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Disconnectedness of the subgraph {F}^{3} for the group {\Gamma}^{3}
Journal of Inequalities and Applications volume 2013, Article number: 283 (2013)
Abstract
In this paper we show that the subgraph {F}^{3} is disconnected and that for all integers m, we find all integers a and b such that (9{m}^{2}4){a}^{2}+4 and 5{b}^{2}\pm 4 are square. It turns out that the set of numbers b comprises the Fibonacci numbers.
Dedication
Dedicated to Professor Hari M Srivastava
1 Introduction
Let \stackrel{\u02c6}{\mathbb{Q}}=\mathbb{Q}\cup \{\mathrm{\infty}\} be the extended rationals, let Γ be the modular group acting on \stackrel{\u02c6}{\mathbb{Q}} as with the upper halfplane \mathcal{H}=\{z\in \mathbb{C}:Imz>0\}:
where a, b, c, and d are rational integers, and let {\Gamma}^{3} denote the group consisting of the cubes of the elements g of Γ, which is the group \{g\in \Gamma :ab+cd\equiv 0(mod3)\}; see [1].
Jones et al. [2] used the notion of the imprimitive action [3–5] for a Γinvariant equivalence relation induced on \stackrel{\u02c6}{\mathbb{Q}} by the congruence subgroup {\Gamma}_{0}(n)=\{g\in \Gamma :c\equiv 0(modn)\} to obtain some suborbital graphs and examined their connectedness and forest properties.
In [6], a {\Gamma}^{3}invariant equivalence relation is introduced by using the subgroup {\Gamma}_{0}^{3}(n)=\{g\in {\Gamma}^{3}:c\equiv 0(modn)\} to obtain suborbital graphs {F}_{u,n}^{3}. There, the connectivity properties of all subgraphs {F}_{u,n}^{3} other than {F}_{1,1}^{3}={F}^{3} are examined.
In this paper we show that the subgraph {F}^{3} is disconnected and give some results, which seem important from the point of view of number theory.
2 Preliminaries
Since {\Gamma}^{3}=\{{g}^{3}:g\in \Gamma \}, it is easily seen that the elements of {\Gamma}^{3} are ones of the forms \left(\begin{array}{cc}3a& b\\ c& 3d\end{array}\right), \left(\begin{array}{cc}a& 3b\\ 3c& d\end{array}\right), and \left(\begin{array}{cc}a& b\\ c& d\end{array}\right) where a, b, c, and d\not\equiv 0(mod3) in the third matrix. Furthermore, {\Gamma}_{\mathrm{\infty}}^{3}<{\Gamma}_{0}^{3}(n)\le {\Gamma}^{3} for each positive integer n, where {\Gamma}_{\mathrm{\infty}}^{3} is the stabilizer of ∞ generated by the element \left(\begin{array}{cc}1& 3\\ 0& 1\end{array}\right), and second inclusion is strict if n>1.
Since the group {\Gamma}^{3} is transitive on \stackrel{\u02c6}{\mathbb{Q}} in [6], any reduced fraction \frac{r}{s} in \stackrel{\u02c6}{\mathbb{Q}} equals g(\mathrm{\infty}) for some g\in {\Gamma}^{3}. Similar to that of [2], we get the following {\Gamma}^{3}invariant equivalence relation on \stackrel{\u02c6}{\mathbb{Q}} by {\Gamma}_{0}^{3}(n) as \frac{r}{s}\sim \frac{x}{y} if and only if {g}^{1}h\in {\Gamma}_{0}^{3}(n), where g=\left(\begin{array}{cc}r& \ast \\ s& \ast \end{array}\right) and h is similar. Furthermore, the above equivalence relation is imprimitive, which means that it is different from the identity relation (a\sim b if and only if a=b) and the universal relation (a\sim b for all a,b\in \stackrel{\u02c6}{\mathbb{Q}}).
From the above defined {\Gamma}^{3}invariant equivalence relation, we can verify that \frac{r}{s}\sim \frac{x}{y} if and only if rysx\equiv 0(modn). The equivalence classes are called blocks, and a block containing the rational \frac{x}{y} is denoted by [\frac{x}{y}].
Although the equivalence relations are resulting almost the same as in [2], the subgraph {F}_{1,1} in [2] is easily shown to be connected, but here we will see that the subgraph {F}_{1,1}^{3} is disconnected. So, using different subgroups changes the characters of the subgraphs.
3 Subgraphs {F}_{u,n}^{3}
The group {\Gamma}^{3} acts on \stackrel{\u02c6}{\mathbb{Q}}\times \stackrel{\u02c6}{\mathbb{Q}} through g:(\alpha ,\beta )\to (g(\alpha ),g(\beta )). The orbits are called suborbital. From the suborbital {O}^{3}(\alpha ,\beta ) containing (\alpha ,\beta ), we can form the suborbital graph {G}^{3}(\alpha ,\beta ) whose vertices are the elements of \stackrel{\u02c6}{\mathbb{Q}} and the edges are the pairs (a,b)\in {O}^{3}(\alpha ,\beta ), which we denote by a\to b and represent as hyperbolic geodesics in ℋ.
Since {\Gamma}^{3} acts transitively on \stackrel{\u02c6}{\mathbb{Q}}, every suborbital contains a pair (\mathrm{\infty},\frac{u}{n}) for some \frac{u}{n} in \stackrel{\u02c6}{\mathbb{Q}}, n\ge 0, (u,n)=1. In this case, we denote the suborbital graph by {G}_{u,n}^{3} for short. From now on, we assume that n>0.
As {\Gamma}^{3} permutes the blocks transitively, all subgraphs corresponding to the blocks are isomorphic, as in [2]. Therefore, we will only consider the subgraph {F}_{u,n}^{3} of {G}_{u,n}^{3} whose vertices form the block [\mathrm{\infty}]=\{\frac{x}{y}\in \stackrel{\u02c6}{\mathbb{Q}}\mid y\equiv 0(modn)\}. The following two results were proved in [6].
Theorem 1 {F}_{u,n}^{3}={F}_{{u}^{\prime},{n}^{\prime}}^{3} if and only if n={n}^{\prime} and u\equiv {u}^{\prime}(mod3n).
Theorem 2 \frac{r}{s}\to \frac{x}{y} is an edge in {F}^{3} (={F}_{1,1}^{3}) if and only if

1.
if r\equiv 0(mod3), then y\equiv \pm s(mod3) and rysx=\pm 1, or

2.
if s\equiv 0(mod3), then x\equiv \pm r(mod3), and rysx=\pm 1, or

3.
if r,s\not\equiv 0(mod3), then x\not\equiv \pm r(mod3), y\not\equiv \pm s(mod3) and rysx=\pm 1.
We can easily get the following lemmas.
Lemma 1 \frac{r}{s}\to \frac{x}{y} is in {F}^{3} if and only if \frac{x}{y}\to \frac{r}{s} is in {F}^{3}.
Lemma 2 [2]
No edges of {F}^{3} cross in ℋ.
4 Disconnectedness of {F}^{3}
Definition 1 For m\in \mathbb{N} and m\ge 2, let {v}_{1},{v}_{2},\dots ,{v}_{m} be a finite sequence of vertices of {F}^{3}. Then the configuration {v}_{1}\to {v}_{2}\to \cdots \to {v}_{m} is called a finite path in {F}^{3}. A subgraph ∧ of {F}^{3} is called connected if every two vertices x and y of ∧ are connected by a finite path in {F}^{3}. Otherwise, we call ∧ disconnected.
Now we give one of our main theorems.
Theorem 3 The graph {F}^{3} is disconnected.
We prove Theorem 3 after giving some theorems, propositions and lemmas as follows.
By Theorem 2, it is easily seen that the graph {F}^{3} is periodic with period 3. That is, if a\to b is in {F}^{3}, then a+3m\to b+3m is in {F}^{3} for all m\in \mathbb{Z}, and therefore, for some m, a+m or b+m (not both) is ∞ or both a+3m and b+3m are in the interval [1,4]. Therefore we can only use the interval [1,4] for our calculations as in Figure 1.
It is clear that T=\left(\begin{array}{cc}1& 5\\ 1& 4\end{array}\right) is in {\Gamma}^{3} and the corresponding transformation T(z)=\frac{z5}{z4} is strictly increasing on [1,4]\cap \mathbb{Q}. Furthermore, it is easily seen that {T}^{m}(\mathrm{\infty})\to {T}^{m}(\frac{1}{1}) is an edge in {F}^{3} for all nonnegative integers m. From this, we get, as an example, a finite path in {F}^{3} as \mathrm{\infty}\to \frac{1}{1}\to \frac{4}{3}\to \frac{11}{8}.
Lemma 3 Let T be as above, then the sequence \{{T}^{m}(1)\} is strictly monotone increasing and T(\frac{1}{0})\to T(\frac{1}{1})\to {T}^{2}(\frac{1}{1})\to \cdots \to {T}^{m}(\frac{1}{1})\to \cdots is an infinite path in {F}^{3} in increasing order.
Proof The conclusion follows from Theorem 2 and from the fact that T(z)=\frac{z5}{z4} is strictly increasing on [1,4)\cap \mathbb{Q}. □
Lemma 4 Let a and b be in ℕ and let 1\le \frac{a}{b}<\frac{5\sqrt{5}}{2}, then \frac{a}{b}<T(\frac{a}{b})<\frac{5\sqrt{5}}{2}.
Proof From \frac{a}{b}<\frac{5\sqrt{5}}{2} we get 2a5b<\sqrt{5}b. Then squaring gives the inequality {a}^{2}+4ab<ab+5{b}^{2}. That is, \frac{a}{b}<\frac{a+5b}{a+4b}=T(\frac{a}{b}). On the other hand, {a}^{2}5ab+5{b}^{2}>0, then it is easily seen that 5{(a4b)}^{2}<{(3a10b)}^{2}. As \frac{a}{b}<2, then taking square roots gives \sqrt{5}(a4b)>3a10b. This shows that T(\frac{a}{b})<\frac{5\sqrt{5}}{2}. □
Proposition 1 Let T be as above and 1\le \frac{a}{b}<\frac{5\sqrt{5}}{2}. Then \frac{a}{b}\to T(\frac{a}{b}) is an edge in {F}^{3} if and only if there exists a natural number u such that {u}^{2}=5{b}^{2}+4 and a=\frac{5b\sqrt{5{b}^{2}+4}}{2}.
Proof Let \frac{a}{b}\to T(\frac{a}{b}) be an edge in {F}^{3}. Then, by using Theorem 2 and Lemma 4, we get {a}^{2}5ab+5{b}^{2}1=0. Since \frac{a}{b}<\frac{5\sqrt{5}}{2}, we have a=\frac{5b\sqrt{5{b}^{2}+4}}{2}. This concludes that \sqrt{5{b}^{2}+4} is an integer u.
Conversely, it is clear to see M=\left(\begin{array}{cc}\frac{5b+\sqrt{5{b}^{2}+4}}{2}& 5b\\ b& \frac{5b+\sqrt{5{b}^{2}+4}}{2}\end{array}\right) is in {\Gamma}^{3} and that M(\frac{1}{0})=\frac{a}{b} and M(\frac{1}{1})=T(\frac{a}{b}). Therefore, by the definition of edges of {F}^{3}, the configuration \frac{a}{b}\to T(\frac{a}{b}) is an edge in {F}^{3}. □
Theorem 4 The positive rational number \frac{x}{y} is in \{{T}^{m}(\frac{1}{0}):m\in \mathbb{N}\}=A if and only if there exists a natural number u such that 5{y}^{2}+4={u}^{2} and x=\frac{5y\sqrt{5{y}^{2}+4}}{2}.
Proof From Proposition 1, the ‘if’ part is clear.
Conversely, we show that under the hypothesis, \frac{x}{y} is in the set A. Since {T}^{m}(\frac{1}{0})=\frac{\frac{5y\sqrt{5{y}^{2}+4}}{2}}{y} for any m\in \mathbb{N}, {lim}_{m\to \mathrm{\infty}}{T}^{m}(\frac{1}{0})=\frac{5\sqrt{5}}{2}. As 1\le \frac{x}{y}<\frac{5\sqrt{5}}{2}, if \frac{x}{y} is not in A, there exists k\in \mathbb{N} such that {T}^{k}(\frac{1}{0})<\frac{x}{y}<{T}^{k+1}(\frac{1}{0}). From this, we get
We know that \frac{\frac{5yu}{2}}{y}\to T(\frac{\frac{5yu}{2}}{y})=1+\frac{y}{3y\frac{3yu}{2}} is an edge in {F}^{3}. From Lemma 2 it must be that T(\frac{\frac{5yu}{2}}{y})=\frac{\frac{5y+u}{2}}{\frac{3y+u}{2}}<\frac{\frac{5a+v}{2}}{\frac{3a+v}{2}}. This gives the inequality vy<au. But, from \frac{\frac{5av}{2}}{a}<\frac{\frac{5yu}{2}}{y}, we arrive at the inequality vy>au, a contradiction. This concludes the proof of the theorem. □
Corollary 1 \frac{1}{0}\to 0+\frac{1}{1}\to 1+\frac{1}{3}\to 1+\frac{3}{8}\stackrel{T}{\to}\cdots \stackrel{T}{\to}1+\frac{{a}_{n}}{{b}_{n}}\stackrel{T}{\to}1+\frac{{b}_{n}}{3{b}_{n}{a}_{n}}\stackrel{T}{\to}\cdots is an infinite path in {F}^{3}, and all vertices of the path are smaller than \frac{5\sqrt{5}}{2}, and the natural numbers x and y in the vertex 1+\frac{x}{y} are such that 5{x}^{2}+4 and 5{y}^{2}+4 are square. Furthermore, {a}_{n}=\frac{3{b}_{n}\sqrt{5{b}_{n}^{2}+4}}{2} and {lim}_{n\to \mathrm{\infty}}(1+\frac{{a}_{n}}{{b}_{n}})=\frac{5\sqrt{5}}{2}, where T is as above.
Proof Lemma 3 and Theorem 4 conclude the proof. □
If we follow the way of the above, we arrive at the following two results without proofs.
Theorem 5 Let T=\left(\begin{array}{cc}1& 5\\ 1& 4\end{array}\right) and \frac{5\sqrt{5}}{2}<\frac{a}{b}\le 2. Then \frac{5\sqrt{5}}{2}<T(\frac{a}{b})<\frac{a}{b} and \frac{a}{b}\to T(\frac{a}{b}) is an edge in {F}^{3} if and only if 5{b}^{2}4 is a square and a=\frac{5b\sqrt{5{b}^{2}4}}{2}.
Corollary 2 3\stackrel{T}{\to}2=1+\frac{1}{1}\stackrel{T}{\to}1+\frac{1}{2}\stackrel{T}{\to}1+\frac{2}{5}\stackrel{T}{\to}\cdots \stackrel{T}{\to}1+\frac{{a}_{n}}{{b}_{n}}\stackrel{T}{\to}1+\frac{{b}_{n}}{3{b}_{n}{a}_{n}}\stackrel{T}{\to}\cdots is an infinite path in {F}^{3} in decreasing order such that all vertices of the path are greater than \frac{5\sqrt{5}}{2}, and the natural numbers x and y in the vertex 1+\frac{x}{y} are such that 5{x}^{2}4 and 5{y}^{2}4 are squares. Furthermore, {a}_{n}=\frac{3{b}_{n}\sqrt{5{b}_{n}^{2}4}}{2} and {lim}_{n\to \mathrm{\infty}}(1+\frac{{a}_{n}}{{b}_{n}})=\frac{5\sqrt{5}}{2}, where T is as above.
Theorem 6 Let k be a natural number and let the vertices {v}_{1},{v}_{2},\dots ,{v}_{k} in [1,3] of {F}^{3} be such that at least one is smaller and one is greater than \frac{5\sqrt{5}}{2}. Then the path \mathrm{\infty}\to {v}_{1}\to {v}_{2}\to \cdots \to {v}_{k} does not occur in {F}^{3}.
Proof If the above situation occurs, since \frac{5\sqrt{5}}{2} is not a vertex in {F}^{3}, there exists 1\le m<k such that {v}_{m}<\frac{5\sqrt{5}}{2}<{v}_{m+1} and {v}_{m}\to {v}_{m+1} is an edge in {F}^{3}.
Since the sequences of the vertices in Corollary 1 and Corollary 2 converge to \frac{5\sqrt{5}}{2}, Lemma 2 gives that there exist naturals m and n such that
Suppose first that n>m. Then, multiplying by {T}^{m} and using Theorem 5, we have \frac{1}{0}\to {T}^{nm}(\frac{3}{1})=\frac{\frac{5b\sqrt{5{b}^{2}4}}{2}}{b} for some b\in \mathbb{N}. Therefore, from Theorem 2, we get b=1. That is, \frac{1}{0}\to \frac{3}{1} is an edge in {F}^{3}, a contradiction.
Suppose now that m>n. Then, again multiplying by {T}^{n}, we have {T}^{mn}(\frac{1}{0})\to \frac{3}{1}. In this case, {T}^{mn}(\frac{1}{0})\ge 2. But, in any case, {T}^{mn}(\frac{1}{0})<\frac{5\sqrt{5}}{2}, a contradiction.
Finally, let m=n. Then \frac{1}{0}\to \frac{3}{1} must be an edge in {F}^{3}, a contradiction. These complete the proof. □
Let S={T}^{1}=\left(\begin{array}{cc}4& 5\\ 1& 1\end{array}\right). Then likewise we do before we give the following five results without a proof.
Lemma 5 Let 3\le \frac{a}{b}<\frac{5+\sqrt{5}}{2}. Then \frac{a}{b}<S(\frac{a}{b})<\frac{5+\sqrt{5}}{2}. Furthermore, \frac{a}{b}\to S(\frac{a}{b}) is an edge in {F}^{3} if and only if \sqrt{5{b}^{2}4} is a natural number and a=\frac{5b+\sqrt{5{b}^{2}4}}{2}.
Corollary 3 3=4\frac{1}{1}\stackrel{S}{\to}4\frac{1}{2}\stackrel{S}{\to}4\frac{2}{5}\stackrel{S}{\to}\cdots \stackrel{S}{\to}4\frac{\frac{3b\sqrt{5{b}^{2}4}}{2}}{b}\stackrel{S}{\to}\cdots is an infinite path in {F}^{3} in increasing order, as seen Figure 2, and the limit of the sequence of vertices is \frac{5+\sqrt{5}}{2}.
Lemma 6 Let \frac{5+\sqrt{5}}{2}<\frac{a}{b}\le 4. Then \frac{5+\sqrt{5}}{2}<S(\frac{a}{b})<\frac{a}{b}. Furthermore, \frac{a}{b}\to S(\frac{a}{b}) is an edge in {F}^{3} if and only if 5{b}^{2}+4 is a square and a=\frac{5b+\sqrt{5{b}^{2}+4}}{2}.
Lemma 7 4\frac{0}{1}\stackrel{S}{\to}4\frac{1}{3}\stackrel{S}{\to}4\frac{3}{8}\stackrel{S}{\to}\cdots \stackrel{S}{\to}4\frac{\frac{3b\sqrt{5{b}^{2}+4}}{2}}{b}\stackrel{S}{\to}\cdots is an infinite path in {F}^{3} in decreasing order, and the limit of the sequence of vertices is \frac{5+\sqrt{5}}{2}.
Theorem 7 Let k be a natural number and let the vertices {v}_{1},{v}_{2},\dots ,{v}_{k}, in [3,4], of {F}^{3} be such that at least one is smaller and one is greater than \frac{5+\sqrt{5}}{2}. Then {v}_{1}\to {v}_{2}\to \cdots \to {v}_{k}\to \mathrm{\infty} does not occur in {F}^{3}.
Proof We conclude the proof as in Theorem 6. □
Proof of Theorem 3 Theorems 6 and 7 conclude that the vertices of {F}^{3} in (\frac{5\sqrt{5}}{2},\frac{5+\sqrt{5}}{2}) are not connected to the vertex ∞. That is, the graph {F}^{3} is disconnected. □
We finally give one of our main results as follows.
Theorem 8 For all natural numbers m, the natural numbers b that make the number (9{m}^{2}4){b}^{2}+4 square are 0,1,3m,9{m}^{2}1,3m(9{m}^{2}1)3m,\dots ,a,b,3mba,\dots .
Proof For the proof, we only use, as above, the interval [1,4]. It is clear that the matrix M=\left(\begin{array}{cc}1& 3m+2\\ 1& 3m+1\end{array}\right) is in {\Gamma}^{3}. Theorem 2 gives that \frac{1}{1}\to M(\frac{1}{1})=\frac{3m+1}{3m} is an edge in {F}^{3}. Since the transformation M(x)=\frac{x+3m+2}{x+3m+1} is an increasing function on [1,4] and \frac{1}{1}<M(\frac{1}{1})=\frac{3m+1}{3m}, we can easily see that, for all k\in \mathbb{N}, {M}^{k}(\frac{1}{1})<{M}^{k+1}(\frac{1}{1}). That is, the sequence \{{M}^{k}(1)\} is an increasing sequence. Furthermore, {M}^{k}(1)=[1;\underset{k\phantom{\rule{0.25em}{0ex}}\mathrm{times}}{\underset{\u23df}{3m,3m,3m,\dots ,3m}}], or
And more, {lim}_{k\to \mathrm{\infty}}{M}^{k}(1)=\frac{3m+2\sqrt{9{m}^{2}4}}{2}. So, for all k\in \mathbb{N}, {M}^{k}(1)<\frac{3m+2\sqrt{9{m}^{2}4}}{2}. If \frac{a}{b}<\frac{3m+2\sqrt{9{m}^{2}4}}{2} we can easily see that T(\frac{a}{b})<\frac{3m+2\sqrt{9{m}^{2}4}}{2}, and furthermore, if \frac{a}{b}\to T(\frac{a}{b})=\frac{a+(3m+2)b}{a+(3m+1)b} is an edge in {F}^{3}, then, by Theorem 2, {a}^{2}(3m+2)b+(3m+2){b}^{2}1=0. Solving the equation, we have a=\frac{(3m+2)b\sqrt{(9{m}^{2}4){b}^{2}+4}}{2}, where we get the sign ‘−’ since for all k\in \mathbb{N}, {M}^{k}(1)<\frac{3m+2\sqrt{9{m}^{2}4}}{2}. Because a is an integer, \sqrt{(9{m}^{2}4){b}^{2}+4} must be an integer. According to Theorem 2, for all k\in \mathbb{N}, {M}^{k}(\frac{1}{1})\to {M}^{k+1}(\frac{1}{1}) is an edge in {F}^{3}. Therefore
is an infinite path γ in {F}^{3}. All denominators of vertices \frac{a}{b} of γ make (9{m}^{2}4){b}^{2}+4 square. We can rewrite γ as
And we conclude that the numbers 0,1,3m,9{m}^{2}1,3m(9{m}^{2}1)3m,\dots ,{a}_{0},{b}_{0},3m{b}_{0}{a}_{0},\dots make (9{m}^{2}4){b}^{2}+4 square.
Let us now show the only nonnegative integers b such that (9{m}^{2}4){b}^{2}+4 is square.
Conversely, suppose that there is a natural number t such that (9{m}^{2}4){t}^{2}+4 is square. Then \frac{{a}_{1}}{{b}_{1}}=\frac{\frac{(3m+2)t\sqrt{(9{m}^{2}4){t}^{2}+4}}{2}}{t} is smaller than \frac{3m+2\sqrt{9{m}^{2}4}}{2} and, from Theorem 2, we get that \frac{{a}_{1}}{{b}_{1}}\to T(\frac{{a}_{1}}{{b}_{1}}) is an edge in {F}^{3}. Suppose, for some k\in \mathbb{N}, {M}^{k}(\frac{1}{1})<\frac{{a}_{1}}{{b}_{1}}<{M}^{k+1}(\frac{1}{1}). Lemma 2 says that {T}^{m}(\frac{1}{1})<\frac{{a}_{1}}{{b}_{1}}<{T}^{m+1}(\frac{{a}_{1}}{{b}_{1}})<{M}^{k+1}(\frac{1}{1}). Therefore, for some y,
From the first inequality, we conclude that t\sqrt{(9{m}^{2}4){y}^{2}+4}>y\sqrt{(9{m}^{2}4){y}^{2}+4}. From the second inequality, we just have t\sqrt{(9{m}^{2}4){y}^{2}+4}<y\sqrt{(9{m}^{2}4){y}^{2}+4}, a contradiction. Consequently, \frac{{a}_{1}}{{b}_{1}} must be in the set \{{M}^{k}(\frac{1}{1}):k\in \mathbb{N}\}. This completes the proof of the theorem. □
From Corollaries 1 and 2, we get the following without a proof.
Corollary 4 The nonnegative integers b such that 5{b}^{2}+4 is square are 0,1,3,8,21,55,144,\dots ,a,b,3ba,\dots .
Corollary 5 The nonnegative integers b making 5{b}^{2}4 square are 1,2,5,13,34,89,\dots ,a,b,3ba,\dots .
From Corollaries 4 and 5, we conclude the following important corollary.
Corollary 6 Let \{{a}_{n}\} and \{{b}_{n}\} be the sequences (0,1,3,\dots ,a,b,3ba,\dots ) and (1,2,5,\dots ,c,d,3dc,\dots ), respectively. Then the sequence ({a}_{1},{b}_{1},{a}_{2},{b}_{2},\dots ,{a}_{n},{b}_{n},\dots ) is the Fibonacci sequence.
Proof Let us see that {a}_{n}+{b}_{n}={a}_{n+1} and {b}_{n}+{a}_{n+1}={b}_{n+1} for all n in ℕ by induction. Suppose that the assertion is true up to the natural number k. Let us show that {a}_{k+1}+{b}_{k+1}={a}_{k+2}. Since {a}_{k+1}=3{a}_{k}{a}_{k1} and {b}_{k+1}=3{b}_{k}{b}_{k1}, {a}_{k+1}+{b}_{k+1}=3({a}_{k}+{b}_{k})({a}_{k1}+{b}_{k1})=3{a}_{k+1}{a}_{k}={a}_{k+2}. The other is similar. □
References
Rankin RA: Modular Forms and Functions. Cambridge University Press, Cambridge; 2008.
Jones GA, Singerman D, Wicks K LMS Lect. Note Ser. 160. The Modular Group and Generalized Farey Graphs 1991, 316–338.
Biggs NL, White AT LMS Lect. Note Ser. In Permutation Groups and Combinatorial Structures. Cambridge University Press, Cambridge; 1979.
Neumann PM: Finite permutation groups and edgecoloured graphs and matrices. In Topics in Group Theory and Computation. Edited by: Curran MPJ. Academic Press, London; 1977.
Sims CC: Graphs and finite permutation groups. Math. Z. 1967, 95: 76–86. 10.1007/BF01117534
Kesicioğlu, Y: Suborbital graphs of Hecke groups {\Gamma}^{3} and {G}_{5}. PhD thesis, Karadeniz Technical University, Trabzon, Turkey (2011)
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Akbaş, M., Kör, T. & Kesicioğlu, Y. Disconnectedness of the subgraph {F}^{3} for the group {\Gamma}^{3}. J Inequal Appl 2013, 283 (2013). https://doi.org/10.1186/1029242X2013283
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DOI: https://doi.org/10.1186/1029242X2013283