Some Enumerations for Parking Functions
ABSTRACT In this paper, let $\mathcal{P}_{n,n+k;\leq n+k}$ (resp. $\mathcal{P}_{n;\leq s}$) denote the set of parking functions $\alpha=(a_1,...,a_n)$ of length $n$ with $n+k$ (respe. $n$)parking spaces satisfying $1\leq a_i\leq n+k$ (resp. $1\leq a_i\leq s$) for all $i$. Let $p_{n,n+k;\leq n+k}=\mathcal{P}_{n,n+k;\leq n+k}$ and $p_{n;\leq s}=\mathcal{P}_{n;\leq s}$. Let $\mathcal{P}_{n;\leq s}^l$ denote the set of parking functions $\alpha=(a_1,...,a_n)\in\mathcal{P}_{n;\leq s}$ such that $a_1=l$ and $p_{n;\leq s}^l=\mathcal{P}_{n;\leq s}^l$. We derive some formulas and recurrence relations for the sequences $p_{n,n+k;\leq n+k}$, $p_{n;\leq s}$ and $p_{n;\leq s}^l$ and give the generating functions for these sequences. We also study the asymptotic behavior for these sequences.

Article: kflaw Preference Sets
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ABSTRACT: In this paper, let $\mathcal{P}_{n;\leq s;k}^l$ denote a set of $k$flaw preference sets $(a_1,...,a_n)$ with $n$ parking spaces satisfying that $1\leq a_i\leq s$ for any $i$ and $a_1=l$ and $p_{n;\leq s;k}^l=\mathcal{P}_{n;\leq s;k}^l$. We use a combinatorial approach to the enumeration of $k$flaw preference sets by their leading terms. The approach relies on bijections between the $k$flaw preference sets and labeled rooted forests. Some bijective results between certain sets of $k$flaw preference sets of distinct leading terms are also given. We derive some formulas and recurrence relations for the sequences $p_{n;\leq s;k}^l$ and give the generating functions for these sequences.07/2008;
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arXiv:0806.0424v1 [math.CO] 3 Jun 2008
Some Enumerations for Parking Functions
PoYi 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 962115M006012
†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 kflaw 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 kflaw 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, noncrossing partitions, nonnesting 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 xparking function if the nondecreasing 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 xparking functions for an arbitrary x. See also [15, 16, 17] for the
explicit formulas and properties for some specified cases of x.
An xparking function (a1,··· ,an) is said to be kleading if a1= k. Let qn,kdenote the
number of kleading ordinary parking functions of length n. Foata and Riordan [5] derived a
generating function for qn,kalgebraically. Recently, Senpeng Eu, Tungshan Fu and ChunJu
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
kflaw preference sets were studied in [7]. Building on work in this paper, we give enumerations
of kflaw 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 SenPeng 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
5