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Remarkable relations between the central binomial series, Eulerian polynomials, and poly-Bernoulli numbers

Beáta Bényi Faculty of Water Sciences, University of Public Service, Baja, HUNGARY [email protected]  and  Toshiki Matsusaka Institute for Advanced Research, Nagoya University, Furo-cho, Chikusa-ku, Nagoya, 464-8601, JAPAN [email protected]
Abstract.

The central binomial series at negative integers are expressed as a linear combination of values of certain two polynomials. We show that one of the polynomials is a special value of the bivariate Eulerian polynomial and the other polynomial is related to the antidiagonal sum of poly-Bernoulli numbers. As an application, we prove Stephan’s observation from 2004.

Key words and phrases:
Central binomial series, Poly-Bernoulli number, Eulerian polynomial
2010 Mathematics Subject Classification:
11B68, 05A05
The second author was supported by JSPS KAKENHI Grant Number 20K14292 and 21K18141.

1. Introduction

The central binomial series is a Dirichlet series defined by

(1.1) ζCB(s)=n=11ns(2nn)(s).\displaystyle\zeta_{CB}(s)=\sum_{n=1}^{\infty}\frac{1}{n^{s}{2n\choose n}}\quad(s\in\mathbb{C}).

Borwein, Broadhurst, and Kamnitzer [6] studied special values ζCB(k)\zeta_{CB}(k) at positive integers and recovered some remarkable connections. A classical evaluation is ζCB(4)=17π43240=1736ζ(4)\zeta_{CB}(4)=\frac{17\pi^{4}}{3240}=\frac{17}{36}\zeta(4). In particular, for k2k\geq 2 Borwein–Broadhurst–Kamnitzer showed that ζCB(k)\zeta_{CB}(k) can be written as a \mathbb{Q}-linear combination of multiple zeta values and multiple Clausen and Glaisher values.

On the other hand, Lehmer [15] proved that for k1k\leq 1, ζCB(k)\zeta_{CB}(k) is a \mathbb{Q}-linear combination of 11 and π/3\pi/\sqrt{3}. For example, we have

ζCB(1)=13π3,ζCB(0)=13+29π3,ζCB(1)=23+29π3,ζCB(2)=43+1027π3.\displaystyle\zeta_{CB}(1)=\frac{1}{3}\frac{\pi}{\sqrt{3}},\quad\zeta_{CB}(0)=\frac{1}{3}+\frac{2}{9}\frac{\pi}{\sqrt{3}},\quad\zeta_{CB}(-1)=\frac{2}{3}+\frac{2}{9}\frac{\pi}{\sqrt{3}},\quad\zeta_{CB}(-2)=\frac{4}{3}+\frac{10}{27}\frac{\pi}{\sqrt{3}}.

He considered the general sum

n=1(2x)2nn(2nn)=2xarcsin(x)1x2(|x|<1)\displaystyle\sum_{n=1}^{\infty}\frac{(2x)^{2n}}{n\binom{2n}{n}}=\frac{2x\arcsin(x)}{\sqrt{1-x^{2}}}\quad(|x|<1)

and its derivatives to derive interesting series evaluations. More precisely, Lehmer provided the following explicit formula for the special values ζCB(k)\zeta_{CB}(k) at negative integers. Define two sequences of polynomials (pk(x))k1(p_{k}(x))_{k\geq-1} and (qk(x))k1(q_{k}(x))_{k\geq-1} by the initial values p1(x)=0,q1(x)=1p_{-1}(x)=0,q_{-1}(x)=1 and the recursion

(1.2) pk+1(x)=2(kx+1)pk(x)+2x(1x)pk(x)+qk(x),qk+1(x)=(2(k+1)x+1)qk(x)+2x(1x)qk(x).\displaystyle\begin{split}p_{k+1}(x)&=2(kx+1)p_{k}(x)+2x(1-x)p^{\prime}_{k}(x)+q_{k}(x),\\ q_{k+1}(x)&=\left(2(k+1)x+1\right)q_{k}(x)+2x(1-x)q^{\prime}_{k}(x).\end{split}

Then for k1k\geq-1, we have

(1.3) n=1(2n)k(2x)2n(2nn)=x(1x2)k+32(x1x2pk(x2)+arcsin(x)qk(x2)).\displaystyle\sum_{n=1}^{\infty}\frac{(2n)^{k}(2x)^{2n}}{\binom{2n}{n}}=\frac{x}{(1-x^{2})^{k+\frac{3}{2}}}\left(x\sqrt{1-x^{2}}p_{k}(x^{2})+\arcsin(x)q_{k}(x^{2})\right).

Consequently,

(1.4) ζCB(k)=13(23)kpk(14)+13(23)k+1qk(14)π3+π3.\displaystyle\zeta_{CB}(-k)=\frac{1}{3}\left(\frac{2}{3}\right)^{k}p_{k}\left(\frac{1}{4}\right)+\frac{1}{3}\left(\frac{2}{3}\right)^{k+1}q_{k}\left(\frac{1}{4}\right)\frac{\pi}{\sqrt{3}}\in{\mathbb{Q}}+{\mathbb{Q}}\frac{\pi}{\sqrt{3}}.

The first few polynomials are: p0(x)=1p_{0}(x)=1, p1(x)=3p_{1}(x)=3, p2(x)=8x+7p_{2}(x)=8x+7; and q0(x)=1q_{0}(x)=1, q1(x)=2x+1q_{1}(x)=2x+1, q2(x)=4x2+10x+1q_{2}(x)=4x^{2}+10x+1.

In 2004, Stephan [23, A098830] observed that the rational part of (1.4) is nothing but (a third of) a sum of poly-Bernoulli numbers of negative indices. Poly-Bernoulli numbers Bn(k)B_{n}^{(k)} are a generalization of classical Bernoulli numbers using polylogarithm functions and were introduced by Kaneko [12]. We will give a precise definition in Section 3.

Conjecture 1.1.

[13, stated by Kaneko, Stephan’s conjecture] For any n0n\geq 0,

(23)npn(14)=k=0nBnk(k).\left(\frac{2}{3}\right)^{n}p_{n}\left(\frac{1}{4}\right)=\sum_{k=0}^{n}B_{n-k}^{(-k)}.

In this article, we connect both polynomials pn(x)p_{n}(x) and qn(x)q_{n}(x) to known numbers and polynomials. More precisely, we prove Stephan’s conjecture (relating this way pn(x)p_{n}(x) to the poly-Bernoulli numbers) using the fact that the polynomial sequence qn(x)q_{n}(x) is a generalization of the classical Eulerian polynomials.

2. The polynomials qn(x)q_{n}(x) and bivariate Eulerian polynomials

Eulerian polynomials have been studied by Euler himself. Since then they have been studied and became classical. Several extensions, generalizations and applications are known today.

Let 𝔖n\mathfrak{S}_{n} denote the set of permutations π=π1π2πn\pi=\pi_{1}\pi_{2}\dots\pi_{n} of [n]={1,2,,n}[n]=\{1,2,\dots,n\}. For each π𝔖n\pi\in\mathfrak{S}_{n}, the excedance set is defined as Exc(π)={i[n]πi>i}\mathrm{Exc}(\pi)=\{i\in[n]\mid\pi_{i}>i\}. We set exc(π)=|Exc(π)|\mathrm{exc}(\pi)=|\mathrm{Exc}(\pi)|. It is well-known that the Eulerian number A(n,k)A(n,k) counts the number of permutations π𝔖n\pi\in\mathfrak{S}_{n} with exc(π)=k\mathrm{exc}(\pi)=k. For instance, A(3,1)=4A(3,1)=4 because there are 44 permutations of {1,2,3}\{1,2,3\} with exc(π)=1\mathrm{exc}(\pi)=1, namely, 132,213,312,321132,213,312,321. A map f:𝔖n0f:\mathfrak{S}_{n}\to{\mathbb{Z}}_{\geq 0} satisfying |{π𝔖nf(π)=k}|=A(n,k)|\{\pi\in\mathfrak{S}_{n}\mid f(\pi)=k\}|=A(n,k) is often called an Eulerian statistic. The map exc\mathrm{exc} is an example of Eulerian statistics. By Foata’s fundamental transformation, it is also known that the number of permutations with kk excedances is the same as the number of permutations with kk descents, or equivalently formulated, with k+1k+1 ascending runs, (see Bóna’s book [5]).

The Eulerian polynomial is defined by

An(x)=π𝔖nxexc(π)=k=0n1A(n,k)xk.A_{n}(x)=\sum_{\pi\in\mathfrak{S}_{n}}x^{\mathrm{exc}(\pi)}=\sum_{k=0}^{n-1}A(n,k)x^{k}.

The generating function of the Eulerian polynomials is given as

n=0An(x)tnn!=1xet(x1)x.\displaystyle\sum_{n=0}^{\infty}A_{n}(x)\frac{t^{n}}{n!}=\frac{1-x}{e^{t(x-1)}-x}.

For a more detailed history and properties on the Eulerian numbers (polynomials) and Eulerian statistics, the articles [5, 9, 19] are good references.

We recall now a generalization of the Eulerian polynomial introduced by Foata–Schützenberger [9, Chapter IV-3]. Here we define a shifted version. Let cyc(π)\mathrm{cyc}(\pi) denote the number of cycles in the disjoint cycle representation of π𝔖n\pi\in\mathfrak{S}_{n}.

Definition 2.1 (Bivariate Eulerian polynomial).

For any integer n0n\geq 0, let F0(x,y)=1F_{0}(x,y)=1 and define

Fn(x,y)=π𝔖nxexc(π)ycyc(π),(n>0).F_{n}(x,y)=\sum_{\pi\in\mathfrak{S}_{n}}x^{\mathrm{exc}(\pi)}y^{\mathrm{cyc}(\pi)},\qquad(n>0).
Example 2.2.

We have F3(x,y)=y3+3xy2+x2y+xyF_{3}(x,y)=y^{3}+3xy^{2}+x^{2}y+xy.

𝔖3\mathfrak{S}_{3} 123=(1)(2)(3)123=(1)(2)(3) 132=(1)(23)132=(1)(23) 213=(12)(3)213=(12)(3) 231=(123)231=(123) 312=(132)312=(132) 321=(13)(2)321=(13)(2)
exc(π)\mathrm{exc}(\pi) 0 11 11 22 11 11
cyc(π)\mathrm{cyc}(\pi) 33 22 22 11 11 22
Table 1. Permutations in 𝔖3\mathfrak{S}_{3} with their weights.

The generating function of the bivariate Eulerian polynomials is given by

(2.1) (x,y;t):=n=0Fn(x,y)tnn!=(1xet(x1)x)y.\displaystyle\mathscr{F}(x,y;t):=\sum_{n=0}^{\infty}F_{n}(x,y)\frac{t^{n}}{n!}=\left(\frac{1-x}{e^{t(x-1)}-x}\right)^{y}.

Savage–Viswanathan [22] derived several identities for the polynomials. Here we recall their recursion formula.

Proposition 2.3.

For n0n\geq 0,

Fn+1(x,y)=(x(1x)ddx+nx+y)Fn(x,y),F_{n+1}(x,y)=\left(x(1-x)\frac{d}{dx}+nx+y\right)F_{n}(x,y),

with the initial value F0(x,y)=1F_{0}(x,y)=1.

Note that by the definition, we have Fn(x,1)=An(x)F_{n}(x,1)=A_{n}(x). Moreover, for y=r2y=r\in{\mathbb{Z}}_{\geq 2}, the polynomials Fn(x,r)F_{n}(x,r) are the rr-Eulerian polynomials originally studied by Riordan [20]. In addition, we have Fn+1(x,1)=(x1)nF_{n+1}(x,-1)=-(x-1)^{n} and Fn+1(1,y)=y(y+1)(y+n)F_{n+1}(1,y)=y(y+1)\cdots(y+n) for any n0n\geq 0.

The surprising fact is however that the values at y=1/ky=1/k, for any positive integer kk, have also nice combinatorial interpretations. Namely, for a sequence 𝐬=(si)i1\mathbf{s}=(s_{i})_{i\geq 1} of positive integers, let the 𝐬\mathbf{s}-inversion sequence of length nn be defined as

In(𝐬)={(e1,,en)n0ei<si for 1in}.\displaystyle I_{n}^{(\mathbf{s})}=\{(e_{1},\ldots,e_{n})\in\mathbb{Z}^{n}\mid 0\leq e_{i}<s_{i}\text{ for }1\leq i\leq n\}.

The ascent statistic on eIn(𝐬)e\in I_{n}^{(\mathbf{s})} is

asc(e)=|{0i<n:eisi<ei+1si+1}|,\displaystyle\text{asc}(e)=\left|\left\{0\leq i<n:\frac{e_{i}}{s_{i}}<\frac{e_{i+1}}{s_{i+1}}\right\}\right|,

with the convention that e0/s0=0e_{0}/s_{0}=0. Then the 𝐬\mathbf{s}-Eulerian polynomials are defined by

En(𝐬)(x)=eIn(𝐬)xasc(e).\displaystyle E_{n}^{(\mathbf{s})}(x)=\sum_{e\in I_{n}^{(\mathbf{s})}}x^{\text{asc}(e)}.

For more properties of the 𝐬\mathbf{s}-Eulerian polynomials, see also Savage–Visontai [21]. Note that for 𝐬=(i)i1=(1,2,3,)\mathbf{s}=(i)_{i\geq 1}=(1,2,3,\dots), the 𝐬\mathbf{s}-Eulerian polynomials are the classical Eulerian polynomials, En(𝐬)=An(x)E_{n}^{(\mathbf{s})}=A_{n}(x). Savage–Viswanathan [22] showed that for 𝐬=((i1)k+1)i1=(1,k+1,2k+1,3k+1,)\mathbf{s}=((i-1)k+1)_{i\geq 1}=(1,k+1,2k+1,3k+1,\dots) where kk is a positive integer, it holds

En(𝐬)(x)=knFn(x,1k).\displaystyle E_{n}^{(\mathbf{s})}(x)=k^{n}F_{n}\left(x,\frac{1}{k}\right).

They called the coefficients in this special case the 1/k1/k-Eulerian numbers. The 1/k1/k-Eulerian numbers play role in the theory of kk-lecture hall polytopes [22] and enumerate certain statistics in kk-Stirling permutations [17]. We now show that for k=2k=2, the En(1,3,5,)(x)=2nFn(x,1/2)E_{n}^{(1,3,5,\dots)}(x)=2^{n}F_{n}(x,1/2) is the same as the qn(x)q_{n}(x) polynomial sequence in Lehmer’s identity.

nn 2nFn(x,1/2)2^{n}F_{n}(x,1/2) qn(x)q_{n}(x)
1-1 - 11
0 11 11
11 11 2x+12x+1
22 2x+12x+1 4x2+10x+14x^{2}+10x+1
33 4x2+10x+14x^{2}+10x+1 8x3+60x2+36x+18x^{3}+60x^{2}+36x+1
44 8x3+60x2+36x+18x^{3}+60x^{2}+36x+1 16x4+296x3+516x2+116x+116x^{4}+296x^{3}+516x^{2}+116x+1
55 16x4+296x3+516x2+116x+116x^{4}+296x^{3}+516x^{2}+116x+1 \cdots
Table 2. The polynomials 2nFn(x,1/2)2^{n}F_{n}(x,1/2) and qn(x)q_{n}(x).
Theorem 2.4.

The generating function

Q(x,t):=n=0qn1(x)tnn!Q(x,t):=\sum_{n=0}^{\infty}q_{n-1}(x)\frac{t^{n}}{n!}

equals (x,1/2;2t)\mathscr{F}(x,1/2;2t), that is, qn(x)=2n+1Fn+1(x,1/2)q_{n}(x)=2^{n+1}F_{n+1}(x,1/2) for any n1n\geq-1.

Proof.

By translating the recursion in (1.2), the generating function Q(x,t)Q(x,t) is characterized by the differential equation

((2xt1)ddt+2x(1x)ddx+1)Q(x,t)=0\left((2xt-1)\frac{d}{dt}+2x(1-x)\frac{d}{dx}+1\right)Q(x,t)=0

and the initial condition Q(x,0)=1Q(x,0)=1. We can check that the function

(1xe2t(x1)x)1/2=(x,1/2;2t)=n=02nFn(x,1/2)tnn!\left(\frac{1-x}{e^{2t(x-1)}-x}\right)^{1/2}=\mathscr{F}(x,1/2;2t)=\sum_{n=0}^{\infty}2^{n}F_{n}(x,1/2)\frac{t^{n}}{n!}

satisfies these conditions by a direct calculation. ∎

The generating function Q(x,t)=(x,1/2;2t)Q(x,t)=\mathscr{F}(x,1/2;2t) tells us that the coefficients of qn(x)q_{n}(x) count perfect matchings with the restriction on the number of matching pairs have odd smaller entries (see [16] and [23, A185411]) and qn(1)=(2n+1)!!q_{n}(1)=(2n+1)!!.

The relation between the polynomials Fn(x,1/2)F_{n}(x,1/2) and qn(x)q_{n}(x) shed light on a proof of Stephan’s conjecture which follows in the next section.

3. The polynomials pn(x)p_{n}(x) and a proof of Stephan’s conjecture

In this section, we focus on the polynomial sequence pn(x)p_{n}(x) in the expression of Lehmer (1.3). We prove the observation of Stephan who noticed a relation of the sequence with the poly-Bernoulli numbers. Poly-Bernoulli numbers were introduced by Kaneko [12] by the polylogarithm function (Lik(z)=m=1zm/mk\mathrm{Li}_{k}(z)=\sum_{m=1}^{\infty}z^{m}/m^{k} for any integer kk) as a generalization of the classical Bernoulli numbers. The poly-Bernoulli numbers Bn(k)B_{n}^{(k)}\in{\mathbb{Q}} are defined by

n=0Bn(k)tnn!=Lik(1et)1et.\sum_{n=0}^{\infty}B_{n}^{(k)}\frac{t^{n}}{n!}=\frac{\mathrm{Li}_{k}(1-e^{-t})}{1-e^{-t}}.

Poly-Bernoulli numbers have attractive properties. In particular, the values with negative indices kk enumerate several combinatorial objects, (see for instance, [3, 4, 8, 11] and the references therein).

As one of the most basic properties, Arakawa and Kaneko [1] showed that

k=0n(1)kBnk(k)=0\sum_{k=0}^{n}(-1)^{k}B_{n-k}^{(-k)}=0

holds for any positive integer nn. Since then, several authors have generalized the formula for the alternating anti-diagonal sum in [14, 18], but not much is known about the anti-diagonal sum in 1.1.

In most of the combinatorial interpretations, the roles of nn and kk are separately significant, hence it is not natural to consider the anti-diagonal sum. However, one of the interpretations, where this is natural, is the set of permutations with ascending-to-max property [10]. A permutation π𝔖n\pi\in\mathfrak{S}_{n} is called ascending-to-max, if for any integer ii, 1in21\leq i\leq n-2

  1. a.

    if π1(i)<π1(n)\pi^{-1}(i)<\pi^{-1}(n) and π1(i+1)<π1(n)\pi^{-1}(i+1)<\pi^{-1}(n) then π1(i)<π1(i+1)\pi^{-1}(i)<\pi^{-1}(i+1), and

  2. b.

    if π1(i)>π1(n)\pi^{-1}(i)>\pi^{-1}(n) and π1(i+1)>π1(n)\pi^{-1}(i+1)>\pi^{-1}(n), then π1(i)>π1(i+1)\pi^{-1}(i)>\pi^{-1}(i+1).

In other words: record a permutation in one-line notation and draw an arrow from value ii to i+1i+1 for each ii. Then, the permutation has the ascending-to-max property if all the arrows starting from the left of nn point forward and all the arrows starting from an element to the right of nn point backward. For instance, 4751836247518362 has the property, but 4138576241385762 has not. It follows from the results of Bényi and Hajnal [2] that the number of permutations π𝔖n+1\pi\in\mathfrak{S}_{n+1} with the ascending-to-max property is given by the anti-diagonal sum bn=k=0nBnk(k)b_{n}=\sum_{k=0}^{n}B_{n-k}^{(-k)}. However, no explicit formula or recursion was known about the sequence bnb_{n}.

Our first result is a recursion for the sequence bnb_{n}.

Proposition 3.1.

The sequence (bn)n0(b_{n})_{n\geq 0} satisfies the recurrence relation b0=1b_{0}=1 and

(3.1) 3bn+1=2bn+k=0n(n+1k)bk+3.\displaystyle 3b_{n+1}=2b_{n}+\sum_{k=0}^{n}{n+1\choose k}b_{k}+3.

In order to prove this theorem, we need some preparations. Recall that by [1, p.163], we have

(3.2) n=0bnxn\displaystyle\sum_{n=0}^{\infty}b_{n}x^{n} =j=0(j!)2x2j(1x)2(12x)2(1(j+1)x)2=1(1x)2j=0fj(21x,21x),\displaystyle=\sum_{j=0}^{\infty}\frac{(j!)^{2}x^{2j}}{(1-x)^{2}(1-2x)^{2}\cdots(1-(j+1)x)^{2}}=\frac{1}{(1-x)^{2}}\sum_{j=0}^{\infty}f_{j}\left(2-\frac{1}{x},2-\frac{1}{x}\right),

where (x)j=x(x+1)(x+2)(x+j1)(x)_{j}=x(x+1)(x+2)\cdots(x+j-1) is the Pochhammer symbol and we put

fj(x,y)=(j!)2(x)j(y)j.f_{j}(x,y)=\frac{(j!)^{2}}{(x)_{j}(y)_{j}}.

By a direct calculation, we have

n=0k=0n\displaystyle\sum_{n=0}^{\infty}\sum_{k=0}^{n} (n+1k)bkxn=k=0bkxkn=0(n+k+1k)xn=k=0bkxk1(1(1x)k+11)\displaystyle{n+1\choose k}b_{k}x^{n}=\sum_{k=0}^{\infty}b_{k}x^{k}\sum_{n=0}^{\infty}{n+k+1\choose k}x^{n}=\sum_{k=0}^{\infty}b_{k}x^{k-1}\left(\frac{1}{(1-x)^{k+1}}-1\right)
=1xx(12x)2j=0fj(31x,31x)1x(1x)2j=0fj(21x,21x).\displaystyle=\frac{1-x}{x(1-2x)^{2}}\sum_{j=0}^{\infty}f_{j}\left(3-\frac{1}{x},3-\frac{1}{x}\right)-\frac{1}{x(1-x)^{2}}\sum_{j=0}^{\infty}f_{j}\left(2-\frac{1}{x},2-\frac{1}{x}\right).

Thus, the desired recursion in (3.1) is equivalent to

(3.3) 2(2x)(1x)2j=0fj(21x,21x)=31x+1x(12x)2j=0fj(31x,31x).\displaystyle\frac{2(2-x)}{(1-x)^{2}}\sum_{j=0}^{\infty}f_{j}\left(2-\frac{1}{x},2-\frac{1}{x}\right)=\frac{3}{1-x}+\frac{1-x}{(1-2x)^{2}}\sum_{j=0}^{\infty}f_{j}\left(3-\frac{1}{x},3-\frac{1}{x}\right).

To prove (3.3), we derive a useful equation.

Lemma 3.2.

For any j0j\in{\mathbb{Z}}_{\geq 0}, we have

(x1)(x2)(fj(x2,y)fj1(x2,y))+(x1)(2x5)fj1(x1,y)\displaystyle(x-1)(x-2)\left(f_{j}(x-2,y)-f_{j-1}(x-2,y)\right)+(x-1)(2x-5)f_{j-1}(x-1,y)
(x1)(xy1)fj(x1,y)(x2)2fj1(x,y)\displaystyle-(x-1)(x-y-1)f_{j}(x-1,y)-(x-2)^{2}f_{j-1}(x,y)
={(x1)(y1)if j=0,0if j>0,\displaystyle=\begin{cases}(x-1)(y-1)&\text{if }j=0,\\ 0&\text{if }j>0,\end{cases}

where we put f1(x,y)=0f_{-1}(x,y)=0.

Proof.

By direct calculation, one can verify it. ∎

Proof of Proposition 3.1.

We prove (3.3). Setting x31/xx\rightarrow 3-1/x and y21/xy\rightarrow 2-1/x and applying Lemma 3.2, we obtain

fj(21x,21x)\displaystyle f_{j}\left(2-\frac{1}{x},2-\frac{1}{x}\right) =(1x)2(12x)(2x)fj(31x,21x)\displaystyle=\frac{(1-x)^{2}}{(1-2x)(2-x)}f_{j}\left(3-\frac{1}{x},2-\frac{1}{x}\right)
1x2x(fj+1(11x,21x)fj(11x,21x)).\displaystyle\quad-\frac{1-x}{2-x}\left(f_{j+1}\left(1-\frac{1}{x},2-\frac{1}{x}\right)-f_{j}\left(1-\frac{1}{x},2-\frac{1}{x}\right)\right).

Summing up both sides over j=0,1,2,j=0,1,2,\dots, we have

j=0fj(21x,21x)=1x2x+(1x)2(12x)(2x)j=0fj(31x,21x).\displaystyle\sum_{j=0}^{\infty}f_{j}\left(2-\frac{1}{x},2-\frac{1}{x}\right)=\frac{1-x}{2-x}+\frac{(1-x)^{2}}{(1-2x)(2-x)}\sum_{j=0}^{\infty}f_{j}\left(3-\frac{1}{x},2-\frac{1}{x}\right).

From Lemma 3.2 again for x31/xx\to 3-1/x and y31/xy\to 3-1/x, we conclude (3.3). ∎

Remark 3.3.

Unfortunately, we could not provide a combinatorial proof for this recurrence, though it would be very interesting to find one using for instance the permutations with the ascending-to-max property.

To relate the sequence (bn)n0(b_{n})_{n\geq 0} to the polynomial sequence (pn(x))n1(p_{n}(x))_{n\geq-1}, we next derive the generating function for pn(x)p_{n}(x).

Proposition 3.4.

We have

P(x,t)\displaystyle P(x,t) :=n=0pn1(x)tnn!=e(1x)t(arcsin(x1/2e(1x)t)arcsin(x1/2))x1/2(1xe2(1x)t)1/2.\displaystyle:=\sum_{n=0}^{\infty}p_{n-1}(x)\frac{t^{n}}{n!}=\frac{e^{(1-x)t}\left(\arcsin(x^{1/2}e^{(1-x)t})-\arcsin(x^{1/2})\right)}{x^{1/2}(1-xe^{2(1-x)t})^{1/2}}.
Proof.

From (1.3), we have

k=0\displaystyle\sum_{k=0}^{\infty} n=1(2n)k1(2x)2n(2nn)tkk!=x(1x2)1/2(x1x2P(x2,t1x2)+arcsin(x)Q(x2,t1x2)).\displaystyle\sum_{n=1}^{\infty}\frac{(2n)^{k-1}(2x)^{2n}}{{2n\choose n}}\frac{t^{k}}{k!}=\frac{x}{(1-x^{2})^{1/2}}\left(x\sqrt{1-x^{2}}P\left(x^{2},\frac{t}{1-x^{2}}\right)+\arcsin(x)Q\left(x^{2},\frac{t}{1-x^{2}}\right)\right).

By applying (1.3) with k=1k=-1 again, the left-hand side equals

n=1(2n)1(2xet)2n(2nn)=xetarcsin(xet)(1x2e2t)1/2.\displaystyle\sum_{n=1}^{\infty}\frac{(2n)^{-1}(2xe^{t})^{2n}}{{2n\choose n}}=\frac{xe^{t}\arcsin(xe^{t})}{(1-x^{2}e^{2t})^{1/2}}.

Thus, Combining with Theorem 2.4, we have

P(x2,t1x2)\displaystyle P\left(x^{2},\frac{t}{1-x^{2}}\right) =etarcsin(xet)x(1x2e2t)1/2arcsin(x)x1x2(x2,12;2t1x2)\displaystyle=\frac{e^{t}\arcsin(xe^{t})}{x(1-x^{2}e^{2t})^{1/2}}-\frac{\arcsin(x)}{x\sqrt{1-x^{2}}}\mathscr{F}\left(x^{2},\frac{1}{2};\frac{2t}{1-x^{2}}\right)
=et(arcsin(xet)arcsin(x))x(1x2e2t)1/2,\displaystyle=\frac{e^{t}\left(\arcsin(xe^{t})-\arcsin(x)\right)}{x(1-x^{2}e^{2t})^{1/2}},

which implies the claim. ∎

We define the sequence ana_{n} as special values of pn(x)p_{n}(x),

(3.4) an=(23)npn(14).\displaystyle a_{n}=\left(\frac{2}{3}\right)^{n}p_{n}\left(\frac{1}{4}\right).

Using the generating function of pn(x)p_{n}(x), we obtain the recurrence formula that the sequence (an)n0(a_{n})_{n\geq 0} satisfies.

Proposition 3.5.

The sequence (an)n0(a_{n})_{n\geq 0} defined in (3.4) satisfies a0=1a_{0}=1 and

3an+1=2an+k=0n(n+1k)ak+3.\displaystyle 3a_{n+1}=2a_{n}+\sum_{k=0}^{n}{n+1\choose k}a_{k}+3.
Proof.

By Proposition 3.4, the generating function for ana_{n} is given by

(3.5) n=0antn+1(n+1)!=32P(14,23t)=6et/2(arcsin(et/2/2)arcsin(1/2))(4et)1/2.\displaystyle\sum_{n=0}^{\infty}a_{n}\frac{t^{n+1}}{(n+1)!}=\frac{3}{2}P\left(\frac{1}{4},\frac{2}{3}t\right)=\frac{6e^{t/2}\left(\arcsin(e^{t/2}/2)-\arcsin(1/2)\right)}{(4-e^{t})^{1/2}}.

Since this function satisfies the differential equation

((4et)ddt2)32P(14,23t)=3et,\left((4-e^{t})\frac{d}{dt}-2\right)\frac{3}{2}P\left(\frac{1}{4},\frac{2}{3}t\right)=3e^{t},

the coefficients ana_{n} satisfy the desired recurrence formula. ∎

In conclusion, we have the main theorem.

Theorem 3.6.

1.1 is true, i.e. for any n0n\geq 0,

(23)npn(14)=k=0nBnk(k).\left(\frac{2}{3}\right)^{n}p_{n}\left(\frac{1}{4}\right)=\sum_{k=0}^{n}B_{n-k}^{(-k)}.
Proof.

Proposition 3.1 and Proposition 3.5 imply the theorem. ∎

In the course of our proof, we obtain two types of generating functions in (3.2) and (3.5) for the sequences (an)n0=(bn)n0(a_{n})_{n\geq 0}=(b_{n})_{n\geq 0}. As a corollary, we have an explicit formula for the anti-diagonal sum, (see [3, p.24]).

Corollary 3.7.
bn=k=0nBnk(k)=(1)n+12j=1n+1(1)jj!{n+1j}(2jj)3j1i=0j13i(2i+1)(2ii).b_{n}=\sum_{k=0}^{n}B_{n-k}^{(-k)}=\frac{(-1)^{n+1}}{2}\sum_{j=1}^{n+1}(-1)^{j}j!\genfrac{\{}{\}}{0.0pt}{}{n+1}{j}\frac{\binom{2j}{j}}{3^{j-1}}\sum_{i=0}^{j-1}\frac{3^{i}}{(2i+1)\binom{2i}{i}}.
Proof.

The result follows from the explicit formula by Borwein–Girgensohn [7] and Theorem 3.6. ∎

As a final remark, we show that the polynomial pn(x)p_{n}(x) can also be expressed in terms of bivariate Eulerian polynomials.

Theorem 3.8.

For any n0n\geq 0, we have

pn(x)=2nk=0n(n+1k)Fnk(x,1/2)Fk(x,1/2).p_{n}(x)=2^{n}\sum_{k=0}^{n}{n+1\choose k}F_{n-k}(x,1/2)F_{k}(x,1/2).
Proof.

Consider

P(x,t)\displaystyle P\left(x,t\right) =e(1x)t(arcsin(x1/2e(1x)t)arcsin(x1/2))x1/2(1xe2(1x)t)1/2\displaystyle=\frac{e^{(1-x)t}\left(\arcsin(x^{1/2}e^{(1-x)t})-\arcsin(x^{1/2})\right)}{x^{1/2}(1-xe^{2(1-x)t})^{1/2}}
=(x,12;2t)1x1/2(1x)1/2(arcsin(x1/2e(1x)t)arcsin(x1/2)).\displaystyle=\mathscr{F}\left(x,\frac{1}{2};2t\right)\frac{1}{x^{1/2}(1-x)^{1/2}}\left(\arcsin(x^{1/2}e^{(1-x)t})-\arcsin(x^{1/2})\right).

Since

ddt1x1/2(1x)1/2(arcsin(x1/2e(1x)t)arcsin(x1/2))\displaystyle\frac{d}{dt}\frac{1}{x^{1/2}(1-x)^{1/2}}\bigg{(}\arcsin(x^{1/2}e^{(1-x)t})-\arcsin(x^{1/2})\bigg{)} =(x,12;2t),\displaystyle=\mathscr{F}\left(x,\frac{1}{2};2t\right),

it holds that

1x1/2(1x)1/2(arcsin(x1/2e(1x)t)arcsin(x1/2))=n=02nFn(x,1/2)tn+1(n+1)!.\frac{1}{x^{1/2}(1-x)^{1/2}}\left(\arcsin(x^{1/2}e^{(1-x)t})-\arcsin(x^{1/2})\right)=\sum_{n=0}^{\infty}2^{n}F_{n}(x,1/2)\frac{t^{n+1}}{(n+1)!}.

Thus, we have

P(x,t)=n=1(2n1k=0n1(nk)Fnk1(x,1/2)Fk(x,1/2))tnn!,\displaystyle P(x,t)=\sum_{n=1}^{\infty}\left(2^{n-1}\sum_{k=0}^{n-1}{n\choose k}F_{n-k-1}(x,1/2)F_{k}(x,1/2)\right)\frac{t^{n}}{n!},

which concludes the proof. ∎

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