ENUMERATION OF PERMUTATIONS BY NUMBER OF ALTERNATING
DESCENTS
SHI-MEI MA AND YEONG-NAN YEH
Abstract. In this paper we present an explicit formula for the number of permutations with
a given number of alternating descents. As an application, we obtain an interlacing property
for the zeros of alternating Eulerian polynomials.
1. Introduction
Let Sn denote the symmetric group of all permutations of [n], where [n] = {1, 2, . . . , n}.
For π = π(1)π(2) · · · π(n) ∈ Sn , we define a descent to be an index i ∈ [n − 1] such that
π(i) > π(i + 1). Let des (π) be the number of descents of π. Then the equations
An (x) =
X
x
des (π)
=
π∈Sn
n−1
X
A(n, k)xk .
k=0
define the Eulerian polynomials An (x) and Eulerian numbers A(n, k). The exponential generating function for An (x) is
A(x, z) =
X
An (x)
n≥1
zn
e(1−x)z − 1
=
.
n!
1 − xe(1−x)z
The numbers A(n, k) satisfy the recurrence relation
A(n + 1, k) = (k + 1)A(n, k) + (n − k + 1)A(n, k − 1).
with the initial conditions A(1, 0) = 1 and A(1, k) = 0 for k ≥ 1 (see [13, A008292]). There is a
large literature devoted to the descent statistic and its variations (see [1, 7, 11] for instance).
As a variation of the descent statistic, the number of alternating descents of a permutation
π ∈ Sn is defined by
altdes (π) = |{2i : π(2i) < π(2i + 1)} ∪ {2i + 1 : π(2i + 1) > π(2i + 2)}|.
We say that π has a 3-descent at index i if π(i)π(i + 1)π(i + 2) has one of the patterns: 132,
213, or 321. Chebikin [3] showed that the alternating descent statistic of permutations in Sn is
equidistributed with the 3-descent statistic of permutations in {π ∈ Sn+1 : π1 = 1}. Then the
equations
n−1
X
X
altdes (π)
b
b k)xk
An (x) =
x
=
A(n,
π∈Sn
k=0
2010 Mathematics Subject Classification. Primary 05A15; Secondary 26C10.
Key words and phrases. Alternating Eulerian polynomials; Derivative polynomials; Zeros.
1
2
S.-M. MA AND Y.-N. YEH
bn (x) and the alternating Eulerian numbers A(n,
b k).
define the alternating Eulerian polynomials A
bn (x) are given as follows:
The first few A
b1 (x) = 1,
A
b2 (x) = 1 + x,
A
b3 (x) = 2 + 2x + 2x2 ,
A
b4 (x) = 5 + 7x + 7x2 + 5x3 ,
A
b5 (x) = 16 + 26x + 36x2 + 26x3 + 16x4 .
A
bn (x) is symmetric.
The bijection π 7→ π c on Sn defined by π c (i) = n + 1 − π(i) shows that A
Chebikin [3] proved the following formulas:
X
n≥1
n
bn (x) z = sec(1 − x)z + tan(1 − x)z − 1 ;
A
n!
1 − x(sec(1 − x)z + tan(1 − x)z)
(1)
n X
k X
n b
b − i, k − j + 1) = (n + 1 − k)A(n,
b k + 1) + (k + 1)A(n,
b k + 2).
A(i, j + 1)A(n
i
i=0 j=0
In recent years, several authors pay attention to the alternating descent statistic and its
associated permutation statistics. The reader is referred to [2, 4, 12] for recent progress on
this subject. For example, Gessel and Zhuang [4] defined an alternating run to be a maximal
consecutive subsequence with no alternating descents. This paper is a continuation of [9]. In
bn (x) in terms of the derivative polynomials Pn (x) defined
Section 2, we express the polynomials A
by Hoffman [6]:
dn
Pn (tan θ) = n tan θ.
dθ
2. An explicit formula
Let D denote the differential operator d/dθ. Set x = tan θ. Then D(xn ) = nxn−1 (1 + x2 ) for
n ≥ 1. Thus Dn (x) is a polynomial in x. Let Pn (x) = Dn (x). Then P0 (x) = x and
Pn+1 (x) = (1 + x2 )Pn0 (x).
Clearly, deg Pn (x) = n + 1. By definition, we have
X
zn
x + tan z
=
,
tan(θ + z) =
Pn (x)
n!
1 − x tan z
n≥0
Let Pn (x) =
Pn+1
k=0
p(n, k)xk . It is easy to verify that
p(n, k) = (k + 1)p(n − 1, k + 1) + (k − 1)p(n − 1, k − 1).
Note that Pn (−x) = (−1)n+1 Pn (x) and xkP2n (x). Thus
b(n+1)/2c
Pn (x) =
X
p(n, n − 2k + 1)xn−2k+1 .
k=0
There is an explicit formula for the numbers p(n, n − 2k + 1).
(2)
(3)
ENUMERATION OF PERMUTATIONS BY NUMBER OF ALTERNATING DESCENTS
3
Proposition 1 ([9, Proposition 1]). For n ≥ 1 and 0 ≤ k ≤ b(n + 1)/2c, we have
X n
i
i
n−i
k
−
,
i!
(−2)
p(n, n − 2k + 1) = (−1)
n − 2k
n − 2k + 1
i
i≥1
where
n
i
is the Stirling number of the second kind.
Now we present the first main result of this paper.
Theorem 2. For n ≥ 1, we have
n
2
bn (x) = (1 − x)n+1 Pn
2 (1 + x )A
1+x
1−x
.
(4)
Proof. It follows from (3) that
X
X
1 + x (z − xz)n
1 + x zn
n+1
= (1 − x)
Pn
(1 − x)
Pn
1 − x n!
1−x
n!
n≥1
n≥1
= (1 + x2 )
2 tan(z − xz)
.
1 − x − (1 + x) tan(z − xz)
Comparing with (1), it suffices to show the following
2 tan(z − xz)
sec(2z − 2xz) + tan(2z − 2xz) − 1
=
.
1 − x(sec(2z − 2xz) + tan(2z − 2xz))
1 − x − (1 + x) tan(z − xz)
(5)
Set t = tan(z − xz). Using the tangent half-angle substitution, we have
sec(2z − 2xz) =
1 + t2
2t
, tan(2z − 2xz) =
.
2
1−t
1 − t2
Then the left hand side of (5) equals
2t
2t(1 + t)
=
.
2
2
1 − t − x(1 + t)
1 − x − (1 + x)t
This completes the proof.
It follows from (2) that
bn+1 (x) = (1−x)n Pn0
2n A
1+x
1−x
b(n+1)/2c
X
=
(n−2k +1)p(n, n−2k +1)(1−x)2k (1+x)n−2k . (6)
k=0
Denote by E(n, k, s) the coefficients xs of (1 − x)2k (1 + x)n−2k . Clearly,
min(b k2 c,s)
E(n, k, s) =
X
j
(−1)
j=0
2k
j
n − 2k
.
s−j
Then we get the following result.
Corollary 3. For n ≥ 2 and 1 ≤ s ≤ n, we have
b + 1, s) = 1
A(n
2n
b(n+1)/2c
X
k=0
(n − 2k + 1)p(n, n − 2k + 1)E(n, k, s).
4
S.-M. MA AND Y.-N. YEH
It follows from (2) and (4) that
bn+1 (x) = (1 + n + 2x + nx2 − x2 )A
bn (x) + (1 − x)(1 + x2 )A
b0 (x)
2A
n
b k) satisfy the recurrence relation
for n ≥ 1. Therefore, the numbers A(n,
b + 1, k) = (k + 1)(A(n,
b k + 1) + A(n,
b k − 1)) + (n − k + 1)(A(n,
b k) + A(n,
b k − 2)),
2A(n
b 0) = 1 and A(1,
b k) = 0 for k ≥ 1.
with initial conditions A(1,
Let π = π(1)π(2) · · · π(n) ∈ Sn . An interior peak in π is an index i ∈ {2, 3, . . . , n − 1} such
that π(i − 1) < π(i) > π(i + 1). Let pk (π) denote the number of interior peaks of π. Let
P
Wn (x) = π∈Sn xpk (π) . Clearly, we have deg Wn (x) = b(n − 1)/2c. It is well known that the
polynomials Wn (x) satisfy the recurrence relation
Wn+1 (x) = (nx − x + 2)Wn (x) + 2x(1 − x)Wn0 (x),
with initial conditions W1 (x) = 1, W2 (x) = 2 and W3 (x) = 4 + 2x. By the theory of enriched
P-partitions, Stembridge [14, Remark 4.8] showed that
4x
2n−1
Wn
=
An (x).
(7)
(1 + x)2
(1 + x)n−1
From [8, Theorem 2], we have
Pn (x) = xn−1 (1 + x2 )Wn (1 + x−2 ).
(8)
Therefore, by (4) and (8), we get a counterpart of (7):
2n−1
2 + 2x2
bn (x).
Wn
=
A
(1 + x)2
(1 + x)n−1
Thus
b(n−1)/2c
bn (x) =
A
1
X
k=0
2n−k−1
W (n, k)(1 + x)n−2k−1 (1 + x2 )k ,
b2n (x) for n ≥ 1.
which implies that (x + 1)kA
3. Zeros of the alternating Eulerian polynomials
It is well known that the Eulerian polynomial An (x) has only real zeros, and the zeros of An (x)
separates that of An+1 (x) (see Bóna [1, p. 24] for instance). Now we present a corresponding
bn (x).
result for A
b2n+1 (x) and A
b2n+2 (x)/(1 + x) are non-real comTheorem 4. For any n ≥ 1, all the zeros of A
bn (x) are all equal to 1. Furthermore,
plexes with multiplicity 1, and the moduli of the zeros of A
bn (x) separates that of A
bn+1 (x). More precisely, let
the sequence of real parts of the zeros of A
n−1
n
n
b2n (x)/(1+x), let {sj ±tj i}
b
{rj ±`j i}j=1 be zeros of A
j=1 be zeros of A2n+1 (x) and let {pj ±qj i}j=1
b2n+2 (x)/(1+x), where −1 < r1 < r2 < · · · < rn−1 < 0, −1 < s1 < s2 < · · · < sn < 0
be zeros of A
and −1 < p1 < p2 < · · · < pn < 0. Then
− 1 < s1 < r1 < s2 < r2 < · · · < rn−1 < sn ,
(9)
− 1 < s1 < p1 < s2 < p2 < · · · < sn < pn .
(10)
bn (x) also separates that of A
bn+1 (x).
Moreover, the sequence of imaginary parts of the zeros of A
ENUMERATION OF PERMUTATIONS BY NUMBER OF ALTERNATING DESCENTS
5
Proof. Set Pen (x) = in−1 Pn (ix). Then
Pen+1 (x) = (1 − x2 )Pen0 (x).
Using [10, Theorem 2], we get that the polynomials Pen (x) have only real zeros, belong to [−1, 1]
and the sequence of zeros of Pen (x) separates that of Pen+1 (x). From [5, Corollary 8.7], we see
that the zeros of the derivative polynomials Pn (x) are pure imaginary with multiplicity 1, belong
to the line segment [−i, i]. In particular, (1 + x2 )kPn (x). Therefore, the polynomials P2n+1 (x)
and P2n+2 (x) have the following expressions:
2
P2n+1 (x) = (1 + x )
n
Y
n
Y
(x + ai ), P2n+2 (x) = x(1 + x ) (x2 + bi ),
2
2
i=1
i=1
where
0 < a1 < b1 < a2 < b2 < · · · < an < bn < 1.
(11)
By (4), we obtain
b2n+1 (x) =
22n A
n
Y
((1 + x)2 + ai (1 − x)2 ),
i=1
2n+1
2
b2n+2 (x) = (1 + x)
A
n
Y
((1 + x)2 + bi (1 − x)2 ).
i=1
Hence
√ √ 2i ai
2i ai
1 − ai
1 − ai
(1 + ai ) x +
+
x+
−
,
1 + ai 1 + ai
1 + ai 1 + ai
i=1
√ √ n
Y
2i bi
1 − bi
2i bi
1 − bi
2n+1 b
+
x+
−
.
2
A2n+2 (x) = (1 + x) (1 + bi ) x +
1 + bi 1 + bi
1 + bi 1 + bi
2n
b2n+1 (x) =
2 A
n
Y
i=1
Set
sj = −
1 − aj
1 − bj
, pj = −
,
1 + aj
1 + bj
By (11), we get (10). Along the same lines, we get (9). Note that
1 − ai
1 + ai
2
√ 2 √ 2
2 ai
1 − bi 2
2 bi
+
=
+
= 1.
1 + ai
1 + bi
1 + bi
bn (x) are all equal to 1, which also implies the interlacing property
Then the moduli of zeros of A
of the imaginary parts of these zeros.
Acknowledgements.
This work was supported by NSFC (11401083) and the Fundamental Research Funds for the
Central Universities (N130423010).
6
S.-M. MA AND Y.-N. YEH
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School of Mathematics and Statistics, Northeastern University at Qinhuangdao, Hebei 066004,
P.R. China
E-mail address: [email protected] (S.-M. Ma)
Institute of Mathematics, Academia Sinica, Taipei, Taiwan
E-mail address: [email protected] (Y.-N. Yeh)