aa r X i v : . [ m a t h . C O ] S e p Abelian Ramsey Length and Asymptotic Lower Bounds ∗ Vincent Jugé † This technical note aims at evaluating an asymptotic lower bound on abelian Ramsey lengths obtained by Tao in [1]. We first provide the minimal amount of background necessary to defineabelian Ramsey lengths, and indicate the lower bound of Tao. We then focus on evaluating thislower bound.
Let A and V be two alphabets. A word on A is a finite sequence a = a a · · · a k of elements of A . The elements a i are called the letters of the word a , and the integer k is the length of a . Forall elements α ∈ A , we denote by | a | α the cardinality of the set { i : a i = α } , i.e. number ofoccurrences of the letter α in the word a . We also denote by A ∗ the set of all words on A . Thewords a i a i +1 · · · a j with ≤ i ≤ j ≤ k , as well as the empty word, are called factors of a .Consider now a word a = a a · a k in A ∗ and a word p = p p . . . p ℓ in V ∗ . We say that a contains p in the abelian sense if there exist non-empty words π , π , . . . , π ℓ in A ∗ such that theconcatenated word π π . . . π ℓ is a factor of a , and such that, for all integers i, j and all letters α ∈ A , if p i = p j , then | π i | α = | π j | α . For instance, the word programmable contains the word aab in the abelian sense, as can be seen by considering the words π = am , π = ma and π = ble .From this point on, we consider the infinite alphabet V = { v i : i ∈ N } , where N is the set ofpositive integers, and we define the Zimin patterns Z i inductively by Z = v and Z i +1 = Z i v i +1 Z i .It turns out that, for all integers i, m ≥ and all alphabets A of cardinality m , there exists aninteger L ab ( m, Z i ) such that all words a ∈ A ∗ with length at least L ab ( m, Z i ) contain the word Z i in the abelian sense.For all integers m ≥ , Tao proves in [1] that L ab ( m, Z i ) ≥ (1 + ε m ( i )) p K ( m, i ) for all i ≥ ,where ε m is a function such that lim + ∞ ε m = 0 and K ( m, i ) is defined as K ( m, i ) = 2 Q i − j =1 S ( m, j ) − ,where S ( m, k ) = ∞ X ℓ =1 T ( m, k, ℓ ) and T ( m, k, ℓ ) = 1 m kℓ X i + ... + i m = ℓ (cid:18) ℓi . . . i m (cid:19) k .Yet, in order to obtain actual lower bounds on L ab ( m, Z i ) , it remains to evaluate the asymp-totical behavior of K ( m, i ) . We evaluate K ( m, i ) up to a multiplicative constant that does notdepend on m or i . More precisely, we prove the following inequalities, which hold for all m ≥ and i ≥ : m i m i +1 ≥ K ( m, i ) ≥ m i m i +1 . ∗ This work is supported by ERC EQualIS (308087). † LSV, CNRS & ENS Cachan, Univ. Paris-Saclay, France Auxiliary inequalities
Before evaluating the lower bound K ( m, i ) , we prove a series of six inequalities that we will usesubsequently. We first study the function f x : y y ln(1 + x/y ) for x, y > . An asymptoticevaluation proves that lim + ∞ f x = x . Furthermore, we compute that f ′′ x ( y ) = − x y ( x + y ) < for y > . It follows that x > f x ( y ) or, equivalenlty, that (1 + x/y ) y < e x for all x, y > . (1)We perform a similar study with the function g : y ( y + 1 /
2) ln(1 + 1 /y ) for y > . Wefind that lim + ∞ g = 1 and that g ′′ ( y ) = y ( y +1) > for y > . It follows that g ( y ) > or,equivalently, that (1 + 1 /y ) y +1 / > e for all y > . (2)Again, we consider the function h : y y ) + ln(2) − ( y −
1) ln(2 π ) for y > , as well asthe real constant λ = π / < . We find that h ′ ( y ) = y − ln(2 π ) < when y ≥ and that exp( h (4)) = π = λ , and it follows that y ≤ λ (2 π ) y − for all y ≥ . (3)Similarly consider the function h : y y ) + ln(2) − ( y −
1) ln(2 π ) for y > . We find that h ′ ( y ) = y − ln(2 π ) < when y ≥ and that exp( h (7)) = π < , and it follows that y ≤ (2 π ) y − for all y ≥ . (4)Then, we set Z ( x ) = √ πx x +1 / e − x for all x ≥ . We prove below that ( a + b )! a ! b ! ≤ Z ( a + b ) Z ( a ) Z ( b ) for all integers a, b ≥ . (5)We study the functions F : ( a, b ) ( a + b )! Z ( a ) Z ( b ) Z ( a + b ) a ! b ! and G : ( a, b ) F ( a + 1 , b ) F ( a, b ) . We computethat G ( a, b ) = (cid:18) a + ba + b + 1 (cid:19) a + b +1 / (cid:18) a + 1 a (cid:19) a +3 / ≥ G ( a, b ) a + b +2 , where G ( a, b ) = 2 a + b + 2( a + b + 1 / a + b +1 a + b + ( a + 3 / aa +1 (by geometric-harmonic inequality) = 1 + b − a + 2 b + 1)( a + b + 1)( a + 1) + (2 a + 3) a ( a + b ) . and since b ≥ , it follows that G ( a, b ) ≥ G ( a, b ) a + b +2 ≥ , i.e. that F ( a, b ) ≤ F ( a + 1 , b ) . Since F ( a, b ) = F ( b, a ) for all a, b ≥ , we derive immediately that F ( a, b ) ≤ F ( a, b + 1) ≤ F ( a + 1 , b + 1) for all integers a, b ≥ . Moreover, Stirling’s approximation formula states that a ! ∼ Z ( a ) when a → + ∞ . This proves that lim α,β → + ∞ F ( α, β ) = 1 , and it follows that F ( a, b ) ≤ for all a, b ≥ ,which is indeed equivalent to the inequality (5).As a corollary, observe that, for all integers i , . . . , i m ≥ and using inequality (5), we alsohave ( i + . . . + i m )! i ! . . . i m ! = m Y j =2 ( i + . . . + i j )!( i + . . . + i j − )! i j ! ≤ m Y j =2 Z ( i + . . . + i j ) Z ( i + . . . + i j − ) Z ( i j ) , from which follows our last auxiliary inequality: ( i + . . . + i m )! i ! . . . i m ! ≤ Z ( i + . . . + i m ) Z ( i ) . . . Z ( i m ) for all integers i , . . . , i m ≥ . (6)2 Evaluating K ( m, i ) We first evaluate T ( m, k, ℓ ) when ℓ = 1 . Here, instead of considering a tuple of non-negativeintegers ( i , . . . , i m ) that sum up to ℓ , we might directly consider the unique integer j ∈ { , . . . , m } such that i j = 1 . Moreover, for each tuple ( i , . . . , i m ) , the multinomial coefficient (cid:0) ℓi ... i m (cid:1) isequal to . It follows that T ( m, k,
1) = m − k P mj =1 m − k , from which we derive the inequalities S ( m, k ) ≥ T ( m, k,
1) = m − k and K ( m, i ) ≤ i − Y j =1 m j − = 2 m i m i +1 .Then, we investigate lower bounds of K ( m, i ) , i.e. upper bounds of T ( m, k + 1 , ℓ ) and of S ( m, k + 1) when m ≥ and k ≥ . Consider some integer ℓ ≥ , and let us write ℓ = am + b ,with a ≥ and ≤ b ≤ m . In addition, let us set V ( m, a, b ) = max i + ... + i m = ℓ (cid:18) ℓi . . . i m (cid:19) and U ( m, a, b ) = 1 m ℓ V ( m, a, b ) .We first observe that T ( m, k + 1 , ℓ ) = 1 m ( k +1) ℓ X i + ... + i m = ℓ (cid:18) ℓi . . . i m (cid:19) k +1 ≤ m ( k +1) ℓ X i + ... + i m = ℓ (cid:18) ℓi . . . i m (cid:19) V ( m, a, b ) k ≤ U ( m, a, b ) k T ( m, , ℓ ) ≤ U ( m, a, b ) k . (by Newton multinomial identity)Since the inequality ( x + 1)!( y − ≥ x ! y ! holds for all integers x ≥ y , it also follows that U ( m, a, b ) = ( am + b )! m am + b ( a + 1)! b a ! m − b .We compute immediately that U ( m, , b ) = m − b b ! = m − if b = 1 , and that U ( m, , b ) ≤ m − if ≤ b ≤ m . When a ≥ , we further compute that U ( m, a, b ) ≤ Z ( am + b ) m am + b Z ( a + 1) b Z ( a ) m − b (using inequality (6)) ≤ √ am + b (2 π ) ( m − / a m/ (1 + b/am ) am (1 + b/am ) b (1 + 1 /a ) ( a +3 / b ≤ √ am (2 π ) ( m − / a m/ (1 + b/am ) am (1 + 1 /a ) b (1 + 1 /a ) ( a +3 / b (since b ≤ m ≤ am ) ≤ √ m (2 aπ ) ( m − / (1 + b/am ) am (1 + 1 /a ) ( a +1 / b ≤ √ m (2 aπ ) ( m − / e b (1 + 1 /a ) ( a +1 / b (using inequality (1)) ≤ √ m (2 aπ ) ( m − / (cid:18) e (1 + 1 /a ) a +1 / (cid:19) b ≤ √ m (2 aπ ) ( m − / . (using inequality (2))3onsequently, since k ≥ , we find that S ( m, k + 1) ≤ ∞ X a =0 m X b =1 U ( m, a, b ) k = U ( m, , k + m X b =2 U ( m, , b ) k + ∞ X a =1 m X b =1 U ( m, a, b ) k ≤ m k + 2( m − m k + m ∞ X a =1 √ m (2 aπ ) ( m − / ! k ≤ m k + 2 mm k + m (2 m ) k/ (2 π ) k ( m − / ζ ( k ( m − / . If we set P ( m, k ) = 1 + 2 m − k + m k +1 (2 m ) k/ (2 π ) k ( m − / ζ ( k ( m − / ,then it follows that S ( m, k + 1) ≤ m k P ( m, k ) and therefore that K ( m, i ) ≥ i − Y j =1 m j − P ( m, j − ≥ P ∞ ( m ) m i m i +1 ,where P ∞ ( m ) is the infinite product ∞ Y j =1 P ( m, j − . It remains to prove that P ∞ ( m ) ≤ .We fisrt assume that ≤ m . For k ≥ , we compute that P ( m, k ) = 1 + 2 m − k + m k +1 (2 m ) k/ (2 π ) k ( m − / ζ ( k ( m − / m − k + m − k (cid:18) m (2 π ) m − (cid:19) k/ ζ ( k ( m − / ≤ m − k + m − k ζ ( k ( m − / (using inequality (4)) ≤ m − k (since ζ ( k ( m − / ≤ ζ (3) ≤ ) ≤ exp(4 m − k ) , (since x ≤ exp( x ) for all x ∈ R )from which we deduce that P ∞ ( m, ≤ , whence ln( P ∞ ( m )) = ∞ X j =1 ln( P ( m, j − ≤ ln( P ( m, ∞ X j =2 m − j ≤ ln(5) + 4 ∞ X j =0 m − − j ≤ ln(5) + 4 m ( m − ≤ ln(5) + 221 ≤ ln(42) . (since ≤ m )Then, we assume that ≤ m ≤ . Again, for k ≥ , we compute that P ( m, k ) = 1 + 2 m − k + m k +1 (2 m ) k/ (2 π ) k ( m − / ζ ( k ( m − / m − k + m (cid:18) m (2 π ) m − (cid:19) k/ ζ ( k ( m − / ≤ m − k + mζ ( k ( m − / λ k (using inequality (3)) ≤ m − k + 3 mλ k ( ζ ( k ( m − / ≤ ζ (3 / ≤ ) ≤ mλ k (since m − ≤ ≤ λ ) ≤ exp(5 mλ k ) (since x ≤ exp( x ) for all x ∈ R )4urthermore, explicit computations in each of the cases m = 4 , m = 5 and m = 6 indicate that Q j =1 P ( m, j − ≤ . Hence, we conclude that ln( P ∞ ( m )) ≤ ln(41) + ∞ X j =5 ln( P ( m, j − ≤ ln(41) + 5 m ∞ X j =5 λ j − ≤ ln(41) + 5 m ∞ X j =0 λ j ≤ ln(41) + 5 mλ − λ ≤ ln(41) + 30 × ≤ ln(42) . (since m ≤ and λ < ) References [1] J. Tao. Pattern occurrence statistics and applications to the Ramsey theory of unavoidablepatterns.