On partitions avoiding 3-crossings

0.31 MB·English

by  Mireille Bousquet-Mélou

ON PARTITIONS AVOIDING 3-CROSSINGS
MIREILLEBOUSQUET-MÉLOUANDGUOCEXIN
6
To Xavier Viennot, on the o

asion of his 60th birthday
0
0
2 Abstra
t. A partition on [n] has a
rossing if there exists i1 < i2 < j1 < j2 su
h
n it2hat i1 and j1 are in the same blo
k, i2 and j2 are in the same blo
k, but i1 and
a are not ikn the same blo
k. Re
entlky, Chen et al. re(cid:28)ned this
lassi
al notion by
J introdu
ing -
rossings,foranyinteger .[nIn] thisnew2terminology,a
lassi
al
rossingis
a2-
rossing. Thenumberof2npartitionsof avoiding -
rossingsiswell-knowntobethe
1 nthCatalannumberCn= n /(n+1). Thisraisesthequestionof
ountingk-non
rossing
2 k≥3 3
partitionsfor . Wepr(cid:0)ove(cid:1)that thesequen
e
ounting -non
rossing partitionsisP-

re
ursive, that is, satis(cid:28)es a linear re
urren
e relation with polynomial
oe(cid:30)
ients. We
] k
O giveexpli
itlysu
hakr≥e
u4rsion. However,we
onje
turethat -non
rossingpartitionsare
notP-re
ursive,for .
C 3
Weobtainsimilarresultsforpartitionsavoidingenhan
ed -
rossings.
.
h
t
a 1. Introdu
tion
m
[n]:= 1,2,...,n
Apartitionof { }isa
olle
tionofnonemptyandmutuallydisjointsubsets
[ [n] [n] [n]
of ,
alled blo
ks, whose union is . The number of partitions of is the Bell number
B B
2 n n
. A well-known re(cid:28)nement of is given by the Stirling number (of the se
ond kind)
v S(n,k) [n] k
. It
ounts partitions of having exa
tly blo
ks.
1
Re
ently another re(cid:28)nement of the Bell numbers by
rossings and nestings has attra
ted
5
5 some interest [8, 14℄. This re(cid:28)nement is based on the standard representationof a partition
P [n] [n] i (i,0)
6 of by a graph, whose vertex set is identi(cid:28)ed with the points ≡ on the
1 i n
0 plane, for ≤ ≤ , and whose edge set
onsists of ar
s
onne
ting the elements that
5
o

ur
onse
utivelyin thesameblo
k(whenea
hblo
kistotallyordered). Forexample,the
0 1478 236 5
standard representation of − − is given by the following graph.
/
h
t
a
m 1 2 3 4 5 6 7 8
:
v k P
Then
rossingsand nestingshaveanaturalde(cid:28)nition. A -
rossing of isa
olle
tionof
i k (i ,j ) (i ,j ) (i ,j ) i <i < <i <j < j < < j
X edges 1 1 , 2 2 , ..., k k su
h that 1 2 ··· k 1 2 ··· k. This
P
r means a subgraph of as drawn as follows.
a
i i i j j j
1 2 k 1 2 k
k
A di(cid:27)erent notion of -
rossing is obtained by representing ea
h blo
k by a
omplete graph
k P k (i ,j ) (i ,j ) (i ,j )
1 1 2 2 k k
[13, p. 85℄. A -nesting of is a
olle
tion of edges , , ..., su
h that
i <i < <i <j <j < <j
1 2 k k k−1 1
··· ··· , as represented below.
Date:January 20,2006.
Keywords and phrases. Partitions,Crossings,D-(cid:28)niteseries.
1
2 MIREILLEBOUSQUET-MÉLOUANDGUOCEXIN
i i i j j j
1 2 k k 2 1
k k k k
Apartitionis -non
rossing ifithasno -
rossings,and -nonnesting ifithasno -nestings.
k k
A variation of -
rossings (nestings),
alled enhan
ed -
rossings (nestings), was also
studied in [8℄. One (cid:28)rst adds a loop to every isolated point in the standard representation
i = j k k
k 1
of partitions. Then by allowing in a -
rossing, we get an enhan
ed -
rossing; in
2
parti
ular,apartitionavoidingenhan
ed -
rossingshaspartsofsize1and2only. Similarly,
i =j k k
k k
by allowing in the de(cid:28)nition of a -nesting, we get an enhan
ed -nesting.
[n]
Chen et al. gave in [8℄ a bije
tion between partitions of and
ertain (cid:16)va
illating(cid:17)
2n k
tableaux of length . Throughthis bije
tion, a partition is -non
rossingif and only if the
k k

orrespondingtableau hasheightlessthan , and -nonnestingif the tableau haswidth less
k k [n]
than . A simple symmetry on tableaux then entails that -non
rossing partitions of
k [n] k n
are equinumerous with -nonnesting partitions of , for all values of and . A se
ond
bije
tion relates partitions to
ertain (cid:16)hesitating(cid:17) tableaux, in su
h a way the size of the
largest enhan
ed
rossing (nesting) be
omes the height (width) of the tableau. This implies
[n] k [n]
that partitions of avoiding enhan
ed -
rossingsare equinumerous with partitions of
k
avoiding enhan
ed -nestings.
C (n) 2 [n]
2
The number of -non
rossing partitions of (Cusua=lly
1alle2dn non
rosksi>ng2parti-
tions [15, 21℄) is well-known to be the Catalan number n n+1 n . For , the
k [n]
number of -non
rossing partitions of is not known, to the exte(cid:0)nt(cid:1)of our knowledge.
k k [n]
However, for any , the number of -non
rossing mat
hings of (that is, partitions in
whi
h all blo
ks have size 2) is known to form a P-re
ursive sequen
e, that is, to satisfy
a linear re
urren
e relation with polynomial
oe(cid:30)
ients [11, 8℄. In this paper, we enumer-
3 [n] 3 [n]
ate -non
rossing partitions of (equivalently, -nonnesting partitions of ). We obtain
C (n)
3
for a (not so simple)
losed form expression (Proposition 7) and a linear re
urren
e
relation with polynomial
oe(cid:30)
ients.
C (n) 3 C (0)=C (1)=
3 3 3
Proposition1. The number of -non
rossingpartitionsis givenby
1, n 0
and for ≥ ,
9n(n+3)C (n) 2 5n2+32n+42 C (n+1)+(n+7)(n+6)C (n+2)=0.
3 3 3
− (1)
(cid:0) (cid:1) (t)= C (n)tn
Equivalently, the asso
iated generating fun
tion C n≥0 3 satis(cid:28)es
d2 d P
t2(1 9t)(1 t) (t)+2t 5 27t+18t2 (t)+10(2 3t) (t) 20=0.
− − dt2C − dtC − C − (2)
n (cid:0) (cid:1)
Finally, as goes to in(cid:28)nity,
39 5√3 9n
C (n) · .
3 ∼ 25 π n7
C (n) n 0
3
The (cid:28)rst few values of the sequen
e , for ≥ , are
1,1,2,5,15,52,202,859,3930,19095,97566,....
A standard study [12, 23℄ of the above di(cid:27)erential equation, whi
h
an be done automati-
C (n) κ9n/n7
3

ally using the Maple pa
kage DEtools, suggests that ∼ for some positive
κ C (n)
3

onstant . However, one needs the expli
it expression of given in Se
tion 2.5 to
κ
prove this statement and (cid:28)nd the value of . The above asymptoti
behaviour is
on(cid:28)rmed
C (n)
3
experimentallybythe
omputationofthe(cid:28)rstvaluesof (afewthousandvalues
anbe
n = 50000

omputed rapidly using the pa
kage Gfun of Maple [20℄). For instan
e, when ,
n7C (n)/9n 1694.9 κ 1695.6
3
then ≃ , while ≃ .
ON PARTITIONS AVOIDING 3-CROSSINGS 3
As dis
ussed in the last se
tion of the paper, the above result might remain isolated,
as there is no (numeri
al) eviden
e that the generating fun
tion of 4-non
rossing partitions
should be P-re
ursive.
3
We obtain a similar result for partitions avoiding enhan
ed -
rossings (or enhan
ed 3-
[n]
nestings). The number of partitions of avoidingenhan
ed2-
rossingsis easilyseen to be
n
the th Motzkin number [22, Exer
ise 6.38℄.
E (n) [n] 3
3
Proposition 2. The number of partitions of having no enhan
ed -non
rossing
E (0)=E (1)=1, n 0
3 3
is given by and for ≥ ,
8(n+3)(n+1)E (n)+ 7n2+53n+88 E (n+1) (n+8)(n+7)E (n+2)=0.
3 3 3
−
Equivalently, the asso
iated g(cid:0)enerating fun
ti(cid:1)on E(t)= n≥0E3(n)tn satis(cid:28)es
d2 d P
t2(1+t)(1 8t) (t)+2t 6 23t 20t2 (t)+6 5 7t 4t2 (t) 30=0.
− dt2E − − dtE − − E −
n (cid:0) (cid:1) (cid:0) (cid:1)
Finally, as goes to in(cid:28)nity,
216 5√3 8n
E (n) · .
3 ∼ 33 π n7
E (n) n 0
3
The (cid:28)rst few values of the sequen
e , for ≥ , are
1,1,2,5,15,51,191,772,3320,15032,71084,....
C (5) E (5) 135 24
3 3
Observethat and di(cid:27)er by 1: this di(cid:27)eren
e
omes from the partition −
15
whi
h has an enhan
ed 3-
rossing but no 3-
rossing (equivalently, from the partition −
24 3
− whi
hhasanenhaEn
(end)3-nκes8tnin,gbutno3-nesting). Again,κthestudyofthedi(cid:27)erential
equation suggests that 3 ∼ n7 for some positive
onstant , but we need the expli
it
E (n) κ
3
expression (33) of to prove this statement and (cid:28)nd the value of . Numeri
ally, we
n=50000 n7E (n)/8n 6687.3 κ 6691.1
3
have found that for , ≃ , while ≃ .
ThestartingpointofourproofofProposition1istheabovementionedbije
tionbetween
k+1 k
partitionsavoiding -
rossingsandva
illatingtableauxofheightatmost . Asdes
ribed
k D
in [8℄, these tableaux
an be easily en
oded by
ertain -dimensional latti
e walks. Let
Zk D n
be a subset of . A -va
illating latti
e walk of length is a sequen
e of latti
e points
p ,p ,...,p D i
0 1 n
in su
h that for all ,
(i) p =p p =p e e =(0,...,0,1,0,...,0)
2i+1 2i 2i+1 2i j j
or − for some
oordinateve
tor ,
(ii) p =p p =p +e e
2i 2i−1 2i 2i−1 j j
or for some .
D Zk Q =Nk
k
We will be interested in twodi(cid:27)erent domains of : the domain of points with
non-negativeinteger
oordinatesandtheWeyl
hamber(withaslight
hangeof
oordinates)
W = (a ,a ,...,a ) Zk :a >a > >a 0 W
k 1 2 k 1 2 k k
{ ∈ ··· ≥ }. Va
illating walks in are related
k+1
to -non
rossingpartitions as follows.
C (n) k
k
Theorem 3 (Chen et al. [8℄). Let denote the number of -non
rossing partitions
[n] C (n) W 2n
k+1 k
of . Then equals the number of -va
illating latti
e walks of length starting
(k 1,k 2,...,0)
and ending at − − .
The proof of Proposition 1 goes as follows: using the re(cid:29)e
tion prin
iple, we (cid:28)rst redu
e
W
k
the enumerationof va
illating walksin the Weyl
hamber to that of va
illating walksin
Q k
k
the non-negative domain . This redu
tion is valid for any . We then fo
us on the
ase
k =2
. We write a fun
tional equation satis(cid:28)ed by a 3-variableseries that
ounts va
illating
Q
2
walks in . This equation is based on a simple re
ursive
onstru
tion of the walks. It is
solvedusinga2-dimensionalversionoftheso-
alledkernelmethod. Thisgivesthegenerating
(t) = C (n)tn
3
fun
tion C as the
onstant term in a
ertain algebrai
series. We then use
C (n)
3
theLagrangeinvPersionformulato(cid:28)ndanexpli
itexpressionof ,andapplythe
reative
teles
oping of [19℄ to obtain the re
urren
e relation. We (cid:28)nally derive from the expression
4 MIREILLEBOUSQUET-MÉLOUANDGUOCEXIN
C (n)
3
of the asymptoti
behaviour of these numbers. The proof of Proposition 2, given in
Se
tion 3, is similar.
3
2. Partitions with no -
rossing
2.1. The refle
tion prin
iple
δ = (k 1,k 2,...,0) λ µ W
k
Let − − and let and be two latti
e points in . Denote by
w (λ,µ,n) q (λ,µ,n) W Q
k k k k
(resp. ) the number of -va
illating (resp. -va
illating) latti
e
n λ µ C (n)
k+1
walksoflength startingat andendingat . ThusTheorem3statesthat equals
w (δ,δ,2n)
k
. The re(cid:29)e
tion prin
iple, in the vein of [10, 24℄, gives the following:
λ µ W W
k k
Proposition 4. For any starting and ending points and in , the number of -
λ µ Q
k
va
illating walks going from to
an be expressed in terms of the number of -va
illating
walks as follows:
w (λ,µ,n)= ( 1)πq (λ,π(µ),n),
k k
− (3)
πX∈Sk
( 1)π π π(µ ,µ ,...,µ )=(µ ,µ ,...,µ )
1 2 k π(1) π(2) π(k)
where − is the sign of and .
= x = x : 1 i < j k
i j
Proof. Consider the set of hyperplanes H { ≤ ≤ }. The re(cid:29)e
tion
(a ,...,a ) x = x
1 k i j
of the point with respe
t to the hyperplane is simply obtained by
a a
i j
ex
hanging the
oordinates and . In parti
ular, the set of (positive) steps taken by
e ,...,e
1 k
va
illating walks, being { }, is invariant under su
h re(cid:29)e
tions. The same holds
for the negative steps, and of
ourse for the (cid:16)stay(cid:17) step, 0. This implies that re(cid:29)e
ting a
Q x = x Q
k i j k
-va
illating walk with respe
t to gives another -va
illating walk. Note that
x =0 e e
i i i
this is not true when re(cid:29)e
ting with respe
t to (sin
e is transformed into − ).
Q w n λ
k
De(cid:28)ne a total ordering on H. Take a -va
illating walk of length going from to
π(µ) m
, and assumeit tou
hesatleastonehyperplanein H. Let be the(cid:28)rst timeit tou
hes
x =x m
i j
ahyperplane. Let bethe smallesthyperplane it tou
hesat time . Re(cid:29)e
t all steps
w m x = x w′ Q
i j k
of after time a
ross ; the resulting walk is a -va
illating walk going from
λ (i,j)(π(µ)) (i,j) i j
to , where denotes the transposition that ex
hanges and . Moreover,
w′ m
also ((cid:28)rst) tou
hes H at time , and the smallest hyperplane it tou
hes at this time is
x =x
i j
.
Q
k
The abovetransformationisasign-reversinginvolutionon thesetof -va
illatingpaths
λ π(µ) π
that go from to , for some permutation , and hit one of the hyperplanes of H. In
the right-hand side of (3), the
ontributions of these walks
an
el out. One is left with the
π
walksthatstaywithintheWeyl
hamber,andthishappensonlywhen istheidentity. The
proposition follows.
(k + 1)
This proposition,
ombined with Theorem 3, gives the number of -non
rossing
n q (δ,π(δ),2n)
k
partitions of as a linear
ombination of the numbers . Hen
e, in order to
(k+1) q (δ,µ,2n)
k

ount -non
rossingpartitions, it su(cid:30)
es to (cid:28)nd a formula for , for
ertain
µ k =2
ending points . This is what we do below for .
2.2. A fun
tional equation
k=2
Let us spe
ialize Theorem 3 and Proposition 4 to . This gives
C (n) = w ((1,0),(1,0),2n)
3 2
= q ((1,0),(1,0),2n) q ((1,0),(0,1),2n).
2 2
− (4)
Q (1,0)
2
From now on, a latti
e walk always means a -va
illating latti
e walk starting at ,
unless spe
i(cid:28)ed otherwise.
ON PARTITIONS AVOIDING 3-CROSSINGS 5
a (n):=q ((1,0),(i,j),n) n (i,j)
i,j 2
Let bethenumberoflatti
ewalksoflength endingat .
Let
F (x,y;t)= a (2n)xiyjt2n
e i,j
i,j,n≥0
X
and
F (x,y;t)= a (2n+1)xiyjt2n+1
o i,j
i,j,n≥0
X
berespe
tivelythegeneratingfun
tionsoflatti
ewalksofevenandoddlength. Theseseries
t Q[x,y]
are power series in with
oe(cid:30)
ients in . We will often work in a slightly larger
Q[x,1/x,y,1/y][[t]] t
ring, namely the ring of power series in whose
oe(cid:30)
ients are Laurent
x y
polynomials in and .
F (x,y;t) F (x,y;t)
e o
Now we (cid:28)nd fun
tional equations for and . By appending to an even
( 1,0) (0, 1) (0,0)
length walk a west step − , or a south step − , or a stay step , we obtain
Q x
2
either an odd length walk (in ), or an even length walk ending on the -axis followed by
y
a south step, or an even length walk ending on the -axis followed by a west step. This

orresponden
e is easily seen to be a bije
tion, and gives the following fun
tional equation:
F (x,y;t)(1+x¯+y¯)t=F (x,y;t)+y¯tH (x;t)+x¯tV (y;t),
e o e e
(5)
x¯ = 1/x y¯ = 1/y H (x;t) V (y;t)
e e
where , , and (resp. ) is the generating fun
tion of even
x y
latti
e walks ending on the -axis (resp. on the -axis).
(1,0) (0,1)
Similarly, by adding to an odd length walk an east step , or a north step , or a
stay step, we obtain an even length walk of positive length. The above
orresponden
e is a
bije
tion and gives another fun
tional equation:
F (x,y;t)(1+x+y)t=F (x,y;t) x.
o e
− (6)
F (x,y;t) F (x,y;t)
e o
Solving equations (5) and (6) for and gives
x t2y¯(1+x+y)H (x;t) t2x¯(1+x+y)V (y;t)
e e
F (x,y;t)= − − ,
e 1 t2(1+x+y)(1+x¯+y¯)
−
x(1+x¯+y¯) x¯V (y;t) y¯H (x;t)
e e
F (x,y;t)=t − − .
o 1 t2(1+x+y)(1+x¯+y¯)
−
(1,0) (0,1)
Sin
e we are mostly interested in even length walks ending at and at , it su(cid:30)
es
H (x;t) V (y;t)
e e
to determine the series and , and hen
e to solve one of the above fun
tional
F (x,y;t)
o
equations. We
hoose the one for , whi
h is simpler.
F (x,y;t) Q
o 2
Proposition 5. The generating fun
tion of -va
illating latti
e walks of odd
(1,0)
length starting from is related to the generating fun
tions of even latti
e walks of the
x y
same type ending on the - or -axis by
K(x,y;t)F(x,y;t)=xy+x2y+x2 xH(x;t) yV(y;t),
− − (7)
F (x,y;t)=tF(x,y;t2) V(y;t)=V (y;t2) H(x;t)=H (x;t2) K(x,y;t)
o e e
where , , , and is the
kernel of the equation:
K(x,y;t)=xy t(1+x+y)(x+y+xy).
− (8)
t
Fromnowon, wewill veryoftenomitthe variable in ournotation. Forinstan
e, wewill
H(x) H(x;t) F(x,y) t
write insteadof . Observethat(7) de(cid:28)nes uniquely asaseriesin with
Q[x,y] y = 0 H(x) = x+t(1+x)F(x,0)

oe(cid:30)
ients in . Indeed, setting shows that while
x=0 V(y)=t(1+y)F(0,y)
setting gives .
6 MIREILLEBOUSQUET-MÉLOUANDGUOCEXIN
2.3. The kernel method
We are going to apply to (7) the obstinate kernel method that has already been used
in [4, 5℄ to solve similar equations. The
lassi
al kernel method
onsists in
oupling the
x y K(x,y)
variables and so as to
an
el the kernel . This gives some (cid:16)missing(cid:17) information
V(y) H(x)
about the series and (see for instan
e [6, 1℄). In its obstinate version, the kernel
methodis
ombinedwithapro
edurethat
onstru
tsandexploitsseveral(related)
ouplings
(x,y)
. This pro
edure is essentially borrowed from [9℄, where similar fun
tional equations
o

ur in a probabilisti

ontext.
x
Let us start with the standard kernel method. First (cid:28)x , and
onsider the kernel as a
y Y t
0
quadrati
polynomial in . Only one of its roots, denoted below, is a power series in :
1 (x¯+3+x)t (1 (1+x+x¯)t)2 4t
Y = − − − −
0 2q(1+x¯)t
(1+x)(1+3x+x2)
=(1+x)t+ t2+
x ···
x
The
oe(cid:30)
ientsofthisseriesareLaurentpolynomialsin ,asiseasilyseenfromtheequation
Y =t(1+x+Y )(1+(1+x¯)Y ).
0 0 0
(9)
y =Y Q[x,x¯][[t]]
0
Setting in (7) gives a valid identity between series of , namely
xH(x)+Y V(Y )=xY +x2Y +x2.
0 0 0 0
Y =x/Y =O(t−1) F(x,Y )
1 0 1
The se
ond root of the kernel is , so that the expression is not
well-de(cid:28)ned.
(X,Y) = (0,0) t K
Now let 6 be a pair of Laurent series in with
oe(cid:30)
ients in a (cid:28)eld
K(X,Y) = 0 K x y
su
h that . Re
all that is quadrati
in and . In parti
ular, the equation
K(x,Y) = 0 X′ Φ(X,Y) = (X′,Y)
admits a se
ond solution . De(cid:28)ne . Similarly, de(cid:28)ne
Ψ(X,Y)=(X,Y′) Y′ K(X,y)=0 Φ Ψ
, where is the se
ond solution of . Note that and are
Y′ =X/Y X′ =Y/X
involutions. Moreover,with the kernelgiven by (8), onehas and . Let
Φ Ψ (x,Y ) 6
0
us examine the a
tion of and on the pair : we obtain an orbit of
ardinality
(Figure 1).
(x,Y )
0
Φ Ψ
(x¯Y ,Y ) (x,Y )
0 0 1
Ψ Φ
(x¯Y ,x¯) (x¯Y ,Y )
0 1 1
Φ Ψ
(x¯Y ,x¯)
1
(x,Y ) Φ Ψ
0
Figure 1. The orbit of under the a
tion of and .
The 6 pairs of power series given in Figure 1
an
el the kernel, and we have framed the
(x,y)
ones that
anbe legally substituted for in the main fun
tional equation(7). Denoting
Y Y H(x) V(y)
0
≡ , we thus obtain three equations relating the unknown series and :
ON PARTITIONS AVOIDING 3-CROSSINGS 7
xH(x)+YV(Y) = xY +x2Y +x2,
(10)
x¯YH(x¯Y)+YV(Y) = x¯Y2+x¯2Y3+x¯2Y2,
(11)
x¯YH(x¯Y)+x¯V(x¯) = x¯2Y +x¯3Y2+x¯2Y2.
(12)
2.4. Positive and negative parts of power series
+
Asimplelinear
ombinationoftheabovethreeequations(namely,(10)−(11) (12))allows
V(Y) H(x¯Y)
us to eliminate the terms and . We are left with:
xH(x)+x¯V(x¯)=x2+(x¯2+x+x2)Y +(x¯3 x¯)Y2 x¯2Y3.
− −
xH(x) x x¯V(x¯)
Sin
e
ontains only positive powers of and
ontains only negativepowers of
x H(x) V(y)
, we have
hara
terizedthe series and .
H(x) V(y) Q
2
Proposition 6. The series and
ounting -va
illating walks of even length
(1,0) x y
starting at and ending on the -axis and on the -axis satisfy
xH(x)=PT x2+(x¯2+x+x2)Y +(x¯3 x¯)Y2 x¯2Y3 ,
x − −
x¯V(x¯)=NT(cid:0)x2+(x¯2+x+x2)Y +(x¯3 x¯)Y2 x¯2Y3(cid:1),
x − −
(cid:0) (cid:1)
PT NT x
x x
where the operator (resp. ) extra
ts positive (resp. negative) powers of in series
Q[x,x¯][[t]]
of .
F (x,y;t) F (x,y;t)
e o
One may then go ba
k to (5(cid:21)6) to obtain expressionsof the series and .
C (n) 3
3
However, our main
on
ern in this note is the number of -non
rossing partitions of
[n] C (n)
3
. Going ba
k to (4), we (cid:28)nd that is determined by the following three equations:
C (n)=q ((1,0),(1,0),2n) q ((1,0),(0,1),2n),
3 2 2
−
q ((1,0),(1,0),2n)=[xtn] H(x)=[x2tn] xH(x),
2
q ((1,0),(0,1),2n)=[ytn] V(y)=[x¯2tn] x¯V(x¯).
2
(t) 3
Using Proposition 6, we obtain the generating fun
tion C of -non
rossingpartitions:
(t)=CT(x¯2 x2) x2+(x¯2+x+x2)Y +(x¯3 x¯)Y2 x¯2Y3 ,
C x − − − (13)
(cid:0) (cid:1)
CT x Q[x,x¯][[t]]
x
wheretheoperator extra
tsthe
onstanttermin ofseriesin . Observethat
Y(x)=xY(x¯) k N ℓ Z
. This implies that for all ∈ and ∈ ,
[xℓ]Y(x)k =[xℓ−k]Y(x¯)k =[xk−ℓ]Y(x)k, CT(x−ℓYk)=CT(xℓ−kYk).
that is, x x
This allows us to rewrite (13) with (cid:16)only(cid:17) six terms:
(t)=1+CT (x¯1 x4)Y +(x¯5 x¯)Y2 (x¯4 x0)Y3 .
C x − − − −
(cid:0) (cid:1)
(t)
The above equation says that C is the
onstant term of an algebrai
fun
tion. By a very
(t)
general theory [16℄, C is D-(cid:28)nite. That is, it satis(cid:28)es a linear di(cid:27)erential equation with
(t)
polynomial
oe(cid:30)
ients. In the next se
tion, we showthat C satis(cid:28)es the equation(2) (or,
equivalently, the P-re
urren
e (1)). Note that this re
urren
e
an be easily guessed using
C (n)
3
the Maple pa
kage Gfun: indeed, the (cid:28)rst 15 values of already yield the
orre
t
re
ursion.
8 MIREILLEBOUSQUET-MÉLOUANDGUOCEXIN
2.5. The Lagrange inversion formula and
reative teles
oping
From now on, several routes lead to the re
urren
e relation of Proposition 1, depending
on how mu
h software one is willing to use. We present here the one that we believe to be
the shortest. Starting from (9), the Lagrangeinversion formula gives
k n n 2j+k
[tn]CT(xℓYk)= a (ℓ,k,j) a (ℓ,k,j)= .
n n
x with n j j+k j ℓ (14)
jX∈Z (cid:18) (cid:19)(cid:18) (cid:19)(cid:18) − (cid:19)
a 0 b a n 1
By
onvention, the binomial
oe(cid:30)
ient b is zero unless ≤ ≤ . Hen
e for ≥ ,
C (n)= a ( 1,1,j) a (4,1,j)+a(cid:0)((cid:1)5,2,j 1) a ( 1,2,j 1) a ( 4,3,j 1)+a (0,3,j 1) .
3 n n n n n n
− − − − − − − − − − −
Z
jX∈ (cid:0) (cid:1)
(15)
j 1 j
Of
ourse, we
ould repla
e all o

urren
es of − by in the above expression, but this
results in a bigger (cid:28)nal formula.
n 1 3 [n]
Proposition 7. For ≥ , the number of -non
rossing partitions of is
n
4(n 1)!n!(2j)!
C (n)= − P(j,n)
3
(j 1)!j!(j+1)!(j+4)!(n j)!(n j+2)!
j=1 − − −
X
with
P(j,n)=24+18n+(5 13n)j+(11n+20)j2+(10n 2)j3+(4n 11)j4 6j5.
− − − −
Proof of the re
urren
e relation of Proposition 1. We (cid:28)nally apply to the above
expression Zeilberger's algorithm for
reative teles
oping [19, Ch. 6℄ (we used the Maple
C (n) n 2
3
pa
kage Ekhad). This gives a re
urren
e relation for the sequen
e : for ≥ ,
9n(n 2)(n 3) 4n2+15n+17 C (n 3) (n 2) 76n4+373n3+572n2+203n 144 C (n 2)
3 3
− − − − − − −
+(n+3) 44n4+(cid:0)189n3+227n2(cid:1)+30n 160 C (n(cid:0) 1)=(n+5)(n+4)(n+3) 4n2+7n+(cid:1) 6 C (n).
3 3
− −
C (0)=C (1)=1
(cid:0) 3 3 (cid:1) (cid:0) (cid:1)
The initial
onditions are . It is then very simple to
he
k that the three
term re
ursion given in Proposition 1 satis(cid:28)es also the above four term re
ursion. This
an
be rephrased by saying that (1) is a right fa
tor of the four term re
ursion obtained via

reative teles
oping. More pre
isely, applying to (1) the operator
(n+1) 4n2+39n+98 (n+6) 4n2+31n+63 N,
−
N (cid:0) n(cid:1) n+1 (cid:0) (cid:1) n
where is the shift operator repla
ing by , gives the four term re
ursion (with
n+3
repla
ed by ).
w ((1,0),(1,0),2n) : n N
2
It is worth noting that though the sequen
e { ∈ } satis(cid:28)es a
2 q ((1,0),(1,0),2n):n N
2
simplere
urren
eoforder ,Mapletellsusthesequen
es{ ∈ }and
q ((1,0),(0,1),2n):n N 3
2
{ ∈ } satisfy more
ompli
ated re
urren
es of order .
2.6. Asymptoti
s
C (n)
3
Finally, wewill derivefrom the expression(15)of the asymptoti
behaviourofthis
k ℓ k 1
sequen
e,asstatedinProposition1. Forany(cid:28)xedvaluesof and ,with ≥ ,we
onsider
a (ℓ,k,j) a (j)
n n
the numbers de(cid:28)ned by (14). For the sake of simpli
ity, we denote them ,
and introdu
e the numbers
n n 2j+k
b (j)= ,
n
j j+k j ℓ
(cid:18) (cid:19)(cid:18) (cid:19)(cid:18) − (cid:19)
a (j) = kb (j)/n. B = b (j) j [ℓ ,n k]
so that n n Let n j n . The sum runs over ∈ + − , where
ℓ = max(ℓ,0) j
+
. Below, we often
onPsider as a real variable running in that interval.
ON PARTITIONS AVOIDING 3-CROSSINGS 9
b (j)
n
The number is then de(cid:28)ned in terms of the Gamma fun
tion rather than in terms of
B N
n
fa
torials. We will show that admits, for any , an expansion of the form
N
c
B =9n i +O(n−N−1) ,
n ni ! (16)
i=1
X
c k ℓ
i
where the
oe(cid:30)
ients depend on and , and explain how to obtain these
oe(cid:30)
ients.
We follow the standards steps for estimating sums of positive terms that are des
ribed, for
b (j)
n
instan
e, in [2, Se
tion 3℄. We begin with a unimodality property of the numbers .
n b (j) j [ℓ ,n k]
n +
Lemma 8 (Unimodality). For large enough, the sequen
e , for ∈ − , is
[2n/3 k/2 1/2,2n/3 k/2 1/3]
unimodal, and its maximum is rea
hed in the interval − − − − .
Proof. One has
b (j) 1 (j+1)(j+k+1)(j ℓ+1)(j+k+ℓ+1)
n
q (j):= = − .
n
b (j+1) (n j)(n j k) 2j+k+1 2j+k+2
n
− − −
j
Ea
h of these three fa
tors is easily seen to be an in
reasing fun
tion of on the relevant
n q (ℓ ) < 1 q (n k 1) > 1 j
n + n 0
interval. Moreover, for large enough, while − − . Let be the
j q (j) 1 b (j)<b (j+1) j <j b (j) b (j+1)
n n n 0 n n
smallestvalueof su
hthat ≥ . Then for and ≥
j j q (x) = 1 x
0 n
for ≥ . We have thus proved unimodality. Solving for helps to lo
ate the
mode: 2n 5+6k
x= +O(1/n).
3 − 12
q (2n/3 k/2 1/2)<1 q (2n/3 k/2 1/3)>1 n
n n
It is then easy to
he
k that − − and − − for
large enough.
The se
ond step of the proof redu
es the range of summation.
ǫ (0,1/6) m
Lemma 9 (A smaller range). Let ∈ . Then for all ,
B = b (j)+o(9nn−m).
n n
|j−2n/X3|≤n1/2+ǫ
j =2n/3 n1/2+ǫ
Proof. Let ± . The Stirling formula gives
b (j)= 3 5/2 9n e−9n2ǫ/2(1+o(1))=o(9nn−m)
n 2 (πn)3/2 (17)
(cid:18) (cid:19)
m
for all (the details of the
al
ulation are given in greater detail below, in the proof of
Lemma 10). Thus by Lemma 8,
b (j) n b (2n/3 n1/2+ǫ)+b (2n/3+n1/2+ǫ) =o(9nn−m).
n n n
≤ −
|j−2n/X3|>n1/2+ǫ (cid:16) (cid:17)
The result follows.
b (j)
n
A (cid:28)rst order estimate of , generalizing (17), su(cid:30)
es to obtain, upon summing over
j B N = 1
n
, an estimate of of the form (16) with . However, numerous
an
ellations o

ur
a (ℓ,k,j) C (n)
n 3
when summing the 6terms in theexpression(15) of . Thisexplains why we
b (n) B
j n
have to work out a longer expansion of the numbers and .
b (n) ǫ (0,1/6) j =2n/3+r r=s√n
j
Lemma 10 (Expansion of ). Let ∈ . Write with and
s nǫ N 1
| |≤ . Then for all ≥ ,
b (j)= n n 2j+k = 3 5/2 9n e−9r2/(2n) N−1ci(s) +O(nN(3ǫ−1/2))
n (cid:18)j(cid:19)(cid:18)j+k(cid:19)(cid:18) j−ℓ (cid:19) (cid:18)2(cid:19) (πn)3/2 i=0 ni/2 !
X
c (s) k ℓ s 3i s c
i i
where is a polynomial in , and , of degree in . Moreover, is an even [odd℄
s i c (s)=1. s
0
fun
tion of if is even [odd℄. In parti
ular, This expansion is uniform in .
10 MIREILLEBOUSQUET-MÉLOUANDGUOCEXIN
Proof. Note that we simply want to prove the existen
e of an expansion of the aboveform.
The
oe(cid:30)
ients
anbe obtained routinely using(preferably) a
omputeralgebrasystem. In
(c ) c = 1
i i≥0 0
what follows, stands for a sequen
e of real numbers su
h that . The a
tual
c (c (s))
i i i≥0
valueof may
hangefromoneformulatoanother. Similarly, denotesasequen
e
s c (s)=1
0
of polynomials in su
h that , having the parity property stated in the lemma.
N 1
We start from the Stirling expansion of the Gamma fun
tion: for all ≥ ,
N−1
c
Γ(n+1)=nn√2πn e−n i +O(n−N) .
ni ! (18)
i=0
X
j =2n/3+r r =s√n N
This gives, for , with , and for any ,
N−1
πn c (s)
Γ(j+1)=2jj e−j i +O(nN(ǫ−1/2))
r 3 i=0 ni/2 ! (19)
X
c (s) i s s
i
for some polynomials of degree in . This estimate is uniform in . Similarly,
N−1
2πn c (s)
Γ(n j+1)=(n j)n−j ej−n i +O(nN(ǫ−1/2)) ,
− − r 3 i=0 ni/2 ! (20)
X
c (s) i s
i
for some polynomials of degree in . Now
nn 3n 9r2 3iri+1
log = log 1 ( 1/2)i
jj(n j)n−j 2j − 4n − i(i+1)ni − −
− i≥2
X (cid:0) (cid:1)
3n 9r2 N−1c (s)
= log i +O(n2ǫ+N(ǫ−1/2)),
2j − 4n − ni/2
i=1
X
c (s) i+2 s (i+2)/(i/2) 6 i 1
i
for some polynomials of degree in . Observe that ≤ for ≥ .
Hen
e
nn = 3ne−9r2/(4n)exp N−1ci(s) +O(n2ǫ+N(ǫ−1/2))
jj(n j)n−j 2j − ni/2 !
− i=1
X
= 3ne−9r2/(4n) N−1ci(s) +O(nN(3ǫ−1/2)) ,
2j ni/2 ! (21)
i=0
X
c (s) 3i s
i
for polynomials of degree in . Putting together (18(cid:21)21), one obtains, uniformly in
s
: n = 3 3ne−9r2/(4n) N−1ci(s) +O(nN(3ǫ−1/2)) ,
(cid:18)j(cid:19) 2√πn 2j i=0 ni/2 ! (22)
X
c (s) 3i s
i
for polynomials of degree in .
Similarly,
n = 3 3n e−9r2/(4n) N−1ci(s) +O(nN(3ǫ−1/2)) ,
(cid:18)j+k(cid:19) 2√πn2j+k i=0 ni/2 ! (23)
X
c (s)
i
with the same degree
ondition on the polynomials . Finally,
N−1
2j+k 3 c (s)
= 22j+k i +O(nN(ǫ−1/2))
(cid:18) j−l (cid:19) r2πn i=0 ni/2 ! (24)
X
c (s) i s
i
for polynomials of degree in . Putting together (22(cid:21)24), we obtain the estimate of
b (j)
n
given in the lemma.
b (j) j j 2n/3 n1/2+ǫ
n
It remains to sum our estimates of for values of su
h that | − |≤ .