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arXiv:0806.0424v1 [math.CO] 3 Jun 2008

Some Enumerations for Parking Functions

Po-Yi Huanga,∗

Jun Mab,†

Jean Yehc,‡

aDepartment of Mathematics, National Cheng Kung University, Tainan, Taiwan

bInstitute of Mathematics, Academia Sinica, Taipei, Taiwan

cDepartment of Mathematics, National Taiwan University, Taipei, Taiwan

Abstract

In this paper, let Pn,n+k;≤n+k (resp.Pn;≤s) denote the set of parking functions

α = (a1,··· ,an) of length n with n+k (respe. n)parking spaces satisfying 1 ≤ ai≤ n+k

(resp. 1 ≤ ai≤ s) for all i. Let pn,n+k;≤n+k= |Pn,n+k;≤n+k| and pn;≤s= |Pn;≤s|. Let

Pl

n;≤sdenote the set of parking functions α = (a1,··· ,an) ∈ Pn;≤ssuch that a1 = l

and pl

n;≤s= |Pl

n;≤s|. We derive some formulas and recurrence relations for the sequences

pn,n+k;≤n+k, pn;≤sand pl

n;≤sand give the generating functions for these sequences. We

also study the asymptotic behavior for these sequences.

Keyword: parking function; leading term; asymptotic behavior

∗Partially supported by NSC 96-2115-M-006-012

†Email address of the corresponding author: majun@math.sinica.edu.tw

‡jean.yh@ms45.url.com.tw

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1Introductioin

Throughout the paper, we let [n] := {1,2,··· ,n} and [m,n] := {m,··· ,n}. Suppose that

n cars have to be parked in m parking spaces which are arranged in a line and numbered 1 to

m from left to right. Each car has initial parking preference ai; if space aiis occupied, the car

moves to the first unoccupied space to the right. We call (a1,··· ,an) preference set. Clearly,

the number of preference sets is mn. If a preference set (a1,··· ,an) satisfies ai ≤ ai+1for

1 ≤ i ≤ n−1, then we say that this preference set is ordered. If all the cars can find a parking

space, then we say the preference set is a parking function. If there are exactly k cars which

can’t be parked, then the preference set is called a k-flaw preference set.

Let n, m, s, and k be four nonnegative integers with 1 ≤ s ≤ m and k ≤ n − 1. Sup-

pose there are m parking spaces. We use Pn,m;≤s;kto denote a set of k-flaw preference sets

(a1,··· ,an) of length n satisfying 1 ≤ ai≤ s for all i. For 1 ≤ l ≤ s, we use Pl

n,m;≤s;kto

denote a set of preference sets (a1,··· ,an) ∈ Pn,m;≤s;ksuch that a1= l. Let Pn,m;=s;k(resp.

Pl

n,m;=s;k) be a set of preference sets (a1,··· ,an) ∈ Pn,m;≤s;k(resp. ∈ Pl

n,m;≤s;k) such that

aj = s for some j. Let pn,m;≤s;k = |Pn,m;≤s;k|, pl

n,m;≤s;k= |Pl

n,m;≤s;k|, pn,m;=s;k= |Pn,m;=s;k|

and pl

n,m;=s;k= |Pl

n,m;=s;k|. For any of the above cases, if the parameter k ( resp. m ) doesn’t

appear, we understand k = 0 ( resp. m = n ); if the parameter m and s are both erased, we

understand s = m = n.

There are some results about parking functions with s = m = n. Riordan introduced

parking functions in [11]. He derived that the number of parking functions of length n is (n+

1)n−1, which coincides with the number of labeled trees on n+1 vertices by Cayley’s formula.

Several bijections between the two sets are known (e.g., see [5, 11, 12]). Furthermore, define

a generating function P(x) =?

inverse of the function ψ(x) = xe−x, i.e., ψ(xP(x)) = x. Riordan concluded that the number

n≥0

(n+1)n−1

n!

xn. It is well known that xP(x) is the compositional

of ordered parking functions is

1

n+1

?2n

n

?, which is also equals the number of Dyck path of

semilength n. Parking functions have been found in connection to many other combinatorial

structures such as acyclic mappings, polytopes, non-crossing partitions, non-nesting partitions,

hyperplane arrangements,etc. Refer to [4, 5, 6, 10, 13, 14] for more information.

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Any parking function (a1,··· ,an) can be redefined that its increasing rearrangement

(b1,··· ,bn) satisfies bi≤ i. Pitman and Stanley generalized the notion of parking functions in

[10]. Let x = (x1,··· ,xn) be a sequence of positive integers. The sequence α = (a1,··· ,an)

is called an x-parking function if the non-decreasing rearrangement (b1,··· ,bn) of α satis-

fies bi ≤ x1+ ··· + xifor any 1 ≤ i ≤ n. Thus, the ordinary parking function is the case

x = (1,··· ,1). By the determinant formula of Gonˇ carove polynomials, Kung and Yan [9]

obtained the number of x-parking functions for an arbitrary x. See also [15, 16, 17] for the

explicit formulas and properties for some specified cases of x.

An x-parking function (a1,··· ,an) is said to be k-leading if a1= k. Let qn,kdenote the

number of k-leading ordinary parking functions of length n. Foata and Riordan [5] derived a

generating function for qn,kalgebraically. Recently, Sen-peng Eu, Tung-shan Fu and Chun-Ju

Lai [1] gave a combinatorial approach to the enumeration of (a,b,··· ,b)-parking functions by

their leading terms.

Riordan [11] told us the relations between ordered parking functions and Dyck paths. Sen-

peng Eu et al. [2, 3] considered the problem of the enumerations of lattice paths with flaws. It

is natural to consider the problem of the enumerations of preference sets with flaws. Ordered

k-flaw preference sets were studied in [7]. Building on work in this paper, we give enumerations

of k-flaw preference sets in [8].

In this paper, we first consider enumerations of parking functions in Pn,m;≤m. When

m ≥ n, Riordan [11] gave a explicit formula pn,m;≤m = (m − n + 1)(m + 1)n−1. We ob-

tain another formula pn,n+k;≤n+k =

?

r0+···+rk=n

?

n

r0,···,rk

? k ?

i=0(ri+ 1)ri−1for any n ≥ 0 and

k ≥ 0 and find that the sequence pn,n+k;≤n+ksatisfies the recurrence relation pn,n+k;≤n+k=

k ?

conclude that pn,m;≤m;n−m= mn−

?

Then, we focus on the problem of enumerations of parking functions in Pn;≤s. We prove that

i=0

?n

i

?pipn−i,n−i+k−1;≤n−i+k−1. When m < n, at least n−m cars can’t find parking spaces. We

?n

m−2

i=0

i

?(i + 1)i−1(m − i − 1)n−ifor any 0 ≤ m ≤ n.

pn;≤s= pn,s;≤s;n−sby a bijection from the sets Pn;≤sto Pn,s;≤s;n−sfor any 1 ≤ s ≤ n. Also we

k+1

?

for any n ≥ k+1, we derive two recurrence relations pn;≤n−k= pn;≤n−k+1−

obtain that pn;≤n−k=

i=0(−1)i?n

i

?(n−i+1)n−i−1(k+1−i)ifor any 0 ≤ k ≤ n−1. Furthermore,

k ?

i=1

?n

i

?pn−i;≤n−kand

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pn;≤n−k= (n+1)n−1−

k ?

i=1

?n

i

?(k−i+1)(k+1)i−1pn−i;≤n−k. Since pn;=n−k+1= pn;≤n−k+1−pn;≤n−k,

we have pn;=n−k= pn;=n−k+1+?n

with pn;=n= nn−1.

k+1

?(n−k)n−k−2−

k ?

i=1

?n

i

?pn−i;=n−kfor any k ≥ 1 and n ≥ k+1,

Motivated by the work of Foata and Riordan in [5] as well as Sen-Peng Eu et al. in [1], we

investigate the problem of the enumerations of some parking functions with leading term l.

We derive the formula pl

n;≤s= sn−1−

l−2

?

i=0

?n−1

i

?(s−i−1)n−i−1pi−

s−2

?

i=l

?n−1

i−1

?(s−i−1)n−ipl

n;≤sto Pn−1;≤sfor

ifor any

1 ≤ l ≤ s ≤ n. We prove that ps

n;≤s= pn−1;≤sby a bijection from the sets Ps

any 1 ≤ s ≤ n−1. Furthermore, for any n ≥ k+1 and l ≤ n−k, we conclude that pl

n;≤n−k=

pl

n;≤n−k+1−

k ?

i=1

?n−1

i

?pl

n;≤n−k+1−pl

n−i;≤n−kand pl

n;≤n−k= pl

n−

k ?

i=1

n;=n= pn−1and pn−k

k ?

?n−1

i

?(k−i+1)(k+1)i−1pl

n−i;≤n−k. Noting

that pl

n;=n−k+1= pl

n;≤n−k, we obtain pn

n;=n−k= pn−1;≤n−kfor any

k ≥ 1. Let k ≥ 1, then pl

n;=n−k= pl

n;=n−k+1+?n−1

k+1

?pl

n−k−1−

i=1

?n−1

i

?pl

n−i;=n−kfor any n ≥ k+2

and l ≤ n − k − 1.

Also we give the generating functions of some sequences. For a fixed k ≥ 0, we define a

generating function Qk(x) =?

any k ≥ 1. Let Q(x,y) =?

Let Rk(x) =

n≥0

pn,n+k;≤n+k

n!

xn, then Q0(x) = P(x), Qk(x) = Qk−1(x)P(x) for

k≥0Qk(x)yk, then Q(x,y) =

pn;≤n−k

n!

xnfor any k ≥ 0, then R0(x) = P(x) and Rk+1(x) = Rk(x) −

P(x)

1−yP(x).

?

n≥k

k+1

?

currence relation, by induction, we prove that Rk(x) = P(x)

i=1

xi

i!Rk+1−i(x) for any k ≥ 1, with initial condition R1(x) = (1 − x)P(x) − 1. Using this re-

k ?

exy−y.

i=0

(−1)i(k+1−i)i

i!

xi−

k−1

?

i=0

(−1)i(k−i)i

i!

xi

for any k ≥ 0. Let R(x,y) =?

Let Hk(x) =

k≥0Rk(x)yk, then R(x,y) =

P(x)−y

?

n≥k

pn;=n−k

n!

xn, then Hk(x) = Hk−1(x) +

xk+1

(k+1)!P(x) −

k ?

i=1

2x2)−x. In fact, we

xi

i!Hk−i(x) for any

k ≥ 2, with initial conditions H0(x) = xP(x) +1 and H1(x) = P(x)(x−1

may prove that for any k ≥ 0,

Hk(x) = P(x)

?

k

?

(−1)i(k − i)i

i!

i=0

(−1)i(k + 1 − i)i

i!

xi−

k+1

?

i=0

(−1)i(k + 2 − i)i

i!

xi

?

−

?k−1

i=0

?

xi−

k

?

i=0

(−1)i(k + 1 − i)i

i!

xi

?

.

Let H(x,y) =?

Let L(x) =?

k≥0Hk(x)yk, then H(x,y) =

p1

n

(n−1)!xnand Tk(x) =

P(x)(exy−1)−y2+y

y(exy−y)

.

n≥1

?

n≥k+1

pn−k

n

(n−1)!xnfor any k ≥ 0, then L(x) and T0(x) sat-

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isfy the differential equations L(x)+xL′(x) = 2xP′(x) and T0(x)+xT′

0(x) = 2xP(x)+x2P′(x),

respectively. Solving these two equations, we have L(x) = x[P(x)]2and T0(x) = xP(x). We

find that Tk(x) = Tk−1(x)+(k+1)k−1

k ?

tions (1−y)

?

and T(x,y)−xP(x) = yT(x,y)+xP(x)[P(xy)−1]−xy[P(xy)]2. The second identity implies

k!

xk+1P(x)−2(k+1)k−2

(k−1)!xkfor any k ≥ 1, equivalently, Tk(x) =

P(x)

i=0

(i+1)i−1

i!

xi+1−

k ?

i=1

2(i+1)i−2

(i−1)!xi. Let T(x,y) =?

T(x,y) + x∂T(x,y)

∂y

= 2xP(x)P(xy)+x2P(xy)P′(x)+x2yP(x)P′(xy)−2xyP′(xy),

k≥0Tk(x)yk, then T(x,y) satisfies the equa-

?

T(x,y) =

xP(xy)[P(x)−yP(xy)]

1−y

.

Let Fk,s(x) =

?

n≥s+1

pn−s

n;≤n−k

(n−1)!xnfor any s ≥ k ≥ 0, then

Fk,s(x) = Ts(x) −

k

?

i=1

(k − i + 1)(k + 1)i−1

i!

xiFk−i,s−i(x)

with the initial condition F0,s(x) = Ts(x) and Fk,s(x) =

k ?

i=0

(−1)i(k+1−i)i

i!

xiTs−i(x). Let Fk(x,y) =

?

s≥k

Fk,s(x)ysfor any k ≥ 0, then

Fk(x,y) = T(x,y) −

k−1

?

i=0

Ti(x)yi−

k

?

i=1

(k − i + 1)(k + 1)i−1

i!

(xy)iFk−i(x,y)

and

Fk(x,y) = T(x,y)

k

?

i=0

(−1)i(k + 1 − i)i

i!

(xy)i−

k−1

?

s=0

k−1−s

?

i=0

(−1)i(k + 1 − i)i

i!

(xy)iTs(x)ys.

Let F(x,y,z) =?

Let Dk,s(x) =

k≥0Fk(x,y)zk, then F(x,y,z) =

pn−s

n;=n−k

(n−1)!xnfor any s ≥ k ≥ 0, then when s = k, we have

T(x,y)−zT(x,yz)

exyz−z

.

?

n≥s+1

Dk,k(x) =

xp(x)

ifk = 0

x[P(x) − 1]

ifk = 1

P(x)

k−1

?

i=0

(−1)i(k−i)i

i!

xi−

k−2

?

i=0

(−1)i(k−1−i)i

i!

xi

ifk ≥ 2

When s ≥ k + 1, Dk,s(x) satisfies the following recurrence relation Dk,s(x) = Dk−1,s(x) +

k ?

Ts−1(x) for any s ≥ 1, equivalently, Dk,s(x) =

xk+1

(k+1)!Ts−k−1(x) −

i=1

xi

i!Dk−i,s−i(x) , with initial conditions D0,0(x) = xP(x) and D0,s(x) =

k+1

?

i=1(−1)i xi

i!Ts−i(x)[(k + 1 − i)i− (k + 2 − i)i].

Let Dk(x,y) =

?

s≥k

Dk,s(x)ysfor any k ≥ 1, then Dk(x,y) satisfies the following recurrence

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