DERIVATIVE POLYNOMIALS AND ENUMERATION OF PERMUTATIONS BY THEIR ALTERNATING DESCENTS
arXiv:1504.02372v1 [math.CO] 9 Apr 2015
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. Moreover, we study the interlacing property of the real parts of the zeros of the generating polynomials of these numbers.
1. Introduction Let Sn denote the symmetric group of all permutations of [n], where [n] = {1, 2, . . . , n}. For a permutation π ∈ Sn , we define a descent to be a position i such that π(i) > π(i + 1). Denote by des (π) the number of descents of π. The classical Eulerian polynomials An (x) are defined by An (x) =
X
x
des (π)
=
n−1 X
A(n, k)xk .
k=0
π∈Sn
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 [2] 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 k)xk b A(n, x = An (x) = k=0
π∈Sn
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
2010 Mathematics Subject Classification. Primary 05A15; Secondary 26C10. Key words and phrases. Alternating Eulerian polynomials; Derivative polynomials; Zeros. 1
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S.-M. MA AND Y.-N. YEH
Chebikin [2] proved that 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)
b k) satisfy the recurrence relation and the numbers A(n, k n X X n i=0 j=0
i
b j + 1)A(n b − i, k − j + 1) = (n + 1 − k)A(n, b k + 1) + (k + 1)A(n, b k + 2). A(i,
In recent years, several authors pay attention to the alternating descent statistic and its associated permutation statistics. The reader is referred to [5, 9] for recent progress on this subject. For example, Gessel and Zhuang [5] defined an alternating run to be a maximal consecutive subsequence with no alternating descents. The purpose of this paper is to present an explicit b k). In Section 2, we express the polynomials A bn (x) in terms of formula for the numbers A(n, the derivative polynomials Pn (x) defined by Hoffman [4]: Pn (tan θ) =
dn tan θ. dθ n
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 D n (x) is a polynomial in x. Let Pn (x) = D n (x). Then P0 (x) = x and Pn+1 (x) = (1 + x2 )Pn′ (x).
(2)
Clearly, deg Pn (x) = n + 1. By definition, we have tan(θ + z) =
X
Pn (x)
n≥0
Let Pn (x) =
Pn+1 k=0
x + tan z zn = , n! 1 − x tan z
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).
The first few terms can be computed directly as follows: P1 (x) = 1 + x2 , P2 (x) = 2x + 2x3 , P3 (x) = 2 + 8x2 + 6x4 , P4 (x) = 16x + 40x3 + 24x5 . Note that Pn (−x) = (−1)n+1 Pn (x) and xkP2n (x). Thus we have the following expression: ⌊(n+1)/2⌋
Pn (x) =
X k=0
p(n, n − 2k + 1)xn−2k+1 .
There is an explicit formula for the numbers p(n, n − 2k + 1).
(3)
DERIVATIVE POLYNOMIALS AND ENUMERATION OF PERMUTATIONS BY THEIR ALTERNATING DESCENTS 3
Proposition 1 ([7, Proposition 1]). For n ≥ 1 and 0 ≤ k ≤ ⌊(n + 1)/2⌋, we have X n i i k i! (−2)n−i p(n, n − 2k + 1) = (−1) − . i n − 2k n − 2k + 1 i≥1
Now we present the first main result of this paper. Theorem 2. For n ≥ 1, we have bn (x) = (1 − x)n+1 Pn 2 (1 + x )A n
2
1+x 1−x
.
(4)
Proof. It follows from (3) that X X 1 + x (z − xz)n 1 + x zn = (1 − x) Pn (1 − x)n+1 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 sec(2z − 2xz) + tan(2z − 2xz) − 1 2 tan(z − xz) = . 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) =
2t 1 + t2 , tan(2z − 2xz) = . 1 − t2 1 − t2
Then the left hand side of (5) equals 2t 2t(1 + t) = . 1 − t2 − x(1 + t)2 1 − x − (1 + x)t This completes the proof.
It follows from (2) that bn+1 (x) = (1−x)n P ′ 2 A n n
1+x 1−x
⌊(n+1)/2⌋
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(⌊ k2 ⌋,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
⌊(n+1)/2⌋
X k=0
(n − 2k + 1)p(n, n − 2k + 1)E(n, k, s).
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S.-M. MA AND Y.-N. YEH
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) . It is well known that the polynomials Wn (x) satisfy the recurrence relation Wn+1 (x) = (nx − x + 2)Wn (x) + 2x(1 − x)Wn′ (x), with initial conditions W1 (x) = 1, W2 (x) = 2 and W3 (x) = 4 + 2x. By the theory of enriched P-partitions, Stembridge [11, Remark 4.8] showed that 2n−1 4x An (x). (7) = Wn 2 (1 + x) (1 + x)n−1 From [6, Theorem 2], we have Pn (x) = xn−1 (1 + x2 )Wn (1 + x−2 ).
(8)
Therefore, combining (4) and (8), we get the counterpart of (7): 2 + 2x2 2n−1 bn (x). Wn A = (1 + x)2 (1 + x)n−1 3. Zeros of the alternating Eulerian polynomials Combining (2) and (4), it is easy to derive that bn+1 (x) = (1 + n + 2x + nx2 − x2 )A bn (x) + (1 − x)(1 + x2 )A b′ (x) 2A n
bn (x) for n ≥ 1. The bijection π 7→ π c on Sn defined by π c (i) = n + 1 − π(i) shows that A b2n (x) for n ≥ 1. It is well known that the classical Eulerian is symmetric. Hence (x + 1)kA polynomials An (x) have only real zeros, and the zeros of An (x) separates that of An+1 (x) (see bn (x). B´ona [1, p. 24] for instance). Now we present the corresponding result for A
b2n+1 (x) and A b2n+2 (x)/(1 + x) are imaginary with Theorem 4. For n ≥ 1, the zeros of A bn (x) are equal to 1. Furthermore, the sequence of multiplicity 1, and the moduli of all zeros of A bn (x) separates that of A bn+1 (x). More precisely, suppose that {rj ± real parts of the zeros of A n−1 n n b2n (x)/(1+x), {sj ±tj i} b ℓj i}j=1 are all zeros of A j=1 are all zeros of A2n+1 (x) and {pj ±qj i}j=1 are b2n+2 (x)/(1+x), where −1 < r1 < r2 < · · · < rn−1 < 0, −1 < s1 < s2 < · · · < sn < 0 all zeros of A and −1 < p1 < p2 < · · · < pn < 0. Then we have
− 1 < s1 < r1 < s2 < r2 < · · · < rn−1 < sn ,
(9)
− 1 < s 1 < p1 < s 2 < p2 < · · · < s n < pn .
(10)
Proof. Define Pen (x) = in−1 Pn (ix). Then
Pen+1 (x) = (1 − x2 )Pen′ (x).
From [8, 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 [3, Corollary 8.7], we see that the zeros of the derivative polynomials Pn (x) are pure imaginary with multiplicity 1, belong
DERIVATIVE POLYNOMIALS AND ENUMERATION OF PERMUTATIONS BY THEIR ALTERNATING DESCENTS 5
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: P2n+1 (x) = (1 + x2 )
n n Y Y (x2 + ai ), P2n+2 (x) = x(1 + x2 ) (x2 + bi ), i=1
i=1
where 0 < a1 < b1 < a2 < b2 < · · · < an < bn < 1.
(11)
Using (4), we get b2n+1 (x) = 2 A 2n
n Y ((1 + x)2 + ai (1 − x)2 ), i=1
b2n+2 (x) = (1 + x) 22n+1 A
Hence
n Y i=1
((1 + x)2 + bi (1 − x)2 ).
√ √ n Y 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 2i bi 1 − bi 1 − bi 2n+1 b + − x+ . 2 A2n+2 (x) = (1 + x) (1 + bi ) x + 1 + bi 1 + bi 1 + bi 1 + bi b2n+1 (x) = 2 A 2n
i=1
Since
√ 2 √ 2 2 ai 1 − ai 2 1 − bi 2 2 bi + = + = 1, 1 + ai 1 + ai 1 + bi 1 + bi bn (x) are equal to 1. Note that the moduli of all zeros of A
sj = −
1 − bj 1 − aj , pj = − . 1 + aj 1 + bj
So (10) is immediate. Along the same lines, one can get (9).
From Theorem 4, we immediately get that the sequence of imaginary parts of the zeros of b bn+1 (x). An (x) also separates that of A References
[1] M. B´ ona, Combinatorics of Permutations, second edition, CRC Press, Boca Raton, FL, 2012. [2] D. Chebikin, Variations on descents and inversions in permutations, Electron. J. Combin. 15 (2008), #R132. [3] G. Hetyei, Tchebyshev triangulations of stable simplicial complexes, J. Combin. Theory Ser. A 115 (2008), 569–592. [4] M.E. Hoffman, Derivative polynomials for tangent and secant, Amer. Math. Monthly 102 (1995) 23–30. [5] I.M. Gessel, Y. Zhuang, Counting Permutations by alternating descents, Electron. J. Combin. 21(4) (2014), #P4.23. [6] S.-M. Ma, Derivative polynomials and enumeration of permutations by number of interior and left peaks, Discrete Math. 312 (2012), 405–412. [7] S.-M. Ma, An explicit formula for the number of permutations with a given number of alternating runs, J. Combin. Theory Ser. A 119 (2012), 1660–1664. [8] S.-M. Ma, Y. Wang, q-Eulerian polynomials and polynomials with only real zeros, Electron. J. Combin. 15 (2008), #R17.
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[9] J.B. Remmel, Generating Functions for Alternating Descents and Alternating Major Index, Ann. Comb., 16 (2012), 625–650. [10] N.J.A. Sloane, The On-Line Encyclopedia of Integer Sequences, http://oeis.org. [11] J. Stembridge, Enriched P-partitions, Trans. Amer. Math. Soc. 349(2): (1997) 763–788. 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)
