A Context-Free Grammar Associated with Fibonacci and Lucas Sequences

. We introduce a context-free grammar G � s ⟶ s + d, d ⟶ s { } to generate Fibonacci and Lucas sequences. By applying the grammar G , we give a grammatical proof of the Binet formula. Besides, we use the grammar G to provide a unifed approach to prove several binomial convolutions about Fibonacci and Lucas numbers, which were given by Hoggatt, Carlitz, and Church. Meanwhile, we also obtain some new binomial convolutions


Introduction
Recall that the Fibonacci sequence F n   and the Lucas sequence L n   are defned through the same recurrence relations: for n ≥ 2, with initial values F 0 � 0, F 1 � 1 and L 0 � 2, L 1 � 1, respectively.
Fibonacci and Lucas numbers have close connections with the golden ratio.Trough this paper, we use α to denote the golden ratio, that is, α � ((1 ). α and β are two roots of the quadratic equation x 2 − x − 1 � 0. Te famous Binet formulas for Fibonacci and Lucas numbers show that, for n ∈ Z, ( Te binomial identities involving Fibonacci and Lucas numbers are studied widely in recent decades.Te study of binomial identities involving Fibonacci and Lucas numbers is beginning from a group of identities of Hoggatt [1].After then, Carlitz [2] and Church and Bicknell [3] enriched the binomial identities family.Te details of these identities will be expanded in the next section.
In this paper, we introduce a context-free grammar to describe Fibonacci and Lucas numbers.Let We fnd that, for n ≥ 0, Here, D is the formal derivative associated with G.By applying this grammar, we give a grammatical framework to prove the binomial convolutions by Hoggatt, Carlitz, Church, and Bicknell and obtain some new binomial convolutions involving Fibonacci and Lucas numbers.
In Section 2, we posed the binomial identities given by Hoggatt, Layman, Carlitz, Church, and Bicknell.In Section 3, we give a new context-free grammar, which is called Fibonacci grammar.Based on the grammar, we give a grammatical expression of Fibonacci and Lucas numbers.As the application of this expression, a grammatical proof of some classic relations associated with Fibonacci and Lucas numbers are given, including the Binet formula.In Section 4, we provide a uniform framework to prove binomial identities by a grammatical manner.We prove all the identities given by Hoggatt, Carlitz, Church, and Bicknell and give some new binomial identities associated with Fibonacci and Lucas numbers.

Binomial Identities of Hoggatt, Carlitz, Church, and Bicknell
In this section, we recall the work of Hoggatt, Carlitz, Church, and Bicknell on binomial identities about Fibonacci and Lucas numbers.
In [1], Hoggatt found that, for n ≥ 1, Carlitz [2] extent Hoggatt's series of identities to more general relations: for 0 ≤ r ≤ m, n ≥ 0 and t ≥ 0, Besides, Church and Bicknell provided several diferent binomial convolutions about Fibonacci and Lucas numbers.In [3], Church and Bicknell showed that, for m, n, r ≥ 0,

Tilings and the Fibonacci Grammar
Te approach of studying combinatorial polynomials by using context-free grammars was introduced by Chen [4].In this decade, many combinatorists have found the relations between combinatorial polynomials and context-free grammars; see [4][5][6][7], for example.A context-free grammar G is a set of substitution rules on a set of variables X.We can defne a formal derivative D associated with a contextfree grammar G as a diferential operator on polynomials or Laurent polynomials in X.In precise, D is a linear operator satisfying the relation which can be in general given as Leibnitz formula For the purpose of combinatorial enumeration, the variables are attached to combinatorial structures by grammatical labelings by Chen and Fu [5].In order to provide a combinatorial expression of Fibonacci numbers, we need a corresponding combinatorial structure.Although many combinatorial interpretations of Fibonacci numbers exist (see exercises 1-9 in [8], p.14, for example), we use the tiling defnition given as below.
For n ≥ 1, a tiling of length n is refer to a tiling of a rectangle with size 1 × n by squares and rectangles with size 1 × 2. Here, we call the rectangle with size 1 × n by an n-board, and call the rectangle with size 1 × 2 by a domino.
For example, there are 5 tilings of length 4 Figure 1: A classical result shows that there are exactly F n+1 tilings with length n for all n ≥ 0. Here, we let F 1 � 1 count the unique empty tiling of 0-board.By counting the number of blocks of a tiling, we give a generating function F n (q) as a qanalogue of Fibonacci numbers: where F(n, k) denotes the number of tilings of an (n − 1)-board with k blocks.Here, we defne F 1 (q) � 1 and F 0 (q) � 0.
Lemma 1.For n ≥ 2, it holds that Proof.Consider the last block of a tiling T of length n.Tere are two classes: ending with a square and ending with a domino.Te frst class corresponds to the function qF n− 1 (q) since the left tiling has size n − 1; and the second class corresponds to the function qF n− 2 (q).Tis completes the proof.Let where q is a constant, that is, q.Since the grammar has close connections between the function F n (q), we call the grammar G as the q-Fibonacci grammar.Let q � 1 in G, the grammar G degenerates to be which is called Fibonacci grammar.

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Theorem 2. Let G be the q-Fibonacci grammar, and D be the formal derivative associated with the grammar G.For n ≥ 2, we have and for n ≥ 0, Proof.It can be seen that ( 15) can be obtained from (14) by setting s � d � 1.To prove (14), we introduce a grammatical labeling of a tiling by labeling the blocks.We label each block of a tiling T by q, and label the last block extra by s if it is a square, and by d if it is a domino.Ten, we defne the weight of the tiling T to be the sum of labelings of all blocks in T, that is, for a tiling T having k blocks, w(T) � sq k when ending at a square and w(T) � dq k when ending at a domino.Ten it is natural to say that since the sum of the weights of tilings ending at a square equals sqF n− 1 (q) and the sum of the weights of tilings ending at a domino equals dqF n− 2 (q).Now, we show by induction that, for n ≥ 2, it holds where T runs over the set of tilings with length n.
� sq, which is equal to the weight of the unique tiling of length 1. Tus, (17) holds for n � 2. Assume that (17) holds for n.To show that (17) is valid for n + 1, we consider the process to generate a tiling of length n + 1 from a tiling of length n.
For a tiling T ending at a domino, the only way to add the length of T is adding a new square in the end of T. In order to label the new tiling consistently, we delete the labeling d for the last domino in old tiling and label the new square by sq.Tis corresponds to the substitution rule d ⟶ sq.
For a tiling T ending at a square, we have two chooses.If we change the last square to a new domino, we change the labeling of the last part from sq to dq.If we add a new square at the end of T, we turn the labeling s from the old last square to the new last square and add a new labeling q to it.Tese two chooses correspond to the substitution rule s ⟶ d + sq.
For example, the frst tiling in Figure 1 is labeled by dq 2 , and the unique corresponding tiling is labeled by sq 3 .
Figure 2 And the second tiling in Figure 1 is labeled by sq 3 , the two corresponding tilings are labeled by sq 4 and dq 3 , respectively Figure 3.
Notice that we can generate all tilings of length n + 1 as above.Tus, where T ′ runs over the set of tilings with length n + 1. Tus (17) holds for n + 1.Now (17) holds for all n ≥ 2 by induction.Tis completes the proof.
□ Theorem 3. Let D 1 be the formal derivative associated with the Fibonacci grammar.For n ≥ 0, and Proof.Equation ( 19) can be deduced from ( 14) by setting q � 1.
As for (20), we have Applying the relation n times repeatedly, we obtain which implies (20).By setting s � d � 1 in (19), we get the following corollary as a grammatical expression about the Fibonacci numbers.
As an application of Fibonacci grammar, we give a grammatical proof of Binet's formula. Equivalently, which implies that Journal of Mathematics Similarly, we have and Combining ( 30) and (32), we obtain By setting s � d � 1 in (33), we obtain Now, we complete the proof.By using the same grammar, we can also generate Lucas numbers in a grammatical manner, whose proof is omitted.

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Following properties of Fibonacci and Lucas numbers are classic and useful in this paper.For the sake of completeness, we provide a grammatical proof.
We have the equation which implies (37) by setting s � d � 1.
Finally, (38) can be deduced from Proof.From (20), it holds that q d q (a) q q s q (b) q q sq (a) q q q sq (b) q q dq (c)  4 Journal of Mathematics Acting the operator D m− n+1 1 on the two hand sides of the above equation, we obtain which is changed to by setting s � d � 1. Tis is the special case of (42 Te last equation holds from (36).Setting s � d � 1 in the above relation, we obtain which equals the right hand side of (42) by (45).□ Lemma 9.For n ≥ 1, Proof.We can verify that it holds that which is reduced to (48) by setting s � d � 1.
Similarly, equations ( 49) and (50) can be obtained by simplifying the expression and the expression in the same manner as above and then setting s � d � 1. □

m-th Order Fibonacci Grammar, Binomial Fibonacci, and Lucas Identities
In this section, we provide a framework to prove the identities involving Fibonacci and Lucas numbers associated with binomial coefcients.
Let G be a context-free grammar with an alphabet X, and let k be a constant for G.We defne the product of the grammar G and k to be the context-free grammar in which each letter a ∈ X corresponds the substitution rule a ⟶ kG(a), denoted as kG.A grammar G on an alphabet X is defned to be linear if for each letter a ∈ X, G(a) is a linear function on X.
Lemma 10.Let G be a linear context-free grammar with an alphabet X, and let k be a constant for G.For n ≥ 0 and each linear function f, it holds that Proof.Because of the linearity of the operator D, it is enough to show (56) holds for every letter a ∈ X.We prove the assertion by induction on n.Te case for n � 1 is evident.Assume that the assertion holds for n.Now, consider the case for n + 1, since which equals k n+1 D n+1 G (a). Tis completes the proof.Consider the following context-free grammar G m : We call G m the m-th order Fibonacci grammar.Let D m be the formal derivative associated with the grammar G m .According to (56), D m is equivalent to D m 1 when acting on a linear functions of s and d.When m � 1, 1-th order Fibonacci grammar is just the Fibonacci grammar.
From (19), it can be verifed that Tus, It should be noticed that D m is not equivalent to D m 1 .For example, One can easy to check that Following assertion is critical for the proof.

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Lemma 11.For n ≥ 1, it holds that (66) Proof.We can verify that (67) where  D is the formal derivative associated with the grammar Notice that  G is as same as the m-th order Fibonacci grammar by setting α 1 � s and α 2 � d.Tus, Tus, As for α 3 , one can verify that Te last equation holds from (38).Tis completes the proof.Now, we begin to proof binomial convolutions about Fibonacci and Lucas numbers.

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Theorem 12.For n ≥ 0, we have Proof.According to (56), Tis deduces that the left hand side of (73) equals the summation: which equals 2 n F mn by (65).Tis completes the proof of (73).Te left hand side of (74) can be calculated by a similar manner.We have 6 Journal of Mathematics (78) By using Gaussian elimination, d 2 − 2sd + s 2 can be represented as the linear combination of α 1 , α 2 and α 3 , namely, Tus, So, Tis completes the proof of (74).
As for (75), we need consider the following grammatical convolution D n m ((3d − s) 2 ).It follows by Leibnitz formula that which reduces by setting s � d � 1 to be Now, let us calculate D n m ((3d − s) 2 ).By using Gaussian elimination, (3d − s) 2 can be represented as the linear combination of α 1 , α 2 and α 3 , namely, Tus, Te last equation holds from (38).Tis completes the proof.

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Theorem 13.For n, m ≥ 1 and r ≥ 0, we have Proof.According to (19), (87) Tus, s�d�1 . ( Let G 4m denote the context-free grammar According to (48)-(50), we have According to (56), where  D is the formal derivative associated with the following grammar Notice that  G is as same as 2m-th Fibonacci grammar D 2m by substituting s, d into A, B, respectively.Tus, Journal of Mathematics Tis deduces that Te last equation holds from (36).Tis completes the proof.

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Theorem 14.For n, m ≥ 1 and r, t ≥ 0, we have and Proof.Consider the grammar G ′ : Ten, it can be easily see that According to Leibnitz formula, the left hand side of (97) can be obtained from D ′ n (aD t 1 (s − d)) by setting a � s � d � 1.Now, the proof of (97) can be reduced to the calculation of In (42), by setting n � r + 1 and k � 1, we obtain And by setting n � r − 1 and k � − 1, we obtain Ten, D ′ can be viewed as the formal derivative associated with F m G r with the alphabet A, B { }.According to (56), which equals F n m F nr+t by (36).Tis complete the proof of (97).
As for (98), we consider the following Leibnitz relation: which equals F n m L nr+t from (37).Tis complete the proof of (98).

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Remark 15.It is easily to see that the technique of proving identities by using the simple Fibonacci grammar can be extended to study corresponding binomial convolutions involving Fibonacci polynomials F n (q), just considering the Fibonacci grammar.Meanwhile, one can extend the Fibonacci grammar to be to get generalized Fibonacci and generalized Lucas numbers, who are defned as the linear recurrence (110) and the initial conditions U 0 � 0, U 1 � 1 and (111) Terefore, there are many identities involving Fibonacci and Lucas numbers who are not in the standard binomial form.For example, Kilic and Tasdemir [9]

Figure 2 :
Figure 2: An example for the action of the substitution rule d ⟶ sq.(a) A tiling labeled by dq 2 .(b) d ⟶ sq.

Figure 3 :
Figure 3: An example for the action of the substitution rule s ⟶ d + sq.(a) A tiling labeled by sq 3 .(b) s ⟶ sq.(c) s ⟶ d.

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Journal of MathematicsBesides, the identities proved here are all from Leibnitz formula, hence in form n k�0 n k  A k B n− k . ) n (B)     A�B�1 +F t− 2 D′ n (A)     A�B�1 provided several binomial double summations in the form for k � 2, 4. It's an interesting question to fnd a universal grammatical proof of these relations.