{p + 4}_{F_{s}} = \frac{1}{F_{s(p+4)}} F_{s(p+4)n} ∗ \frac{1}{F_{s(p+2)}} (−1)^{s} (−1)^{ns} F_{s(p+2)n} ∗ \binom{n}{p}_{F_{s}}
= \frac{1}{F_{s(p+4)}F_{s(p+2)}} F_{s(p+4)n} ∗ (−1)^{s(n+1)} F_{s(p+2)n} ∗ \binom{n}{p}_{F_{s}},

as wanted.

Some examples are

\binom{n + 1}{4}_{F_{s}} = \frac{1}{F_{4s}} ∑_{i=0}^{n} (−1)^{si} F_{4s(n−i)}\binom{i}{2}_{F_{s}}.  (77)

\binom{n + 2}{6}_{F_{s}} = \frac{1}{F_{6s}F_{4s}} ∑_{i=0}^{n} ∑_{j=0}^{i} (−1)^{s(j+1)} F_{4sj}F_{6s(i−j)}\binom{n − i}{2}_{F_{s}}.  (78)

In the following corollary we will deal with k given sequences (a_{n})_{0} , (a_{n})_{1} , ⋯, (a_{n})_{k} and the convolution of them (a_{n})_{0} ∗ (a_{n})_{1} ∗ ⋯ ∗ (a_{n})_{k}, which we will denote as ∗_{j=0}^{k} (a_{n})_{j}.

Corollary 17. Let p ∈ ℕ be given. The following identities hold (a)

\binom{n + p}{2p}_{F_{s}} = (−1)^{spn} ∗_{j=0}^{p−1} \frac{(−1)^{sj(n+1)}}{F_{2s(p−j)}} F_{2s(p−j)n}.  (79)

(b)

\binom{n + p − 1}{2p − 1}_{F_{s}} = ∗_{j=0}^{p−1} \frac{(−1)^{sj(n+1)}}{F_{s(2p−1−2j)}} F_{s(2p−1−2j)n}.  (80)

Proof. (a) According to (50) and (19) we have that

z^{p}Z (\binom{n}{2p}_{F_{s}}) = \frac{z^{p+1}}{(z − (−1)^{sp}) ∏_{j=0}^{p−1} (z^{2} − (−1)^{sj} L_{2s(p−j)}z + 1)},

or (by using (51))

Z (\binom{n + p}{2p}_{F_{s}}) = \frac{z}{z − (−1)^{sp}} ∏_{j=0}^{p−1} \frac{z}{z^{2} − (−1)^{sj} L_{2s(p−j)}z + 1},

from where (79) follows (by using (16) and the convolution theorem).

(b) According to (50) and (20) we have that

z^{p−1}Z (\binom{n}{2p − 1}_{F_{s}}) = \frac{z^{p}}{∏_{j=0}^{p−1} (z^{2} − (−1)^{sj} L_{s(2p−1−2j)}z + (−1)^{s})}

= ∏_{j=0}^{p−1} \frac{z}{z^{2} − (−1)^{sj} L_{s(2p−1−2j)}z + (−1)^{s}} ,

from where (80) follows (with (51), (16) and the convolution theorem).

\binom{n + 1}{2}_{F_{s}} = \frac{1}{F_{2s}} ∑_{j=0}^{n} (−1)^{s(n−j)} F_{2sj}.  (81)

\binom{n + 2}{4}_{F_{s}} = \frac{1}{F_{4s}F_{2s}} ∑_{i=0}^{n} ∑_{j=0}^{i} (−1)^{s(j+1)} F_{2sj}F_{4s(i−j)}.  (82)

In the following corollary we present decompositions of some sequences of s-Fibonomials as linear combinations of certain Fibonacci sequences.

Corollary 18. We have the following decompositions of the s-Fibonomials:

\binom{n + p}{2p}_{F_{s}}  (83)

= ∑_{j=0}^{p−1} \frac{1}{F_{2s(p−j)} ∏_{i=0,i≠j}^{p−1} ((−1)^{sj} L_{2s(p−j)} − (−1)^{si} L_{2s(p−i)})} ∑_{t=0}^{n} (−1)^{s(p(n−t)+j(t+1))} F_{2s(p−j)t}. (p ≥ 1)

\binom{n + p − 1}{2p − 1}_{F_{s}}  (84)

= ∑_{j=0}^{p−1} \frac{(−1)^{js(n+1)}}{F_{s(2p−1−2j)} ∏_{i=0,i≠j}^{p−1} ((−1)^{sj} L_{s(2p−1−2j)} − (−1)^{si} L_{s(2p−1−2i)})}F_{s(2p−1−2j)n}. (p ≥ 1)

\binom{n + p − 1}{2p}_{F_{s}} = ∑_{t=0}^{n} ∑_{j=0}^{p−1} \begin{matrix}(−1) & s(p(n−t)+jt)+1 \\ ∏ & {p−1 \atop i=0,i≠j} & ((−1)^{sj} L_{2s(p−j)} − (−1)^{si} L_{2s(p−i)}) \end{matrix} ×  (85)

× (F_{2s(p−j)t+1} + \frac{F_{2s(p−j)−1} − L_{2s(p−j)}}{F_{2s(p−j)}} F_{2s(p−j)t}) . (p ≥ 2)

\binom{n + p − 2}{2p − 1}_{F_{s}} = ∑_{j=0}^{p−1} \begin{matrix}(−1) & s(jn+1)+1 \\ ∏ & {p−1 \atop i=0,i≠j} & ((−1)^{sj} L_{s(2p−1−2j)} − (−1)^{si} L_{s(2p−1−2i)}) \end{matrix} ×  (86)

× (F_{s(2p−1−2j)n+1} + \frac{F_{s(2p−1−2j)−1} − L_{s(2p−1−2j)}}{F_{s(2p−1−2j)}} F_{s(2p−1−2j)n}) . (p ≥ 2)

\binom{n + p − 2}{2p}_{F_{s}} = ∑_{t=0}^{n} ∑_{j=0}^{p−1} \frac{(−1)^{s(p(n−t)+j(t+1))+1} L_{2s(p−j)}}{∏_{i=0,i≠j}^{p−1} ((−1)^{sj} L_{2s(p−j)} − (−1)^{si} L_{2s(p−i)})} ×  (87)

× (F_{2s(p−j)t+1} + \frac{F_{2s(p−j)−1} − \frac{L_{4s(p−j)}+1}{L_{2s(p−j)}}}{F_{2s(p−j)}} F_{2s(p−j)t}) . (p ≥ 3)

\binom{n + p − 3}{2p − 1}_{F_{s}} = ∑_{j=0}^{p−1} \frac{(−1)^{sj(n+1)+1} L_{s(2p−1−2j)}}{∏_{i=0,i≠j}^{p−1} ((−1)^{sj} L_{s(2p−1−2j)} − (−1)^{si} L_{s(2p−1−2i)})} ×  (88)

× (F_{s(2p−1−2j)n+1} + \frac{F_{s(2p−1−2j)−1} − \frac{L_{2s(2p−1−2j)}+(−1)^{s}}{L_{s(2p−1−2j)}}}{F_{s(2p−1−2j)}} F_{s(2p−1−2j)n}) . (p ≥ 3)

\binom{n + p − 3}{2p}_{F_{s}} = ∑_{t=0}^{n} ∑_{j=0}^{p−1} \begin{matrix}(−1) & s(p(n−t)+jt)+1 & (L_{4s(p−j)} + 1) \\ ∏ & {p−1 \atop i=0,i≠j} & ((−1)^{sj} L_{2s(p−j)} − (−1)^{si} L_{2s(p−i)}) \end{matrix} ×  (89)

× (F_{2s(p−j)t+1} + \frac{F_{2s(p−j)−1} − \frac{L_{6s(p−j)}+L_{2s(p−j)}}{L_{4s(p−j)}+1}}{F_{2s(p−j)}} F_{2s(p−j)t}) . (p ≥ 4)

\binom{n + p − 4}{2p − 1}_{F_{s}}  (90)

= ∑_{j=0}^{p−1} \begin{matrix}(−1) & s(jn+1)+1 & (L_{2s(2p−1−2j)} + (−1)^{s}) \\ ∏ & {p−1 \atop i=0,i≠j} & ((−1)^{sj} L_{s(2p−1−2j)} − (−1)^{si} L_{s(2p−1−2i)}) \end{matrix} ×

× (F_{s(2p−1−2j)n+1} + \frac{F_{s(2p−1−2j)−1} − \frac{L_{3s(2p−1−2j)}+(−1)^{s}L_{s(2p−1−2j)}}{L_{2s(2p−1−2j)}+(−1)^{s}}}{F_{s(2p−1−2j)}} F_{s(2p−1−2j)n}) . (p ≥ 4)

Proof. The proof is similar in all the four cases. It depends on an adequate partial fractions decomposition. We show all the steps of the proof only for the case (a), and indicate the corresponding decompositions used in the remaining cases. For the case (a) we use that for p ≥ 1 one has the following partial fractions decompositions:

\frac{z^{p}}{∏_{j=0}^{p−1} (z^{2} − (−1)^{sj} L_{2s(p−j)}z + 1)}  (91)

= ∑_{j=0}^{p−1} \frac{1}{∏_{i=0,i≠j}^{p−1} ((−1)^{sj} L_{2s(p−j)} − (−1)^{si} L_{2s(p−i)})} \frac{z}{z^{2} − (−1)^{sj} L_{2s(p−j)}z + 1},

and

\frac{z^{p}}{∏_{j=0}^{p−1} (z^{2} − (−1)^{js} L_{s(2p−1−2j)}z + (−1)^{s})}  (92)

= ∑_{j=0}^{p−1} \frac{1}{∏_{i=0,i≠j}^{p−i} ((−1)^{sj} L_{s(2p−1−2j)} − (−1)^{si} L_{s(2p−1−2i)})} \frac{z}{z^{2} − (−1)^{js} L_{s(2p−1−2j)}z + (−1)^{s}} .

We begin by noting that with (50) and (19) we can write

z^{p}Z (\binom{n}{2p}_{F_{s}}) = \frac{z^{p+1}}{(z − (−1)^{sp}) ∏_{j=0}^{p−1} (z^{2} − (−1)^{sj} L_{2s(p−j)}z + 1)}
= \frac{z}{z − (−1)^{sp}} \frac{z^{p}}{∏_{j=0}^{p−1} (z^{2} − (−1)^{sj} L_{2s(p−j)}z + 1)},

and then, by using (91) we have

z^{p}Z (\binom{n}{2p}_{F_{s}})

= \frac{z}{z − (−1)^{sp}} ∑_{j=0}^{p−1} \frac{1}{∏_{i=0,i≠j}^{p−1} ((−1)^{sj} L_{2s(p−j)} − (−1)^{si} L_{2s(p−i)})} \frac{z}{z^{2} − (−1)^{sj} L_{2s(p−j)}z + 1}

\frac{z}{z − (−1)^{sp}} ∑_{j=0}^{p−1} \frac{(−1)^{sj}}{F_{2s(p−j)} ∏_{i=0,i≠j}^{p−1} ((−1)^{sj} L_{2s(p−j)} − (−1)^{si} L_{2s(p−i)})} \frac{F_{2s(p−j)} (−1)^{sj} z}{z^{2} − (−1)^{sj} L_{2s(p−j)}z + 1}.

Thus, according to (51), convolution theorem and (16) we have that

\binom{n + p}{2p}_{F_{s}}
= (−1)^{spn} ∗ ∑_{j=0}^{p−1} \frac{(−1)^{sj}}{F_{2s(p−j)} ∏_{i=0,i≠j}^{p−1} ((−1)^{sj} L_{2s(p−j)} − (−1)^{si} L_{2s(p−i)})} (−1)^{sjn} F_{2s(p−j)n}
= ∑_{t=0}^{n} (−1)^{sp(n−t)} ∑_{j=0}^{p−1} \frac{1}{F_{2s(p−j)} ∏_{i=0,i≠j}^{p−1} ((−1)^{sj} L_{2s(p−j)} − (−1)^{si} L_{2s(p−i)})} (−1)^{sj(t+1)} F_{2s(p−j)t}
= ∑_{j=0}^{p−1} \frac{1}{F_{2s(p−j)} ∏_{i=0,i≠j}^{p−1} ((−1)^{sj} L_{2s(p−j)} − (−1)^{si} L_{2s(p−i)})} (−1)^{s(p(n−t)+j(t+1))} ∑_{t=0}^{n} F_{2s(p−j)t},

which proves (83). Also, by using (50), (20) and (92) we have that

z^{p−1}Z (\binom{n}{2p − 1}_{F_{s}})

= \frac{z^{p}}{∏_{j=0}^{p−1} (z^{2} − (−1)^{sj} L_{s(2p−1−2j)}z + (−1)^{s})}

= ∑_{j=0}^{p−1} \frac{1}{∏_{i=0,i≠j}^{p−i} ((−1)^{sj} L_{s(2p−1−2j)} − (−1)^{si} L_{s(2p−1−2i)})} \frac{z}{z^{2} − (−1)^{sj} L_{s(2p−1−2j)}z + (−1)^{s}}

= ∑_{j=0}^{p−1} \frac{1}{∏_{i=0,i≠j}^{p−i} ((−1)^{sj} L_{s(2p−1−2j)} − (−1)^{si} L_{s(2p−1−2i)})} \frac{(−1)^{js}}{F_{s(2p−1−2j)}} \frac{F_{s(2p−1−2j)} (−1)^{js} z}{z^{2} − (−1)^{sj} L_{s(2p−1−2j)}z + (−1)^{s}}

from where (by using (51) and (16))

\binom{n + p − 1}{2p − 1}_{F_{s}}

= ∑_{j=0}^{p−1} \frac{1}{∏_{i=0,i≠j}^{p−i} ((−1)^{sj} L_{s(2p−1−2j)} − (−1)^{si} L_{s(2p−1−2i)})} \frac{(−1)^{js}}{F_{s(2p−1−2j)}} (−1)^{jsn} F_{s(2p−1−2j)n}

= ∑_{j=0}^{p−1} \frac{(−1)^{js(n+1)}}{F_{s(2p−1−2j)} ∏_{i=0,i≠j}^{p−i} ((−1)^{sj} L_{s(2p−1−2j)} − (−1)^{si} L_{s(2p−1−2i)})}F_{s(2p−1−2j)n},

which proves (84).

For the case (b), use the partial fractions decomposition (valid for p ≥ 2):

\frac{z^{p−2}}{∏_{j=0}^{p−1} (z^{2} − (−1)^{sj} L_{2s(p−j)}z + 1)}

= − ∑_{j=0}^{p−1} \frac{1}{∏_{i=0,i≠j}^{p−1} ((−1)^{sj} L_{2s(p−j)} − (−1)^{si} L_{2s(p−i)})} \frac{z − (−1)^{sj} L_{2s(p−j)}}{z^{2} − (−1)^{sj} L_{2s(p−j)}z + 1},

and

\frac{z^{p−2}}{∏_{j=0}^{p−1} (z^{2} − (−1)^{sj} L_{s(2p−1−2j)}z + (−1)^{s})}

= ∑_{j=0}^{p−1} \begin{matrix}(−1) & s+1 \\ ∏ & {p−1 \atop i=0,i≠j} & ((−1)^{sj} L_{s(2p−1−2j)} − (−1)^{si} L_{s(2p−1−2i)}) \end{matrix} \frac{z − (−1)^{s} L_{s(2p−1−2j)}}{z^{2} − (−1)^{sj} L_{s(2p−1−2j)}z + (−1)^{s}} .

For the case (c), use the partial fractions decomposition (valid for p ≥ 3):

\frac{z^{p−3}}{∏_{j=0}^{p−1} (z^{2} − (−1)^{sj} L_{2s(p−j)}z + 1)} = ∑_{j=0}^{p−1} \frac{1}{∏_{i=0,i≠j}^{p−1} ((−1)^{sj} L_{2s(p−j)} − (−1)^{si} L_{2s(p−i)})} ×

×\frac{(−1)^{sj+1} L_{2s(p−j)}z + L_{4s(p−j)} + 1}{z^{2} − (−1)^{sj} L_{2s(p−j)}z + 1},

and

\frac{z^{p−3}}{∏_{j=0}^{p−1} (z^{2} − (−1)^{sj} L_{s(2p−1−2j)}z + (−1)^{s})} = ∑_{j=0}^{p−1} \frac{1}{∏_{i=0,i≠j}^{p−1} ((−1)^{sj} L_{s(2p−1−2j)} − (−1)^{si} L_{s(2p−1−2i)})} ×

×\frac{(−1)^{sj+1} L_{s(2p−1−2j)}z + L_{2s(2p−1−2j)} + (−1)^{s}}{z^{2} − (−1)^{sj} L_{s(2p−1−2j)}z + (−1)^{s}} ,

Finally, For the case (d) use the partial fractions decomposition (valid for p ≥ 4):

\frac{z^{p−4}}{∏_{j=0}^{p−1} (z^{2} − (−1)^{sj} L_{2s(p−j)}z + 1)} = − ∑_{j=0}^{p−1} \frac{1}{∏_{i=0,i≠j}^{p−1} ((−1)^{sj} L_{2s(p−j)} − (−1)^{si} L_{2s(p−i)})} ×

×\frac{(L_{4s(p−j)} + 1) z − (−1)^{sj} (L_{6s(p−j)} + L_{2s(p−j)})}{z^{2} − (−1)^{sj} L_{2s(p−j)}z + 1},

and

\frac{z^{p−4}}{∏_{j=0}^{p−1} (z^{2} − (−1)^{sj} L_{s(2p−1−2j)}z + (−1)^{s})}

= (−1)^{s+1} ∑_{j=0}^{p−1} \frac{1}{∏_{i=0,i≠j}^{p−1} ((−1)^{sj} L_{s(2p−1−2j)} − (−1)^{si} L_{s(2p−1−2i)})} ×

×\frac{(L_{2s(2p−1−2j)} + (−1)^{s}) z − (−1)^{sj} (L_{3s(2p−1−2j)} + (−1)^{s} L_{s(2p−1−2j)})}{z^{2} − (−1)^{sj} L_{s(2p−1−2j)}z + (−1)^{s}}.

Some examples are the following:

\binom{n + 2}{4}_{F_{s}} = \frac{1}{L_{4s} + (−1)^{s+1} L_{2s}} ∑_{t=0}^{n} (\frac{F_{4st}}{F_{4s}} − (−1)^{s(t+1)} \frac{F_{2st}}{F_{2s}} ) .  (93)

\binom{n + 1}{4}_{F_{s}} = \frac{1}{(−1)^{s} L_{2s} − L_{4s}} ∑_{t=0}^{n} ( \begin{matrix}F_{4st+1} + \frac{F_{4s−1}−L_{4s}}{F_{4s}} F_{4st} \\ + (−1)^{st+1} (F_{2st+1} + \frac{F_{2s−1}−L_{2s}}{F_{2s}} F_{2st}) \end{matrix} ) .  (94)

\binom{n}{3}_{F_{s}} = \frac{(−1)^{s+1}}{L_{3s} − (−1)^{s} L_{s}} (F_{3sn+1} + \frac{F_{3s−1} − L_{3s}}{F_{3s}} F_{3sn} + (−1)^{sn+1} (F_{sn+1} + \frac{F_{s−1} − L_{s}}{F_{s}} F_{sn})) .

(95)

\binom{n}{5}_{F_{s}} = \frac{−L_{5s}}{(L_{5s} − (−1)^{s} L_{3s}) (L_{5s} − L_{s})} (F_{5sn+1} + \frac{F_{5s−1} − \frac{L_{10s}+(−1)^{s}}{L_{5s}}}{F_{5s}} F_{5sn})

+\frac{(−1)^{s(n+1)+1} L_{3s}}{((−1)^{s} L_{3s} − L_{5s}) ((−1)^{s} L_{3s} − L_{s})} (F_{3sn+1} + \frac{F_{3s−1} − \frac{L_{6s}+(−1)^{s}}{L_{3s}}}{F_{3s}} F_{3sn})

+\frac{−L_{s}}{(L_{s} − L_{5s}) (L_{s} − (−1)^{s} L_{3s})} (F_{sn+1} + \frac{F_{s−1} − \frac{L_{2s}+(−1)^{s}}{L_{s}}}{F_{s}} F_{sn}) .  (96)

\binom{n + 1}{6}_{F_{s}} = ∑_{t=0}^{n} ( \begin{matrix}\frac{(−1)^{s(n−t)+1} L_{6s}}{(L_{6s} − (−1)^{s} L_{4s}) (L_{6s} − L_{2s})} & (F_{6st+1} + \frac{F_{6s−1} − \frac{L_{12s}+1}{L_{6s}}}{F_{6s}} F_{6st}) \\ +\frac{(−1)^{s(n+1)+1} L_{4s}}{((−1)^{s} L_{4s} − L_{6s}) ((−1)^{s} L_{4s} − L_{2s})} (F_{4st+1} + \frac{F_{4s−1} − \frac{L_{8s}+1}{L_{4s}}}{F_{4s}} F_{4st}) \\ +\frac{(−1)^{s(n−t)+1} L_{2s}}{(L_{2s} − L_{6s}) (L_{2s} − (−1)^{s} L_{4s})} (F_{2st+1} + \frac{F_{2s−1} − \frac{L_{4s}+1}{L_{2s}}}{F_{2s}} F_{2st}) \end{matrix} ) .

(97)

5 Final remarks

We can write (53) (with G = F and m_{1} = ⋯ = m_{l} = 0) as

(−1)^{s+1} ∑_{i=0}^{p} ∑_{j=0}^{i} (−1)^{\frac{(sj+2(s+1))(j+1)}{2}} \binom{p + 1}{j}_{F_{s}}F_{t_{1}s(i−j)}^{k_{1}} ⋯ F_{t_{l}s(i−j)}^{k_{l}}\binom{n + p − i}{p}_{F_{s}} = F_{t_{1}sn}^{k_{1}} ⋯ F_{t_{l}sn}^{k_{l}},

(98)

where p = t_{1}k_{1} +⋯+t_{l}k_{l}. It is possible to see (98) as a linear system in the p indeterminates \binom{n+p−i}{p}_{F_{s}}, i = 1, 2, …, p, with m equations, where m is the number of terms F_{t_{1}sn}^{k_{1}} ⋯ F_{t_{l}sn}^{k_{l}} we can form such that t_{1}k_{1} + ⋯ + t_{l}k_{l} = p (of course we refer to non-trivial terms and up to natural equivalences).

Conjecture 19. Any s-Fibonomial \binom{n+k}{p}_{F_{s}}, k = 0, 1, …, p − 1, can be written as a linear

combination of homogeneous terms F_{t_{1}sn}^{k_{1}}F_{t_{2}sn}^{k_{2}} ⋯ F_{t_{l}sn}^{k_{l}} where t_{1}k_{1} + t_{2}k_{2} + ⋯ + t_{l}k_{l} = p.

In the simplest case p = 2 we have two homogeneous terms, namely F_{sn}^{2} and F_{2sn}, and (98) can be solved for \binom{n}{2}_{F_{s}} and \binom{n+1}{2}_{F_{s}} to obtain

\binom{n}{2}_{F_{s}} = \frac{(−1)^{s}}{2} (\frac{F_{sn}^{2}}{F_{s}^{2}} − \frac{F_{2sn}}{F_{2s}} ) .  (99)

\binom{n + 1}{2}_{F_{s}} = \frac{1}{2} (\frac{F_{sn}^{2}}{F_{s}^{2}} + \frac{F_{2sn}}{F_{2s}} ) .  (100)

In the case p = 3 we have three homogeneous terms F_{3sn}, F_{2sn}F_{sn} and F_{sn}^{3}, and (98) can be solved for \binom{n}{3}_{F_{s}}, \binom{n+1}{3}_{F_{s}} and \binom{n+2}{3}_{F_{s}} to obtain

\binom{n}{3}_{F_{s}} = \frac{1}{2} (−1)^{s+1} \frac{F_{2sn}F_{sn}}{F_{2s}F_{s}} + \frac{1}{3} (−1)^{s} \frac{F_{3sn}}{F_{3s}} + \frac{1}{6} (−1)^{s} \frac{F_{sn}^{3}}{F_{s}^{3}} .  (101)

\binom{n + 1}{3}_{F_{s}} = \frac{(−1)^{s+1}}{3L_{s}} \frac{F_{3sn}}{F_{3s}} + \frac{(−1)^{s}}{3L_{s}} \frac{F_{sn}^{3}}{F_{s}^{3}} .  (102)

\binom{n + 2}{3}_{F_{s}} = \frac{1}{2} \frac{F_{2sn}F_{sn}}{F_{2s}F_{s}} + \frac{1}{3} \frac{F_{3sn}}{F_{3s}} + \frac{1}{6} \frac{F_{sn}^{3}}{F_{s}^{3}} .  (103)

In the case p = 4 we have 5 homogeneous terms

F_{4sn} , F_{3sn}F_{sn} , F_{sn}^{4} , F_{2sn}^{2} , F_{2sn}F_{sn}^{2}.  (104)

The corresponding system (98) can be solved for the s-Fibonomials \binom{n−i}{4}_{F_{s}}, i = 0, 1, 2, 3, to obtain

\binom{n}{4}_{F_{s}} = −\frac{1}{4} \frac{F_{4sn}}{F_{4s}} + \frac{L_{3s}}{10L_{s}F_{s}^{2}} \frac{F_{3sn}F_{sn}}{F_{3s}F_{s}} + \frac{(−1)^{s+1}}{10F_{s}^{2}} \frac{F_{2sn}^{2}}{F_{2s}^{2}} − \frac{1}{4} \frac{F_{2sn}F_{sn}^{2}}{F_{s}^{2}F_{2s}} .  (105)

\binom{n + 1}{4}_{F_{s}} = \frac{(−1)^{s} F_{s}}{4F_{3s}} \frac{F_{4sn}}{F_{4s}} + \frac{(−1)^{s}}{10F_{s}^{2}} \frac{F_{3sn}F_{sn}}{F_{3s}F_{s}} + \frac{(−1)^{s+1}}{10F_{s}^{2}} \frac{F_{2sn}^{2}}{F_{2s}^{2}} + \frac{(−1)^{s+1} F_{s}}{4F_{3s}} \frac{F_{2sn}F_{sn}^{2}}{F_{s}^{2}F_{2s}} . (106)

\binom{n + 2}{4}_{F_{s}} = \frac{(−1)^{s+1} F_{s}}{4F_{3s}} \frac{F_{4sn}}{F_{4s}} + \frac{(−1)^{s}}{10F_{s}^{2}} \frac{F_{3sn}F_{sn}}{F_{3s}F_{s}} + \frac{(−1)^{s+1}}{10F_{s}^{2}} \frac{F_{2sn}^{2}}{F_{2s}^{2}} + \frac{(−1)^{s} F_{s}}{4F_{3s}} \frac{F_{2sn}F_{sn}^{2}}{F_{s}^{2}F_{2s}} . (107)

\binom{n + 3}{4}_{F_{s}} = \frac{1}{4} \frac{F_{4sn}}{F_{4s}} + \frac{L_{3s}}{10L_{s}F_{s}^{2}} \frac{F_{3sn}F_{sn}}{F_{3s}F_{s}} + \frac{(−1)^{s+1}}{10F_{s}^{2}} \frac{F_{2sn}^{2}}{F_{2s}^{2}} + \frac{1}{4} \frac{F_{2sn}F_{sn}^{2}}{F_{s}^{2}F_{2s}} .  (108)

However, in this case a linear dependence relation appears, namely:

5F_{sn}^{4} + 3F_{2sn}^{2} − 4F_{3sn}F_{sn} = 0,  (109)

so the way of representing the s-Fibonomials as linear combinations of homogeneous terms stated in our conjecture should be not unique.

It remains (for a future work) to have a proof of the conjecture above and to identify what kind of linear dependencies exist among the homogeneous terms.

6 Acknowledgments

I would like to thank the anonymous referees for their careful reading of the original version and their valuable observations.

References

[1] Arthur T. Benjamin and Sean S. Plott, A combinatorial approach to Fibonomial coefficients, Fibonacci Quart. 46/47 (2008/2009), 7–9. See also Errata: 48 (2010), 276.

[2] L. Carlitz, Generating functions for powers of certain sequence of numbers, Duke Math. J. 29 (1962), 521–537.

[3] Urs Graf, Applied Laplace Transforms and z-Transforms for Scientists and Engineers. A Computational Approach using a ‘Mathematica’ Package, Birkhäuser, 2004.

[4] H. W. Gould, Generalization of Hermite’s divisibility theorems and the Mann-Shanks primality criterion for s-Fibonomial arrays, Fibonacci Quart. 12 (1974), 157–166.

[5] V. E. Hoggatt, Jr. Fibonacci numbers and generalized binomial coefficients, Fibonacci Quart. 5 (1967), 383–400.

[6] A. F. Horadam, Generating functions for powers of a certain generalised sequence of numbers, Duke Math. J. 32 (1965), 437–446.

[7] Thomas Koshy, Fibonacci and Lucas Numbers with Applications, John Wiley & Sons, Inc. 2001.

[8] C. Pita, More on Fibonomials, Proceedings of the XIV Conference on Fibonacci Numbers and Their Applications (2010), to appear.

[9] J. Riordan, Generating functions for powers of Fibonacci numbers, Duke Math. J. 29 (1962), 5–12.

[10] J. Seibert and P. Trojovský, On some identities for the Fibonomial coefficients, Math. Slovaca 55 (2005), 9–19.

[11] J. Seibert and P. Trojovský, On sums of certain products of Lucas numbers, Fibonacci Quart. 44 (2006), 172–180.

[12] A. G. Shannon, A method of Carlitz applied to the k-th power generating function for Fibonacci numbers, Fibonacci Quart. 12 (1974), 293–299.

[13] I. Strazdins, Lucas factors and a Fibonomial generating function, in G. E. Bergum, A. N. Philippou and A. F. Horadam, eds., Applications of Fibonacci Numbers, Vol. 7, Kluwer Academic Publishers (1998), 401–404.

[14] S. Vajda, Fibonacci and Lucas Numbers, and the Golden Section, Dover, 1989.

[15] Robert Vilch, Z Transform Theory and Applications, D. Reidel Publishing Company, 1987.

2010 Mathematics Subject Classification: Primary 11Y55; Secondary 11B39. Keywords: s-Fibonomial coefficients, Z transform.

(Concerned with sequence \underline{A000032}, \underline{A000045}, \underline{A001076}, \underline{A001654}, \underline{A001906}, \underline{A010048}, \underline{A034801}, \underline{A034802}, \underline{A047946}, \underline{A083564}, \underline{A092521}, and \underline{A156085}.)

Received October 19 2010; revised version received March 9 2011. Published in Journal of Integer Sequences, March 26 2011.

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