We prove that the Eulerian polynomial satisfies certain polynomial congruences. Furthermore, these congruences characterize the Eulerian polynomial

Iijima, Kazuki

Sasaki, Kyouhei

Takahashi, Yuuki

Yoshinaga, Masahiko

English

Hokkaido University Collection of Scholarly and Academic Papers

EULERIAN POLYNOMIALS AND POLYNOMIAL CONGRUENCES

Contributions to Discrete Mathematics (E-Journal, University of Calgary)

Volume 14, Number 1, Pages 46–54ISSN 1715-0868EULERIAN POLYNOMIALS AND POLYNOMIALCONGRUENCESKAZUKI IIJIMA, KYOUHEI SASAKI, YUUKI TAKAHASHI,AND MASAHIKO YOSHINAGAAbstract. We prove that the Eulerian polynomial satisfies certainpolynomial congruences. Furthermore, these congruences characterizethe Eulerian polynomial.1. IntroductionThe Eulerian polynomial A`(x) (` ≥ 1) was introduced by Euler in thestudy of sums of powers [5]. In this paper, we define the Eulerian polynomialA`(x) as the numerator of the rational function(1.1) F`(x) =∞∑k=1k`xk =(xddx)` 11− x =A`(x)(1− x)`+1 .The first few examples are A1(x) = x,A2(x) = x + x2, A3(x) = x + 4x2 +x3, A4(x) = x + 11x2 + 11x3 + x4, etc. The Eulerian polynomial A`(x)is a monic of degree ` with positive integer coefficients. Write A`(x) =∑`k=1A(`, k)xk. The coefficient A(`, k) is called an Eulerian number.In the 1950s, Riordan [10] discovered a combinatorial interpretation ofEulerian numbers in terms of descents and ascents of permutations, andCarlitz [3] defined q-Eulerian numbers. Since then, Eulerian numbers areactively studied in enumerative combinatorics. (See [4, 11, 8].)Another combinatorial application of the Eulerian polynomial was foundin the theory of hyperplane arrangements [15, 14]. The characteristic poly-nomial of the so-called Linial arrangement [9] can be expressed in terms ofthe root system generalization of Eulerian polynomials introduced by Lamand Postnikov [6]. The comparison of expressions in [9] and [15] yields thefollowing.Received by the editors June 15, 2017, and in revised form January 13, 2019.2000 Mathematics Subject Classification. 05A15, 05E99.Key words and phrases. Eulerian polynomials, Hyperplane arrangements.The authors thank Mr. Rei Yoshida for several discussions about the contents of thispaper. The authors also thank Prof. Richard Stanley for noticing [12]. M. Yoshinaga waspartially supported by JSPS KAKENHI Grant Number JP25400060, JP15KK0144, andJP16K13741.c©2019 University of Calgary46EULERIAN POLYNOMIALS AND POLYNOMIAL CONGRUENCES 47Theorem 1.1 ([15, Proposition 5.5]). For `,m ≥ 2, the Eulerian polynomialA`(x) satisfies the following(1.2) A`(xm) ≡(1 + x+ x2 + · · ·+ xm−1m)`+1A`(x) mod (x− 1)`+1.The purpose of this paper is two-fold. First, we give a direct and simplerproof of Theorem 1.1. Second, we prove the converse of the above theo-rem. Namely, the congruence (1.2) characterizes the Eulerian polynomial asfollows.Theorem 1.2. Let f(x) be a monic of degree `. Then, f(x) = A`(x) if andonly if the congruence (1.2) holds for some m ≥ 2. (See Theorem 5.1.)The remainder of this paper is organized as follows. After recalling classi-cal results on Eulerian polynomials in §2, we briefly describe in §3 the proofof the congruence (1.2) in [15] that is based on the expression of character-istic polynomials of Linial hyperplane arrangements. In §4, we give a directproof of the congruence. In §5, we give the proof of Theorem 1.2.Remark. The right-hand side of (1.2) is discussed also in [12, Proposition2.5].2. Brief review of Eulerian polynomialsIn this section, we recall classical results on the Eulerian polynomial andthe Eulerian numbers A(`, k). By definition (1.1), the Eulerian polynomialA`(x) satisfies the relationA`(x)(1− x)`+1 = xddxA`−1(x)(1− x)` ,which yields the following recursive relation.(2.1) A(`, k) = k ·A(`− 1, k) + (`− k + 1) ·A(`− 1, k − 1).Consider the coordinate change w = 1x . Then, the Euler operator istransformed as x ddx = −w ddw . The direct computation using the relation11− 1w= 1− 11−w yields x`+1A`( 1x) = A`(x). Equivalently, A(`, k) = A(`, `+1− k).Definition (1.1) is also equivalent to(2.2) A`(x) = (1− x)`+1 ·∞∑k=0k`xkand(2.3)∞∑k=0k`xk = A`(x) ·∞∑k=0(−x)k(−`− 1k).48 K. IIJIMA, K. SASAKI, Y. TAKAHASHI, AND M. YOSHINAGAThen, (2.2) yields(2.4) A(`, k) =k∑j=0(−1)j(`+ 1j)(k − j)`,and (2.3) yields(2.5) k` =∑`j=1A(`, j)(k + `− j`).Note that both sides of (2.5) are polynomials of degree ` in k, and it holdsfor any k > 0. Hence the equality holds at the level of polynomials in t.Thus, we have(2.6) t` =∑`j=1A(`, j)(t+ `− j`),which is called the Worpitzky identity [13]. Using the shift operator S :t 7−→ t− 1 (see §3), the Worpitzky identity can be written as(2.7) t` = A`(S)(t+ ``).Next, we consider exponential generating series of A`(x), and describe rela-tions with Bernoulli numbers. First, using (2.2), we have∞∑`=0A`(x)`!t` =∞∑`=0t``!(1− x)`+1∞∑n=0n`xn= (1− x)∞∑n=0xnent(1−x)=1− x1− xet(1−x) .(2.8)Replacing x by −1 in (2.8), we have(2.9)∞∑`=0A`(−1)`!t` =21 + e2t.Recall that the Bernoulli polynomial B`(x) is defined by(2.10)∞∑`=0B`(x)`!t` =textet − 1 ,(B0(x) = 1, B1(x) = x−12 , B2(x) = x2−x+16 , B3(x) = x3−32x2+12x,B4(x) =x4 − 2x3 + x2 − 130 , · · · ) and the constant term B`(0) is called the Bernoullinumber. Replacing x by 0 and t by at with a ∈ C in (2.10), we have(2.11)∞∑`=0B`(0)`!(at)` =ateat − 1 .EULERIAN POLYNOMIALS AND POLYNOMIAL CONGRUENCES 49Using the identity 2te2t+1= 2te2t−1 − 4te4t−1 , the Bernoulli number B`(0) can beexpressed as(2.12) B`(0) =`2`(1− 2`)A`−1(−1).There is another relation between Eulerian polynomials and Bernoulli poly-nomials. Let ` > 0. Using (2.5) and the famous formula∑N−1x=0 x` =B`+1(N)−B`+1(0)`+1 , we have(2.13) B`+1(N)−B`+1(0) = (`+ 1) ·∑`k=1A(`, k)(`+N − k`+ 1).With the shift operator S, (2.13) can also be expressed as(2.14) B`+1(t)−B`+1(0) = (`+ 1)A`(S)(t+ ``+ 1).(This formula appeared in [13, page 209] as “The second form of Bernoullifunction.”)3. Background on hyperplane arrangementsIn this section, we recall the proof of the congruence (1.2) presented in[15].LetA = {H1, . . . ,Hk} be a finite set of affine hyperplanes in a vector spaceV . We denote the set of all intersections of A by L(A) = {∩S | S ⊂ A}.The set L(A) is partially ordered by reverse inclusion, which has a uniqueminimal element 0ˆ = V . The characteristic polynomial of A is defined byχ(A, q) =∑X∈L(A)µ(X)qdimX ,where µ is the Mo¨bius function on L(A), defined byµ(X) ={1, if X = 0ˆ−∑Y <X µ(Y ), otherwise.Let V = {(x0, x1, . . . , xn) ∈ R`+1 |∑xi = 0} ⊂ R`+1. For integers 0 ≤ i <j ≤ ` and s ∈ Z, denote by Hij,s the affine hyperplane {(x0, . . . , x`) ∈ V |xi − xj = s}.Let m ≥ 1 be a positive integer. The arrangementLm = {Hij,s | 0 ≤ i < j ≤ `, 1 ≤ s ≤ m}is called the (extended) Linial arrangement (of type A`). The Linial ar-rangement has several intriguing enumerative properties [9]. Postnikov andStanley [9] (see also [1]) gave the following expression for the characteristicpolynomial χ(Lm, t).(3.1) χ(Lm, t) =(1 + S + S2 + · · ·+ Smm+ 1)`+1t`,50 K. IIJIMA, K. SASAKI, Y. TAKAHASHI, AND M. YOSHINAGAwhere S acts on a function f(t) by Sf(t) = f(t − 1) (naturally Skf(t) =f(t−k)) as the shift operator. Using the Worpitzky identity (2.7), (3.1) canbe written as(3.2) χ(Lm, t) =(1 + S + S2 + · · ·+ Smm+ 1)`+1A`(S)(t+ ``).On the other hand, using the lattice points interpretation of the Worpitzkyidentity, the following formula was obtained in [15](3.3) χ(Lm, t) = A`(Sm+1)(t+ ``).The formulas (3.2) and (3.3) imply that the operator(3.4)(1 + S + S2 + · · ·+ Smm+ 1)`+1A`(S)−A`(Sm+1)annihilates the polynomial(t+``)of degree `, which means that (3.4) is divis-ible by (S − 1)`+1 (see [15, Prop. 2.8]). Hence the congruence (1.2) follows.4. Direct proof of the congruence4.1. Special case: m = 2. We first handle the case m = 2. By consider-ing F`(x) + F`(−x), it is easily seen that the formal power series F`(x) =∑∞k=1 k`xk satisfies(4.1) F`(x)− 2`+1F`(x2) = −F`(−x).Using the Eulerian polynomial, (4.1) can be written as(4.2) (1 + x)`+1 ·A`(x)− 2`+1 ·A`(x2) = − (1− x)`+1 ·A`(−x),which implies the congruence (1.2) for m = 2.Remark. Substituting formally x = 1 into (4.1), we obtain the formula“F`(1) =A`(−1)2`+1(2`+1−1) .” Then, (2.12) implies “F`(1) = −B`+1(0)`+1 ”, whichgives the correct value ζ(−`) = −B`+1(0)`+1 of the Riemann zeta function for` ≥ 1.4.2. General case. Let m ≥ 2. Denote by ζm = e2pi√−1/m the primitivemth root of unity. We will use the following fact(4.3)m−1∑j=1ζjkm ={m− 1, if m|k,−1, if m 6 |k,for k ∈ Z.Using definition (1.1) (or (2.2)), the polynomialA`(xm)−(1 + x+ x2 + · · ·+ xm−1m)`+1A`(x)EULERIAN POLYNOMIALS AND POLYNOMIAL CONGRUENCES 51can be expressed as(1− xm)`+1∞∑k=1k`xmk −(1 + x+ x2 + · · ·+ xm−1m)`+1(1− x)`+1∞∑k=1k`xk=(1− xmm)`+1·{m ·∞∑k=1(mk)`xmk −∞∑k=1k`xk}.It is enough to show that(4.4) P (x) := (1 + x+ · · ·+ xm−1)`+1{m ·∞∑k=1(mk)`xmk −∞∑k=1k`xk}becomes a polynomial. Applying (4.3), we haveP (x) = (1 + x+ · · ·+ xm−1)`+1∞∑k=1m−1∑j=1ζjkm k`xk=m−1∏i=1(1− ζimx)`+1 ·∞∑k=1m−1∑j=1k`(ζjmx)k=m−1∑j=1 ∏1≤i≤m−1i 6=j(1− ζimx)`+1 ·A`(ζjmx),which is a polynomial in x. This completes the proof of Theorem 1.1.Remark. The congruence (1.2) is not optimal when ` is even. Indeed, if `is even, the congruence (1.2) holds modulo (1 − x)`+2, which follows fromthe symmetry A(`, k) = A(`, `+ 1− k) and A`(−1) = 0.5. A characterization of the Eulerian polynomialIn this section, we prove the following.Theorem 5.1. Let f(x) = x`+a1x`−1 + · · ·+a` ∈ C[x] be a monic complexpolynomial of degree ` > 0. Then, the following are equivalent.(a) f(x) = A`(x).(b) For any m ≥ 2, f(x) satisfies the congruence (1.2). Namely,(5.1) f(xm) ≡(1 + x+ · · ·+ xm−1m)`+1f(x) mod (1− x)`+1is satisfied.(c) The congruence for m = 2 holds, namely,f(x2) ≡(1 + x2)`+1f(x) mod (1− x)`+1is satisfied.(d) There exists an integer m ≥ 2 such that the congruence (5.1) holds.52 K. IIJIMA, K. SASAKI, Y. TAKAHASHI, AND M. YOSHINAGAProof. (a) =⇒ (b) is nothing but Theorem 1.1. The implications (b) =⇒(c) =⇒ (d) are obvious.Let us assume (d). We shall prove (a). Choose an integer m ≥ 2 suchthat (5.1) is satisfied. There exists a polynomial g(x) ∈ C[x] that satisfies(5.2) f(xm)−(1 + x+ · · ·+ xm−1m)`+1· f(x) = (1− x)`+1 · g(x).Note that deg g = m`+m− `− 2 < (m− 1)(`+ 1). Dividing this equationby (1− xm)`+1, we have(5.3)f(xm)(1− xm)`+1 −1m`+1· f(x)(1− x)`+1 =g(x)(1 + x+ · · ·+ xm−1)`+1 .We expand g(x)(1+···+xm−1)`+1 into a partial fraction,(5.4)g(x)(1 + · · ·+ xm−1)`+1 =m−1∑j=1Rj(x)(1− ζjmx)`+1,where Rj(x) ∈ C[x] with degRj(x) ≤ `. As is well known in the theoryof formal power series [11], there exist polynomials α(t), βj(t) ∈ C[t] withdegα(t),deg βj(t) ≤ ` such that(5.5)f(x)(1− x)`+1 =∞∑k=0α(k)xk, andRj(ζ−jm x)(1− x)`+1 =∞∑k=0βj(k)xk.Then, the right hand side of (5.3) isRHS of (5.3) =m−1∑j=1∑k≥0βj(k)ζjkm xk=m−1∑j=1m−1∑r=0∞∑q=0βj(qm+ r)ζj(qm+r)m xqm+r=∞∑q=0m−1∑r=0m−1∑j=1ζjrmβj(qm+ r)xqm+r.(5.6)On the other hand, the left hand side of (5.3) isLHS of (5.3) =∑k≥0α(k)xkm − 1m`+1∑k≥0α(k)xk=∞∑q=0(α(q)− 1m`+1α(mq))xmq −∞∑q=0m−1∑r=1α(mq + r)m`+1xmq+r.(5.7)EULERIAN POLYNOMIALS AND POLYNOMIAL CONGRUENCES 53Comparison of (5.6) and (5.7) givesm−1∑j=1βj(qm) = α(q)− 1m`+1α(mq)m−1∑j=1βj(qm+ 1)ζjm = −1m`+1α(mq + 1)...m−1∑j=1βj(qm+m− 1)ζj(m−1)m = −1m`+1α(mq +m− 1),(5.8)for any q ≥ 0. Since both sides of (5.8) are polynomials in q, we have thefollowing polynomial identities.m−1∑j=1βj(t) = α(tm)− 1m`+1α(t)m−1∑j=1βj(t)ζjm = −1m`+1α(t)...m−1∑j=1βj(t)ζj(m−1)m = −1m`+1α(t).(5.9)By summing up all identities in (5.9), we obtain a functional equationα(tm)=1m`α(t).This relation is satisfied only by the polynomial of the form α(t) = c0 · t`,where c0 ∈ C. Again comparing (1.1) and (5.5), f(x) = c0 · A`(x). Sincef(x) is a monic, we have f(x) = A`(x). References1. Christos A. Athanasiadis, Extended Linial hyperplane arrangements for root systemsand a conjecture of Postnikov and Stanley, J. Algebraic Combin. 10 (1999), no. 3,207–225. MR 17231842. Matthias Beck and Sinai Robins, Computing the continuous discretely, Undergradu-ate Texts in Mathematics, Springer, New York, 2007, Integer-point enumeration inpolyhedra. MR 22719923. L. Carlitz, q-Bernoulli and Eulerian numbers, Trans. Amer. Math. Soc. 76 (1954),332–350. MR 00605384. 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MR 378041515. , Worpitzky partitions for root systems and characteristic quasi-polynomials,Tohoku Math. J. (2) 70 (2018), no. 1, 39–63. MR 3772805Department of Mathematics, Hokkaido University, North 10, West 8,Kita-ku, Sapporo 060-0810, JAPANE-mail address: k.iijima1203@gmail.comDepartment of Mathematics, Hokkaido University, North 10, West 8,Kita-ku, Sapporo 060-0810, JAPANE-mail address: kyo.sasaki2308@gmail.comDepartment of Mathematics, Hokkaido University, North 10, West 8,Kita-ku, Sapporo 060-0810, JAPANE-mail address: sin3xcos4x@gmail.comDepartment of Mathematics, Hokkaido University, North 10, West 8,Kita-ku, Sapporo 060-0810, JAPANE-mail address: yoshinaga@math.sci.hokudai.ac.jp

Eulerian polynomials and polynomial congruences

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Eulerian polynomials and polynomial congruences

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