OOn the number of fixed points of the map γ Niccol`o Castronuovo
Dipartimento di Matematica, Universit`a di Bologna, Bologna, 40126, ITALY
Abstract
We recursively define a sequence { F n,k } n,k ∈ N and we prove thatsuch sequence contains only the symbols { , } . We investigate somenumber-theoretic properties of such sequence and of the way it can begenerated. The number F n can be interpreted as the number of fixedpoints of semilength n of the map γ introduced in [2]. Our resultspartially answers conjectures posed in [4]. In this paper we consider an infinite (0 , F defined in the followingway. Let F := [ F n,k ] n ≥ k ≥ be the doubly-infinite matrix all of whose entriesare equal to 0 . Apply to F the following step:Step 0 Set F , = 1 . For all i ≥ F the following step:Step i For each pair ( n, k ) such that the entry F n,k changes its value in Step i − , increase F n + k,k and F n +1 − k, n +1 − k by 1 .F is the matrix obtained in this way.We say that an entry F n,k of matrix F is created at Step i if F n,k > F it changes its value during Step i. It is trivial to verify that F ,k > k = 0 and that F n,k = 0if k > n. Hence the matrix F is lower triangular and { F n,k } n ≥ k ≥ can bethought as a doubly-indexed sequence.The matrix F is related to the map γ, a bijection defined over the set ofDyck words of semilength n. This map and its properties are introduced in[2] and further studied in [3] and [4]. This last paper, in particular, dealswith the characterization of the fixed points of γ. a r X i v : . [ m a t h . N T ] F e b ore precisely, F n,k is equal to the number of Dyck words of semilength n, with principal prefix of length k and fixed under the action of γ. The factthat there is at most one of such words (see [2]) implies that the matrix F is a0 − n, F n := (cid:80) k F n,k , is the total number of Dyck words of semilength n fixed by γ (see [2] for the main definitions).The reason for which we do not reintroduce the definition of the map γ is that the sequence { F n,k } n,k can be defined implicitly as above (see [4]).Hence all the results of the paper can be stated in a number-theoretic formwithout appealing to the original definition of F n . The first few rows of the matrix are reported below (the elements abovethe main diagonal are all zeros and are not indicated). F = 10 10 1 10 1 0 10 1 1 0 10 1 0 0 1 10 1 1 1 0 0 10 1 0 0 0 0 0 10 1 1 0 1 0 1 1 10 1 0 1 1 0 0 1 0 10 1 1 0 0 1 0 0 0 0 10 1 0 0 0 0 0 0 1 0 1 10 1 1 1 1 0 1 0 0 0 0 0 10 1 0 0 1 0 0 0 0 0 1 1 0 10 1 1 0 0 0 1 1 0 0 1 0 1 1 10 1 0 1 0 1 0 1 0 0 0 1 0 0 0 10 1 1 0 1 0 0 1 1 0 0 0 0 0 0 0 10 1 0 0 1 0 0 0 0 0 0 0 1 1 0 0 1 1 . . . The first values of the sequence { F n } n ≥ are1 , , , , , , , , , , , , , , , , , . . . We study the matrix F and the sequence F n , and investigate their prop-erties.Our results partially answer the following two conjectures posed to theauthor by Cori [4]. 2 onjecture 1.1. F n ≥ for all n > . Conjecture 1.2. lim n →∞ F n = ∞ . In particular we answer in the affirmative Conjecture 1.1 and give someresults toward the solution to Conjecture 1.2. F and a free subsemigroup of SL (3 , Z ) . Identify the entry F i,j of the matrix F with the integer vector with coor-dinates ( i, j ) in the Z × Z lattice plane. It follows immediately from thedefinition of the matrix F, that, if i, j > , F i,j > i, j ) can be reached iteratively applying to the vector (1 ,
1) the followingaffine transformations (in arbitrary order)ˆ G : (cid:18) xy (cid:19) → (cid:18) (cid:19) · (cid:18) xy (cid:19) , and ˆ S : (cid:18) xy (cid:19) → (cid:18) − − (cid:19) · (cid:18) xy (cid:19) + (cid:18) (cid:19) . The group generated by the affine transformations ˆ G and ˆ S can be iden-tified with the subgroup (cid:104) S, G (cid:105) of SL (3 , Z ) generated by matrices G = and S = − − . In this identification, the actions of ˆ S and ˆ G on the lattice point ( x, y ) ∈ Z correspond to the actions of S and G, respectively, on the lattice point( x, y, ∈ Z . Hence we can consider directly the action of the group over the set Z . If w ∈ < S, G >, we denote by w ( x, y ) the image ofthe vector ( x, y ) under this action.Now we study some properties of the group < S, G > and of matrices S and G. Notice that the group generated by S and G is not free. In fact we have( GS − ) = . If we restrict our attention to the monoid generatedby S and G, it is free. 3 heorem 2.1. The monoid H generated by S and G is free. The action of H on Z is free i.e. for all ( x, y ) ∈ Z if w ( x, y ) = w (cid:48) ( x, y ) , with w, w (cid:48) ∈ H, then w = w (cid:48) . For the proof it will be useful a version of the so called
Ping-Pong lemma or Table-Tennis lemma for semigroups (see e.g. [5][p. 188]), that we reporthere.
Lemma 2.2.
Let Γ be a group acting on a set X. Assume that there exist γ , γ ∈ Γ and X , X ⊆ X such that X ∩ X = ∅ , γ ( X ∪ X ) ⊆ X and γ ( X ∪ X ) ⊆ X . Then the semigroup generated in Γ by γ and γ is free. Now we proceed to the proof of Theorem 2.1.
Proof.
For the first part, we apply the previous lemma with γ := G, γ := S and Γ := H. As described above, H acts in the standard way on the set X = Z , and, more generally, on R . Notice that the only fixed point underthe action of H on R is ( − . , , hence any line through this point is mappedonto another such line. Moreover, every point of the line y = x + is fixedby S and every point of the line y = 0 is fixed by G. We consider the following disjoint subsets of X : X := { ( x, y ) ∈ Z | x > − , < y < x } X := { ( x, y ) | x > − , x < y < x + 12 } Those subsets are depicted in the figure below. xy X X
4t is trivial to verify that G ( X ∪ X ) ⊆ X , S ( X ∪ X ) ⊆ X . Now we prove the second part of the theorem. We want to show that ∀ ( x, y ) ∈ Z , if w ( x, y ) = w (cid:48) ( x, y ) , with w, w (cid:48) ∈ H, then w = w (cid:48) . We proceed by induction on the minimum of the lengths of w and w (cid:48) , thoughtas words in the letters S and G,m := min {| w | , | w (cid:48) |} . If m = 0 then one of the two words is the identity. The equation w ( x, y ) =( x, y ) , with w different from the identity is clearly impossible since both S and G increase the abscissa of the point on which they act. Suppose theassertion true for all values of m up to N. If m = N + 1 and the first letterof w and w (cid:48) is the same, e.g. the letter G, we have w = G ˆ w, w (cid:48) = G ˆ w (cid:48) and G ˆ w ( x, y ) = G ˆ w (cid:48) ( x, y ) . This implies ˆ w ( x, y ) = ˆ w (cid:48) ( x, y ) and hence w = w (cid:48) by the inductive hypothesis. If m > w = G ˆ w and w (cid:48) = S ˆ w (cid:48) , then w ( x, y ) ∈ X and w (cid:48) ( x, y ) ∈ X . Since X ∩ X = ∅ it is impossible that w ( x, y ) = w (cid:48) ( x, y ) . This concludes the proof.As recalled above, there is a bijection between H and the set { ( i, j ) | i, j > , F i,j (cid:54) = 0 } . This bijection maps an element w ∈ H to the pair w (1 , . Hencethe previous theorem leads to the following corollary.
Corollary 2.3.
The matrix F is a 0-1 matrix. The following lemma will be useful in the sequel.
Lemma 2.4.
The matrices S and G satisfy the following identities for all iG i = i
00 1 00 0 1 ,S i = i + 1 − i i i − (2 i − i ,S i G = i + 1 1 i i i , SG ) i = a ( i ) b ( i ) c ( i ) d ( i ) a ( i − b ( i )0 0 1 where • a ( n ) satisfies a ( n ) = 4 a ( n − − a ( n − with a (0) = 1 and a (1) = 3 (it is, up to a shift, sequence A001835 in [7]). • b ( n ) satisfies b ( n ) = 4 b ( n − − b ( n − with b (0) = 0 and b (1) = 1 (itis sequence A001353 in [7]). • c ( n ) satisfies c ( n ) = 5 c ( n − − b ( n −
2) + c ( n − with c (1) = 1 and c ( k ) = 0 for k ≤ (it is sequence A061278 in [7]). • d ( n ) satisfies d ( n ) = 4 d ( n − − d ( n − with d (0) = 0 and d (1) = 2 (it is sequence A052530 in [7]).Proof. The assertions are easily provable by induction.Now consider the subset of Z × Z whose points have equal positive coor-dinates A = { ( k, k ) | k ∈ Z k ≥ } . Clearly A = ∪ i ≥ { S i (1 , } ∪ { (0 , } . Hence, F k,k = 1if k ≥ . Moreover, by the previous lemma, ∪ i ≥ G i ( A ) = { ( kd, d ) | k, d ∈ N , k > } . This implies the following proposition.
Proposition 2.5. If n ≥ and k divides n then F n,k = 1 . The previous proposition implies that F n ≥ τ ( n ) , where τ is the number-of-divisors function (see e.g. [1]). We will see belowthat the previous inequality can be substantially improved.6 A modular recursion
Now we prove a lemma that show how the entries of the matrix F are relatedto each others through a modular recursion of the indices. Lemma 3.1.
Consider the matrix F. Then F n,k = F n − k,h , (1) where h is the remainder of n − k + 1 in the division by k, i.e. h ≡ k mod 2 n − k + 1 and ≤ h ≤ n − k. Equation (1) together with the initial condition F , = 1 characterizes the matrix F. Proof.
As recalled above, if F n,k = 1 then 1 ≤ k ≤ n or k = n = 0 . The lemma is clearly true for the entries of the form F n,n , n ≥ , i.e. thoseobtained applying only the operation S to (1 , . In fact, in this case, F n,n = 1 ,h = 0 , n − k = 0 and F , = 1 . Now choose a pair ( n, k ) with k < n.
We want to show that ( n, k ) = S i G ( n − k, h ) , i ≥ . By Lemma 2.4, ( n, k ) = S i G ( x, y ) if and only if n =(2 i + 1) x + y + i and k = 2 ix + y + i. Hence n − k = x, n − k + 1 = 2 x + 1and h = y. Note that F n,k = 1 with k < n if and only if ( n, k ) = w (1 ,
1) where w ∈ H is a non-empty word with at least one letter equal to G. Let w = S i l GS i l − G . . . S i GS i , where l j ≥ ≤ j ≤ l. Set( x, y ) := S i l − G . . . S i GS i (1 , . Then ( n, k ) = S i l G ( x, y ) . Thus F n,k = F x,y . F Theorem 4.1.
For every n ≥ and for every x such that • ≤ x ≤ n − and • there exists a divisor h of x with x + 1 | n − h + 1 , it holds F n,n − x = 1 . roof. If x = 0 the proposition is trivial. Let x ≥ F n,n − x = F x,h where h ≡ n − x mod 2 x + 1 . Now,by Proposition 2.5, we have that F x,h = 1 if h | x. Moreover h ≡ n − x mod 2 x + 1 if and only if 2 x + 1 | n − x − h. Since 2 x + 1 is odd this lastcondition is equivalent to 2 x + 1 | n − x − h and this in turn is equivalentto 2 x + 1 | n − h + 1 . Hence, if h | x and 2 x + 1 | n − h + 1 , we have F n,n − x = 1 . We denote by a n the number of entries F n,n − x of row n of the matrix F such that 0 ≤ x ≤ n − h of x with 2 x +1 | n − h +1 . Clearly a n ≤ F n , for every n. Theorem 4.2.
We have a n ≥ max { τ ( n ) , τ (2 n − , τ o ( n + 1) } , more precisely a n ≥ τ ( n ) + τ (2 n −
1) + τ o ( n + 1) − − δ n ≡ − δ n ≡− for every n ≥ , where τ is the number-of-divisors function and τ o is thenumber-of-odd-divisors function.Proof. Let n − x be a divisor of n. Then n = ( n − x ) j, where j is an integer.Then 2 x + 1 = 2 n − nj + 1 . Since nj = n − x is a divisor of n, it divides also x, hence F n,n − x is one of those entries counted by a n . Hence a n ≥ τ ( n ) . Noticethat, in this case, x = 0 or n − x ≤ n . From the previous theorem, taking h = 1 , it follows, in particular, that F n,n − x = 1 if 2 x + 1 is a divisor of 2 n − . Hence a n ≥ τ (2 n −
1) and x = n − x +1 ≤ n − . In the last case, x ≤ n − and n − x ≥ n +12 . As a consequence a n ≥ τ ( n )+ τ (2 n − − , where the 2 in the right-hand side takes into accountthe fact that F n,n and F n, have been counted two times.Similarly, taking h = x in the previous theorem, it follows that F n,n − x = 1when x is such that 2 x + 1 | n − x + 1 . This is equivalent to 2 x + 1 | n + 2which, since 2 x + 1 is odd, is in turn equivalent to 2 x + 1 | n + 1 . Hence F n,n − x = 1 if 2 x + 1 is an odd divisor of n + 1 and a n ≥ τ o ( n + 1) . Here x = n or 2 x + 1 ≤ n +12 . In the last case x ≤ n − and n − x ≥ n +14 . Notice that if 2 x + 1 divides n + 1 and 2 n − n + 1) − (2 n −
1) = 3 hence x = 0 or x = 1 . If x = 1 , x + 1 = 3 divides n + 1 if andonly if it divides also 2 n − . a n ≥ τ ( n ) + τ (2 n −
1) + τ o ( n + 1) − − δ n ≡ − δ n ≡− where, in the right hand side, the 3 takes into account the fact that F n,n hasbeen counted three times and F n, has been counted two times, the δ n ≡ takes into account the fact that F n, n has been counted two times if n is evenand δ n ≡− takes into account the fact that F n,n − has been counted twotimes if n + 1 is divisible by 3.Since a number has 1 as its only odd divisor if and only if it is a power of2, the previous theorem shows that Conjecture 1.1 is proved for every n > p of the form p = 2 q − p − F n ≥ n (cid:54) = 2 q − p of the form 2 q − Mersenne prime (see e.g.[1]). The Mersenne primes p such that 2 p − p has the form 2 q − , then also q is prime. Theorem 4.3.
The number a n is one plus the number of solutions of theDiophantine equation n = 2 xyz + yz + x + y, with y, z ≥ and x ≥ . (2) Moreover, a n = 1 + (cid:88) ≤ j
Let n be fixed. The number of elements of the form F n,n − x with x (cid:54) = 0 , such that there exists an h with h | x and 2 x + 1 | n − h + 1 is equal to thenumber of solutions h, k, j ≥ n − h + 1 = j (2 hk + 1) (3)i.e. 2 n = 2 jhk + 2 h + j − . In this equation j must be odd and hence j = 2ˆ j + 1 . So we get the equation n = 2ˆ jhk + hk + h + ˆ j, with h, k ≥ j ≥ . (4)Hence a n is equal to the number of solutions to this equation increased byone since we have to take into account the case with x = 0 . n − ˆ j = h ((2ˆ j + 1) k + 1) . Hence, if n and ˆ j ≥ h, k ≥ d =(2ˆ j + 1) k + 1 > n − ˆ j congruent to one mod 2ˆ j + 1 . Now we prove Conjecture 1.1.
Theorem 4.4. F n ≥ for all n > . Proof.
As recalled above, the assertion follows from the previous results forevery n > p of the form p = 2 q − , q a prime, suchthat 2 p − p be a number with these properties such that F p = 2 . Since F n ≥ a n for every n and since a n ≥ a p = 2 . Moreover a p ≥ D ( p ) + D ( p − . Clearly D ( p ) = 1 . Hence D ( p −
1) = 0 . We want to show that it is impos-sible. The integer p − d > d ≡ p − r p i where p is a prime with p ≡ − , i = 0 or i = 1and r ≥ . In fact, if between the prime factors of p − − , then p − . Since p = 2 q − , we have2(2 q − −
1) = 3 r p i which implies p = 2 , i = 1 and 2 q − − r = 1 . The only solutions to theprevious equation in positive integers q and r are q = 3 and r = 1 . In fact, in2002 , Mih˘ailescu proved that the only solution to the Diophantine equation x a − y b = 1 , with x, y, a, b > x = 3 , , y = 2 , a = 2 , b = 3 , thus solving thecelebrated Catalan’s conjecture (see [6], the solution of the particular case ofthis conjecture with x = 2 and y = 3 is attributed to Gersonides).If q = 3 and r = 1 , we get p = 7 , whereas we are considering a number p > . This concludes the proof. 10
Other properties of the matrix F In this section we investigate further properties of matrix F. Theorem 5.1.
Matrix F has periodic diagonals. In particular, the a -thsubdiagonal has period a + 1 . Matrix F has periodic columns. In particular, column a has period a .Proof. To prove the first part of the theorem, fix a ∈ N and consider theelements F n,n − a , n > a, of the matrix F. These elements constitute the a -thsubdiagonal. By Lemma 3.1 we have F n,n − a = 1 if and only if F a,h where h ≡ n − a mod 2 a + 1 . Since a is fixed, this proves that the a -th subdiagonalis periodic with period 2 a + 1 . To prove the second part, fix a ∈ N and consider the elements F n,a , n ≥ a. These elements constitutes the a -th column of F. By Lemma 3.1 F n,a = 1 ifand only if F n − a,h where h ≡ a mod 2 n − a + 1 . If n is sufficiently large,this implies h = a and hence the a -th column has period a. Theorem 5.2.
For each quadruple ( k, d, i, j ) ∈ N , k > , we have F kd + id + j ( i +1)(2( k − d +1) ,d + j (2( k − d +1) = 1 . Proof.
Consider the set A := { ( kd, d ) | k, d ∈ N , k > } . We have F x,y = 1for all ( x, y ) ∈ A by Proposition 2.5. By Lemma 3.1, we have ∪ i,j ≥ G i S j ( A ) = { kd + id + j ( i + 1)(2( k − d + 1) , d + j (2( k − d + 1)) | k, d, j, l ∈ N , k > } . This concludes the proof.The previous theorem implies the following Corollary.
Corollary 5.3.
For each t ∈ N , F t +2 , t +2 = 1 and F t +4 , t +2 = 1 . Proof.
For the first part, substitute i = 0 , d = 1 , j = 1 and k − t in theprevious theorem. For the second part, by Lemma 3.1, we get F j +4 , j +2 = 1if and only if F j +2 ,h = 1 where h ≡ j + 2 mod 6 j + 4 i.e. h = 2 j + 2 . Since F j +2 , j +2 = 1 by the first part, we get the assertion.11 Conjectures about the matrix F In this section we formulate others conjectures about F and explain theirrelation with Conjecture 1.1. To this aim we need to introduce the notion oftrack vector.Denote by φ the map that associates the pair of integers ( n, k ) the pair( n − k, h ) where h ≡ k mod 2 n − k + 1 and 0 ≤ h ≤ n − k. It followsfrom the proof of Lemma 3.1 that φ ( n, k ) = ( n − k, h ) if and only if thereexists an i ∈ N such that S i G ( n − k, h ) = ( n, k ) . As a consequence, ( n, k ) = S i l GS i l − G . . . S i G ( m, m ) , where i , . . . , i l , m ∈ N , if and only if φ i l ( n, k ) =( m, m ) . The number of operations of the form S i G needed to get a given elementis related to the breadth of an element. Here we recall the definition ofbreadth and of track vector. Following [4], the track vector of an element( n, k ) is defined as the vector ( i + 1 , . . . , i l + 1) where( n, k ) = S i l GS i l − G . . . S i GS i (1 , . In this case, the breadth of ( n, k ) is equal to l. Since S i (1 ,
1) = ( i , i ) , thebreadth of ( n, k ) is equal to the number of times it is necessary to apply themap φ to ( n, k ) to get an entry of the form ( m, m ) . Theorem 6.1.
The elements of F appearing in Theorem 4.1 with x (cid:54) = 0 , i.e. the elements F n,n − x , x (cid:54) = 0 , such that there exists an h with h | x and x + 1 | n − x + 1 are precisely the elements in row n with track vector ofthe form ( a, (1) p , b ) , with a, b, p ≥ . Moreover, if n is fixed, the number of elements F n,n − x , x (cid:54) = 0 , such thatthere exists an h with h | x and x + 1 | n − x + 1 is equal to the number ofelements in row n with track vector ((1) p , a, (1) q ) , where p, q ≥ , a ≥ . Proof.
By the previous observations and by the proof of Theorem 4.1, wehave that the elements F n,n − x , x (cid:54) = 0 , such that there exists an h with h | x and 2 x + 1 | n − x + 1 are precisely the elements of the form S b − G p +1 S a − (1 , a, b, p ≥ a, (1) p , b ) . As in the previous Sections, we denote by a n the number of elements inrow n of the form described in the proposition. It follows from Theorem 4.3that a n − n, r ) withtrack vector ((1) p , a, (1) q ) is equal to ap (2 q + 2) + a ( q + 1) − p (2 q + 1) = 2 ap ( q + 1) + a ( q + 1) − p ( q + 1) + p. n is fixed and we substitute q + 1 = ˆ q the number of such elements is equalto the number of solutions of the equation n = 2 ap ˆ q + a ˆ q − p ˆ q + p = 2 p ˆ q ( a −
1) + ( a − q + ˆ q + p. Set ˆ a = a − . We get the equation n = 2 p ˆ q ˆ a + ˆ a ˆ q + ˆ q + p, where ˆ a, ˆ q ≥ , and p ≥ . This equation coincides with Equation 2.
Corollary 6.2.
The number of elements in row n with track vector of theform ( a, (1) p , b ) , with a, b, p ≥ is equal to the number of elements in row n with track vector of the form ((1) p , a, (1) q ) , where p, q ≥ , a ≥ . Moreoverthis common value is a n − . We conjecture that the sequence a n tends to infinity. Conjecture 6.3. a n → ∞ , more precisely a n ≥ (cid:98) log( n ) (cid:99) − . The inequality of the conjecture originates from numerical evidences. No-tice that it is not true that a n ≥ (cid:98) log( n ) (cid:99) . In fact a = 8 but (cid:98) log(18007) (cid:99) =9 . Notice that the previous conjecture implies Conjecture 1.1 and Theorem4.4.Moreover, by the paper [4], it follows that the set of elements in row n withtrack vector ((1) p , a, (1) q ) , where p, q ≥ , a ≥ elementary partitions , which in turn are a subset of the set ofpartitions with n subpartitions. Hence, if s n is the number of partitions with n subparitions, the previous conjecture implies also that s n → ∞ . Noticethat sequence { s n } n ∈ N is sequence A116473 in [7], where it is reported thatit is conjectures that s n → ∞ . Thus Conjecture 6.3 would imply also thisconjecture present in [7].Another conjecture suggested by strong numerical evidences is the fol-lowing. Conjecture 6.4.
Let c n be the number of elements in row n with breadth . Then c n → ∞ . We conclude this Section with a conjecture about the possible positionsof elements created at Step i inside the matrix F. Conjecture 6.5.
Consider the elements of matrix F that are created atStep i. Let r i be the maximal index of a row of matrix F containing suchan element. Then { r j +1 } j ∈ N is sequence A061278 in [7] and { r j } j ∈ N issequence A001571 in [7]. Moreover, if i is even, row r i of F contains two lements created at Step i. On the other hand, if i is odd, row r i contains oneelement created at Step i. Example 6.6.
Let i = 6 . The maximal row containing an element created atStep 6 is row 35. Notice that 35 is the third element (avoiding the first zero)of sequence A001571 in [7]. Moreover, row 35 of F contains two elementscreated at Step 6, precisely F , and F , . F n We conclude the paper improving the best known upper bound for F n . Corollary 5 in [2] states that F n ≤ min { n, φ (2 n + 1) } where φ is the Eulertotient function, see e.g. [1]. In fact F n ≤ n since the matrix F is lowertriangular. Moreover, it is shown in [2], that if F n,k = 1 then gcd( k, n + 1) =1 (this can also be shown easily using the recursive definition of F used inthis paper). Hence F n ≤ φ (2 n + 1) . It is possible to slightly improve this bound in the following way. Considerthe same notation of the proof of Theorem 2.1. Since G (1 ,
1) = (2 , ∈ X ,S (1 ,
1) = (2 , ∈ X , G ( X ∪ X ) ⊆ X and S ( X ∪ X ) ⊆ X we have thatevery element ( n, k ) , n > , such that F n,k = 1 is contained in X ∪ X . Inparticular, k < n + or k > n + . Hence F n ≤ n − (cid:18) n − (cid:18) n (cid:19) − (cid:19) = 5 n − . Thus we get F n ≤ min (cid:26) n − , φ (2 n + 1) (cid:27) . References [1] T.M. Apostol.
Introduction to Analytic Number Theory . UndergraduateTexts in Mathematics. Springer New York, 1998.[2] M. Barnabei, F. Bonetti, N. Castronuovo, and R.Cori. Some permuta-tions on Dyck words.
Theoretical Computer Science , 635:51–63, 2016.[3] N. Castronuovo, R. Cori, and S. Labb´e. A permutation on words in atwo letter alphabet. In
Combinatorics on Words - 11th InternationalConference, WORDS 2017, Montreal, QC, Canada, September 11-15,2017, Proceedings , Lecture Notes in Computer Science, pages 240–251.Springer, 2017. 144] R. Cori, A. Frosini, G. Palma, E. Pergola, and S. Rinaldi. On doublysymmetric dyck words.
Under review .[5] P. de la Harpe.
Topics in Geometric Group Theory . Chicago Lectures inMathematics. University of Chicago Press, 2000.[6] R. Schoof.