Two integer sequences related to Catalan numbers

0.15 MB·English

by  Michel Lassalle

Two integer sequences related to Catalan numbers
Michel Lassalle
Centre National de la Recherche Scientifique
2
Institut Gaspard-Monge, Universit´e de Marne-la-Vall´ee
1
0 77454 Marne-la-Vall´ee Cedex, France
2 [email protected]

n http://igm.univ-mlv.fr/~lassalle
a
J

0
1 Abstract

] We prove the following conjecture of Zeilberger. Denoting by C the Catalan
O n
number,defineinductively A by (−1)n−1A = C + n−1(−1)j 2n−1 A C and
C n n n j=1 2j−1 j n−j
. an = 2An/Cn. Then an (hence An) is a positive integPer. (cid:0) (cid:1)
h
t
a
m 1 Introduction
[

4 Let z be an indeterminate. For any nonnegative integer r, we denote by Cr(z) the
v Narayana polynomial defined by C (z) = 1 and
5 0
2
r
2
4 Cr(z) = N(r,k)zk−1,
9. Xk=1
0
0 with the Narayana numbers N(r,k) given by
1
:
v 1 r r
i N(r,k) = .
X r(cid:18)k −1(cid:19)(cid:18)k(cid:19)
r
a
There is a rich combinatorial litterature on this subject. Here we shall only refer to [14,
18], [17, Exercice 36] and references therein.
We have C (1) = C , the ordinary Catalan number, since
r r
r
1 2r
C = = N(r,k).
r
r+1(cid:18)r (cid:19)
Xk=1
Moreover[4]theNarayanapolynomialC (z)canbeexpressedintermsofCatalannumbers
r
as
r −1
C (z) = zm(z +1)r−2m−1 C . (1)
r m
(cid:18) 2m (cid:19)
mX≥0
1
The polynomial
(z +1)C (z)−C (z) = (1−r)zr−1 +...
r r+1
can be expressed, in a unique way, in terms of the monic polynomials zmC (z) with
r−2m+1
degree r −m. As a consequence, we may define the real numbers A (r) by
m
r −1
(z +1)C (z)−C (z) = (−z)m A (r)C (z). (2)
r r+1 m r−2m+1
(cid:18)2m−1(cid:19)
mX≥1
Note that the real numbers {A (r),m ≥ 1} depend on r and that we have A (r) = 1.
m 1
By relation (1), the left-hand side may be written as
r −1
− zk(z +1)r−2k C .
k
(cid:18)2k −1(cid:19)
Xk≥0
Similarly the right-hand side becomes
r −1 r −2m
(−1)mzm+n(z +1)r−2m−2n A (r)C .
m n
(cid:18)2m−1(cid:19)(cid:18) 2n (cid:19)
m≥X1,n≥0
Comparing coefficients of zk(z +1)r−2k on both sides yields
r −1 r −1 r −2m
−C = (−1)m A (r)C ,
k m n
(cid:18)2k −1(cid:19) (cid:18)2m−1(cid:19)(cid:18) 2n (cid:19)
mX+n=k
with r ≥ 2k. This is easily transformed to
n
2n−1
C = (−1)j−1 A (r)C .
n j n−j
(cid:18)2j −1(cid:19)
Xj=1
In other words, A := A (r) is an integer independent of r, given by the recurrence
n n
formula
n−1
2n−1
(−1)n−1A = C + (−1)j A C . (3)
n n j n−j
(cid:18)2j −1(cid:19)
Xj=1
This relation can be chosen as a definition, equivalent to (2).
In Section 2 we prove
Theorem 1. The integers {A ,n ≥ 2} are positive and increasing.
n
The values of A for 1 ≤ n ≤ 14 are given by
n
1,1,5,56,1092,32670,1387815,79389310,5882844968,548129834616,62720089624920,
8646340208462880,1413380381699497200,270316008395632253340.
2
At the time the positivity of A was only conjectured, we asked Doron Zeilberger for
n
an advice. He suggested to consider also the numbers a = 2A /C which, by an easy
n n n
transformation of (3), are inductively defined by
n−1
n−1 n+1 a
(−1)n−1a = 2+ (−1)j j .
n
(cid:18)j −1(cid:19)(cid:18)j +1(cid:19)n−j +1
Xj=1
In Section 3 we prove the following conjecture of Zeilberger.
Theorem 2. The numbers {a ,n ≥ 2} are increasing positive integers. They are odd if
n
and only if n = 2k −2 for some k ≥ 2.
The values of a for 1 ≤ n ≤ 16 are given by
n
2,1,2,8,52,495,6470,111034,2419928,65269092,2133844440,83133090480,
3805035352536,202147745618247,12336516593999598,857054350280418290.
Both sequences seem to be new.
Of course since 2A = a C , Theorem 1 is an obvious consequence of Theorem 2.
n n n
However we begin by a direct proof of Theorem 1, which has its own interest. This proof
describes the algebraic framework of our method (the theory of symmetric functions), and
it yields a generating function for the A ’s.
n
The A ’s are also worth a separate study in view of their connection with probability
n
theory, which was recently noticed. Actually Novak [15] observed, as an empirical evi-
dence, that the integers (−1)n−1A are precisely the (classical) cumulants of a standard
n
semicircular random variable. Independently Josuat-Verg`es [9] defined a q-semicircular
law, which specializes to the standard semicircular law at q = 0. He described combinato-
rial properties of its (classical) cumulants, which are polynomials in q having (−1)n−1A
n
as their constant terms. It follows from his work that A may be obtained by a weighted
n
enumeration of connected matchings (equivalently, fixed-point free involutions on 1,...,2n
such that no proper interval is stable). Finding a similar bijective proof for a would be
n
very interesting.
2 Properties of A
n
ThissectionisdevotedtoaproofofTheorem1. WeshallusetworemarksofKrattenthaler
and Lascoux. Define
C 1
C(z) = n zn = zn,
(2n)! n!(n+1)!
Xn≥0 Xn≥0
A
A(z) = (−1)m−1 m zm−1.
(2m−1)!
mX≥1
3
Krattenthaler observed that the definition (3) is equivalent with
d
A(z)C(z) = 2 C(z). (4)
dz
Therefore we have only to show that, if we write
d
C′(z)/C(z) = log(C(z)) = c zm−1,
m
dz
mX≥1
the coefficient c has sign (−1)m−1.
m
Let S denote the ring of symmetric functions [13, Section 1.2]. Consider the classical
bases of complete functions h and power sums p . Denote
n n
H(z) = h zn, P(z) = p zn−1
n n
Xn≥0 Xn≥1
their generating functions. It is well known [13, p. 23] that P(z) = H′(z)/H(z).
Since the complete symmetric functions h are algebraically independent, they may
n
be specialized in any way. More precisely, for any sequence of numbers {c ,n ≥ 0} with
n
c = 1, there is a homomorphism from S into the ring of real numbers, taking h into
0 n
c . Under the extension of this ring homomorphism to formal power series, the image of
n
H(z) is the generating function of the c ’s. By abuse of notation, we write h = c and
n n n
H(z) = c zn.
n≥0 n
TherPefore it is sufficient to prove the following statement.
Lemma 1. When the complete symmetric functions are specialized to
1
h = ,
n
n!(n+1)!
i.e. when H(z) = C(z), the coefficients of P(−z) are all positive.
Lascoux observed an empirical evidence for the following more general result (which
gives Lemma 1 when x = 2).
Lemma 2. Let x be a positive real number and (x) = k (x+i−1) denote the classical
k i=1
rising factorial. When the complete symmetric functionQs are specialized to
1
h = ,
n
n!(x)
n
the coefficients of P(−z) are all positive.
Proof. With this specialization we have
1 zn
H(z) = = F (x;z),
0 1
(x) n!
Xn≥0 n
4
the so called confluent hypergeometric limit function. Since
d 1
F (x;z) = F (x+1;z),
0 1 0 1
dz x
we have
F (x+1;z)
0 1
P(z) = .
x F (x;z)
0 1
The classical Gauss’s continued fraction [20, p. 347] gives the following expression for the
right-hand side
F (x+1;z) 1
0 1
= .
x F (x;z) z
0 1
x+
z
(x+1)+
z
(x+2)+
(x+3)+ ...
This continued fraction may be written as a Taylor series by iterating the usual binomial
formula
k +k −1
(1+f z)−k1 = f k2 1 2 (−z)k2,
1 1
(cid:18) k (cid:19)
kX2≥0 2
where fk2 is itself of the form (c (1 + f z))−k2. This classical method ( [12, Exercise
1 1 2
4.2], [16]) yields
1 −z ki k +k −1
i i+1
P(z) = ,
x (cid:18)(x+i−1)(x+i)(cid:19) (cid:18) k (cid:19)
(k1,kX2,k3,...)Yi≥1 i+1
where the k ’s are nonnegative integers. This proves the lemma.
i
In the previous formula, the contributions with k = 0 and k > 0 are necessarily
i i+1
zero. Hence for n ≥ 2 we have
(−1)n−1 1 ki k +k −1
i i+1
p = , (5)
n
x (cid:18)(x+i−1)(x+i)(cid:19) (cid:18) k (cid:19)
κX∈CnYi≥1 i+1
with C the set of sequences κ = (k ,k ,...,k ) of l ≤ n − 1 (strictly) positive integers
n 1 2 l
summing to n−1. There are 2n−2 such sequences (compositions of n−1). For instance
1 −1 1 1
p = , p = , p = + .
1 x 2 x2(x+1) 3 x2(x+1)2(x+2) x3(x+1)2
To each κ = (k ,k ,...,k ) ∈ C let us associate two elements of C defined by
1 2 l n n+1
κ = (k ,k ,...,k +1) and κ = (k ,k ,...,k ,1). We have
1 1 2 l 2 1 2 l
C = {κ ,κ }.
n+1 1 2
κ[∈Cn
5
Denoting g ,g ,g the respective contributions to (5) of κ,κ ,κ , we have easily
κ κ1 κ2 1 2
−(x+n−1)(x+n)g ≥ g (i = 1,2),
κi κ
the equality occuring for κ = (1,1,...,1) and i = 2. Summing all contributions to (5),
we obtain
−(x+n−1)(x+n)p ≥ 2p .
n+1 n
Since (4) implies
A = (−1)n−12(2n−1)!p | , (6)
n n x=2
substituting x = 2, the above estimate yields A > A , which achieves the proof of
n+1 n
Theorem 1.
Incidentally we have shown that the generating function of the A ’s is given by
n
A 2
(−1)m−1 m zm−1 = .
(2m−1)! z
mX≥1 2+
z
3+
z
4+
5+ ...
3 Properties of a
n
This section is devoted to the proof of Theorem 2. The following lemma plays a crucial
role.
Lemma 3. When the complete symmetric functions are specialized to h = 1/((x) n!)
n n
with x > 0, we have
d
zP(z)2 +z P(z)+xP(z) = 1.
dz
Equivalently for any integer n ≥ 1, we have
n
(n+x)p = − p p . (7)
n+1 r n−r+1
Xr=1
Proof. In view of
F (x+1;z)
0 1
P(z) = ,
x F (x;z)
0 1
we have
d 1 x
P(z)2+ P(z) = F (x+1;z)2+ F (x;z) F (x+2;z)− F (x+1;z)2 ,
dz x2 F (x;z)2 0 1 x+10 1 0 1 0 1
0 1 (cid:16) (cid:17)
hence
d z F (x+2;z) F (x+1;z)
zP(z)2 +z P(z)+xP(z) = 0 1 + 0 1 .
dz x(x+1) F (x;z) F (x;z)
0 1 0 1
6
We have
z
F (x;z)− F (x+1;z) = F (x+2;z),
0 1 0 1 0 1
x(x+1)
since
zk 1 1 z zk−1 1
− = .
k! (cid:18)(x) (x+1) (cid:19) x(x+1)(k −1)!(x+2)
k k k−1
Substituting x = 2, we obtain a second (inductive) proof of Theorem 1.
Theorem 3. The integers {A ,n ≥ 2} are positive, increasing and given by
n
n
n−r +1 2n+1
A = A A .
n+1 r n−r+1
n+2 (cid:18)2r −1(cid:19)
Xr=1
The numbers {a ,n ≥ 2} are positive, increasing and given by
n
n
1 1 n+1 n+1
a = a a .
n+1 r n−r+1
2 n+1(cid:18)r +1(cid:19)(cid:18)r −1(cid:19)
Xr=1
Proof. We substitute (6) into (7). Then we substitute A = a C /2. In both cases we
n n n
proceed by induction.
The positive integer
2 n+1 n+1 n n n n
c = = −
n,r
n+1(cid:18)r+1(cid:19)(cid:18)r−1(cid:19) (cid:18)r −1(cid:19)(cid:18)r(cid:19) (cid:18)r −2(cid:19)(cid:18)r +1(cid:19)
is known [7, 8] to be the number of walks of n unit steps on the square lattice (i.e. either
up, down, right or left) starting from (0,0), finishing at (2r −n−1,1) and remaining in
the upper half-plane y ≥ 0. One has c = c , which corresponds to symmetry with
n,r n,n−r+1
respect to the y-axis.
We need some arithmetic results about c . Given a rational number r, there exist
n,r
unique integers a,b,m with a and b odd such that r = 2ma/b. The integer m is called
the 2-adic valuation of r and denoted ν (r). A classical theorem of Kummer (appeared
2
in 1852, see [19] for an inductive proof) states that the 2-adic valuation of a+b is a
a
nonnegative integer, equal to the number of “carries” necessary to add a and b(cid:0)in b(cid:1)ase 2.
Lemma 4. (i) If n is even, c is even.
n,r
(ii) If n is odd and r even, c is a multiple of 4.
n,r
Proof. (i) Since n+1 is odd, we have
n+1 n+1
ν (c ) = 1+ν +ν ≥ 1.
2 n,r 2 2
(cid:18)(cid:18)r+1(cid:19)(cid:19) (cid:18)(cid:18)r−1(cid:19)(cid:19)
7
(ii) Since r +1 is odd, we have
n n+1
ν (c ) = 1+ν +ν .
2 n,r 2 2
(cid:18)(cid:18)r(cid:19)(cid:19) (cid:18)(cid:18)r−1(cid:19)(cid:19)
Since r − 1 and n − r + 2 are odd, at least one carry is needed to add them in base 2.
Thus ν ( n+1 ) ≥ 1.
2 r−1
(cid:0) (cid:1)
We also consider
2
1 2p
c = .
2p−1,p
p(cid:18)p+1(cid:19)
Lemma 5. (i) If p is even, c is a multiple of 8.
2p−1,p
(ii) If p is odd and p 6= 2k −1 for any k ≥ 1, c is a multiple of 4.
2p−1,p
(iii) If p = 2k −1 for some k ≥ 1, c is odd.
2p−1,p
Proof. (i) We have
2
4p 2p−1
c = .
2p−1,p (p+1)2(cid:18) p (cid:19)
Since p+1 is odd, we have
2p−1
ν (c ) = 3+2ν ≥ 3.
2 2p−1,p 2
(cid:18)(cid:18) p (cid:19)(cid:19)
(ii) Since p is odd, we have
2p
ν (c ) = 2ν .
2 2p−1,p 2
(cid:18)(cid:18)p+1(cid:19)(cid:19)
We may write the binary representations of p+1 and p−1 as
p+1 =c c ...c 10...00,
k+m k+m−1 k+2
p−1 =c c ...c 01...10.
k+m k+m−1 k+2
At least one carry is needed to add them, except if c = c = ... = c = 0, i.e.
k+m k+m−1 k+2
p+1 = 2k, which is excluded. Therefore ν ( 2p ) ≥ 1.
2 p+1
(iii) When p+1 = 2k, the binary represe(cid:0)ntat(cid:1)ions of p+1 and p−1 are 10...00 and
01...10. Therefore ν ( 2p ) = 0.
2 p+1
(cid:0) (cid:1)
Our proof of Theorem 2 also needs the following result.
Theorem 4. The numbers {A ,n ≥ 3} and {C ,n ≥ 3} have the same parity. They are
n n
odd if and only if n = 2k −1 for some k ≥ 2.
8
Proof. It is known [5, 11] that the Catalan numbers {C ,n ≥ 3} are odd if and only if
n
n = 2k −1 for some k ≥ 2. Therefore we have only to prove the first statement.
We proceed by induction on n. We use (3) to express (−1)nA +C , as a signed sum
n n
of the terms
2n−1
y = A C
j j n−j
(cid:18)2j −1(cid:19)
for 1 ≤ j ≤ n−1. Two cases must be considered.
Case 1: n = 2m for some m. Then y and y are obviously odd. On the other hand,
1 n−1
the contributions {y ,2 ≤ j ≤ n−2} are even, because the conditions n = 2m, j = 2k−1
j
and n−j = 2l −1 cannot be simultaneously satisfied if j 6= 1 or n−j 6= 1.
Case 2: n 6= 2m for any m. Then all contributions {y ,1 ≤ j ≤ n−1} are even. Indeed
j
if A and C are odd, we have j = 2k −1 and n−j = 2l −1 with k,l ≥ 2, since j and
j n−j
n−j cannot be equal to 1. But then 2n−1 is even because we have
2j−1
(cid:0) (cid:1)
2n−1 a+b
= ,
(cid:18)2j −1(cid:19) (cid:18) a (cid:19)
with a = 2k+1 − 3 and b = 2l+1 − 2. Then the binary representations of a and b are
1...1101 (k +1 digits) and 1...1110 (l +1 digits). Since k,l ≥ 2, at least one carry is
needed to add them. Summing all contributions, A and C have the same parity.
n n
Proof of Theorem 2. We proceed by induction on n. Our inductive hypothesis is
twofold. We assume that for any m ≤ n, the positive number a is an integer, and
m
that a is odd if and only if m = 2k −2 for some k ≥ 2.
m
If n is even, say n = 2p, a is an integer because by Theorem 3 we have
n+1
p
1
a = c a a ,
2p+1 2p,r r 2p−r+1
2
Xr=1
and we apply Lemma 4 (i). Moreover a is even, since a or a is even for each r.
2p+1 r 2p−r+1
If n is odd, say n = 2p+1, we have
p
1 1
a = c a a + c a2 . (8)
2p+2 2 2p+1,r r 2p−r+2 4 2p+1,p+1 p+1
Xr=1
Each term of the form c or a a is an even integer. This results from Lemma 4
2p+1,r r 2p−r+2
(ii) if r is even, and from the fact that a and a are even, if r is odd. We are left to
r 2p−r+2
consider the term
1
w = c a2 .
p 4 2p+1,p+1 p+1
This is an integer whenever a is even. If a is odd, by our induction hypothesis, one
p+1 p+1
has p+1 = 2k−2 for some k ≥ 2, and we apply Lemma 5 (i). Summing all contributions,
a is an integer.
2p+2
9
It remains to show that a is odd if and only if 2p+2 = 2k−2, i.e. p+1 = 2k−1−1,
2p+2
for some k. We have seen that all terms on the right hand side of (8) are even, except
possibly for w .
p
If p+1 is even, w is even because of Lemma 5 (i). If p+1 is odd and p+1 6= 2k−1 for
p
any k, w is even by Lemma 5 (ii) and because a is even. Finally the only possibility
p p+1
for w to be odd occurs when p + 1 = 2k − 1 for some k. Then in view of Theorem 4,
p
A and C are odd. Hence a2 /4 = (A /C )2 is also odd. Applying Lemma 5
p+1 p+1 p+1 p+1 p+1
(iii), w is odd.
p
4 Extension to type B
Generalized Narayana numbers have been introduced in [6, Section 5.2] in the context of
the non-crossing partition lattice for the reflection group associated with a root system.
Ordinary Narayana polynomials correspond to a root system of type A. For a root system
of type B, generalized Narayana polynomials are defined [6, Example 5.8] by W (z) = 1
0
and
r 2
r
W (z) = zk.
r
(cid:18)k(cid:19)
Xk=0
For their combinatorial study we refer to [2, 3] and references therein.
We have W (1) = W , the central binomial coefficient, since
r r
r 2
2r r
W = = .
r
(cid:18)r (cid:19) (cid:18)k(cid:19)
Xk=0
Moreover [3, equation (2.1)] the Narayana polynomial W (z) may be expressed in terms
r
of central binomial coefficients as
r
W (z) = zm(z +1)r−2m W . (9)
r m
(cid:18)2m(cid:19)
mX≥0
Exactly as in Section 2 the polynomial
(z +1)W (z)−W (z) = −2rzr +...
r r+1
may be expressed, in a unique way, in terms of the monic polynomials zmW (z) with
r−2m+1
degree r −m+1. As a consequence, we may define the real numbers B (r) by
m
r
(z +1)W (z)−W (z) = (−z)m B (r)W (z). (10)
r r+1 m r−2m+1
(cid:18)2m−1(cid:19)
mX≥1
Note that the real numbers {B (r),m ≥ 1} depend on r and that we have B (r) = 2.
m 1
Using relation(9), andparallelingthecomputationinSection1, thisdefinitioniseasily
shown to be equivalent to
n
2n−1
W = (−1)j−1 B (r)W .
n j n−j
(cid:18)2j −1(cid:19)
Xj=1
10