On a Number Pyramid Related to the Binomial, Deleham, Eulerian, MacMahon and

Stirling number triangles

Ghislain R. Franssens Belgian Institute for Space Aeronomy Ringlaan 3 B-1180 Brussels BELGIUM

               ghislain.franssens@oma.be

Abstract

We study a particular number pyramid b_{n,k,l} that relates the binomial, Deleham, Eulerian, MacMahon-type and Stirling number triangles. The numbers b_{n,k,l} are generated by a function B^{c}(x, y, t), c ∈ ℂ, that appears in the calculation of derivatives of a class of functions whose derivatives can be expressed as polynomials in the function itself or a related function. Based on the properties of the numbers b_{n,k,l}, we derive several new relations related to these triangles. In particular, we show that the number triangle T_{n,k}, recently constructed by Deleham (Sloane’s \underline{A088874}), is generated by the Maclaurin series of sech^{c} t, c ∈ ℂ. We also give explicit expressions and various partial sums for the triangle T_{n,k}. Further, we find that e_{2p}^{m}, the numbers appearing in the Maclaurin series of cosh^{m} t, for all m ∈ ℕ, equal the number of closed walks, based at a vertex, of length 2p along the edges of an m-dimensional cube.

1 Introduction

In this work we study a function B^{c}(x, y, t), the c-th power of B(x, y, t) defined in Eq. (3.1), that plays a central role in the calculation of derivatives, of a class of functions whose derivatives can be expressed as polynomials in the function itself or a related function. The construction of these polynomials, in terms of the function B^{c}(x, y, t), is treated in a separate paper [3]. Here we focus on B^{c}(x, y, t) as a generating function in its own right, and derive from it some interesting number-theoretic results.

We show that the function B^{c}(x, y, t) generates a number pyramid b_{n,k,l}, of which various partial sums are closely related to some important number triangles, including the binomial coefficients \binom{n}{k}, a number triangle T_{n,k} recently constructed by Deleham [11, \underline{A088874}], the Eulerian numbers A_{n,k} [2], a particular kind of MacMahon numbers B_{n,k} [5, p. 331], and Stirling numbers of the first kind s(n, k) [1, p. 824, 24.1.3].

We derive several new expressions related to these triangles. For the triangles A_{n,k} and B_{n,k}, we obtain new generating functions. We show in particular that the so far unstudied triangle T_{n,k} is generated by the Maclaurin series of sech^{c} t, for all c ∈ ℂ. The numbers T_{n,k} are thus as fundamental for sech^{c} t as the Euler numbers E_{n} are for sech t [1, p. 804, 23.1.2]. We give explicit expressions and various partial sums for the numbers T_{n,k}.

Moreover, the special cases c = m ∈ ℤ_{+} and c = −m ∈ ℤ_{−} give rise to a particular generalization of the Euler numbers E_{n}, here denoted E_{n}^{m} and called “multinomial Euler numbers”, and a generalization of even parity numbers e_{n} (defined in Eq. (2.2)), here denoted e_{n}^{m} and called “even multinomial parity numbers”, respectively. The E_{n}^{m} are generated by the Maclaurin series of sech^{m} t (so E_{n}^{1} = E_{n}) and the e_{n}^{m} by the Maclaurin series of cosh^{m} t (so e_{n}^{1} = e_{n}). Obviously, E_{2p+1}^{m} = 0 and e_{2p+1}^{m} = 0, for all p ∈ ℕ, because sech^{m} t and cosh^{m} t are even functions of t. We obtain explicit formulas for the numbers E_{2p}^{m} and e_{2p}^{m}, as well as relations between them. The numbers e_{2p}^{m} turn out to have as combinatorial interpretation, the number of closed walks, based at a vertex, of length 2p along the edges of an m-dimensional cube.

2 Notation and definitions

1. Define the sets of positive odd and even integers ℤ_{o,+} and ℤ_{e,+}, the negative odd and even integers ℤ_{o,−} and ℤ_{e,−}, the odd integers ℤ_{o} ≜ ℤ_{o,−} ∪ ℤ_{o,+} and the even integers ℤ_{e} ≜ ℤ_{e,−} ∪ {0} ∪ ℤ_{e,+}, the positive integers ℤ_{+} ≜ ℤ_{o,+} ∪ ℤ_{e,+} and negative integers ℤ_{−} ≜ ℤ_{o,−}∪ℤ_{e,−}, the natural numbers ℕ ≜ {0}∪ℤ_{+} and the integers ℤ ≜ ℤ_{−}∪{0}∪ℤ_{+}. Let ℤ_{+,n} ≜ {1, 2, ..., n}, ℤ_{−,n} ≜ {−n, −(n − 1), ..., −1}, ℕ_{n} ≜ {0} ∪ ℤ_{+,n} and denote by ℂ the complex numbers.

2. Define

δ_{condition} ≜ \{ \begin{matrix}1, & if condition is true; \\ 0, & if condition is false, \end{matrix}  (2.1)

and for all n ∈ ℤ the even and odd parity numbers

e_{n} ≜ δ_{n∈ℤ_{e}},  (2.2)
o_{n} ≜ δ_{n∈ℤ_{o}}.  (2.3)

3. Denote the n-th derivative with respect to x by D_{x}^{n}.

4. We define 0^{n} ≜ δ_{n=0}, for all n ∈ ℕ, and z^{0} ≜ 1, for all z ∈ ℂ.

5. Let n ∈ ℕ and z ∈ ℂ. Denote by

z^{(n)} ≜ δ_{n=0} + δ_{n>0} z(z + 1)(z + 2)...(z + (n − 1)),  (2.4)
= \frac{Γ(z + n)}{Γ(z)},  (2.5)
= ∑_{k=0}^{n}(−1)^{n−k}s (n, k) z^{k},  (2.6)

the rising factorial polynomial (Pochhammer’s symbol). In particular, 0^{(n)} = δ_{n=0} and m^{(n)} = (m − 1 + n)!/(m − 1)! for m ∈ ℤ_{+}. Also, denote by

z_{(n)} ≜ δ_{n=0} + δ_{n>0} z(z − 1)(z − 2)...(z − (n − 1)),  (2.7)
= \frac{Γ(z + 1)}{Γ(z + 1 − n)},  (2.8)
= ∑_{k=0}^{n} s (n, k) z^{k},  (2.9)

the falling factorial polynomial. In particular, 0_{(n)} = δ_{n=0} and m_{(n)} = (m!/(m − n)!) δ_{n≤m} for m ∈ ℤ_{+}. In Eqs. (2.6) and (2.9), s (n, k) are Stirling numbers of the first kind. We have z_{(n)} = (−1)^{n} (−z)^{(n)}.

6. We will need

\frac{1}{(1 − z)^{c}} = ∑_{n=0}^{+∞} c^{(n)} \frac{z^{n}}{n!} ,  (2.10)

(1 + z)^{c} = ∑_{n=0}^{+∞} c_{(n)} \frac{z^{n}}{n!} ,  (2.11)

being absolutely and uniformly convergent series for all z ∈ {z ∈ ℂ : |z| < 1} and for all c ∈ ℂ. We have for all n ∈ ℕ and for all a, b ∈ ℂ,

(a + b)^{(n)} = ∑_{k=0}^{n} \binom{n}{k}a^{(n−k)}b^{(k)},  (2.12)

(a + b)_{(n)} = ∑_{k=0}^{n} \binom{n}{k}a_{(n−k)}b_{(k)}.  (2.13)

In particular, for a = c = −b, we get the orthogonality relations, for all n ∈ ℕ and for all c ∈ ℂ,

∑_{k=0}^{n} \binom{n}{k}c^{(n−k)} (−c)^{(k)} = δ_{n=0},  (2.14)

∑_{k=0}^{n} \binom{n}{k}c_{(n−k)} (−c)_{(k)} = δ_{n=0}.  (2.15)

7. With m, n ∈ ℕ and K ≜ {k_{1}, k_{2}, ..., k_{m} ∈ ℕ}, define |K| ≜ k_{1}+k_{2}+...+k_{m}, # (K) ≜ m and \binom{n}{K} ≜ n!/ (k_{1}!k_{2}!...k_{m}!), expressions that are used in the last section.

3 The generating function B^{c}(x, y, t)

For all x, y, t ∈ ℂ define

B(x, y, t) ≜ \{ \begin{matrix}\frac{x−y}{xe^{− \frac{x−y}{2} t}−ye^{+ \frac{x−y}{2} t}} , & if x ≠ y; \\ \frac{1}{1−xt}, & if x = y. \end{matrix}  (3.1)

Proposition 3.1. The Maclaurin series of the c-th power of B(x, y, t), for all c ∈ ℂ, is given by

B^{c}(x, y, t) = ∑_{n=0}^{+∞} 2^{−n}B_{n}(x, y; c)\frac{t^{n}}{n!},  (3.2)

and converges absolutely and uniformly for |t| < |\frac{ln x−ln y}{x−y} |. For all n ∈ ℕ,

B_{n}(x, y; c) = ∑_{k=0}^{n} B_{n,k}(c)x^{n−k}y^{k},  (3.3)

with the coefficients B_{n,k}(c) satisfying, for all k ∈ ℕ_{n},

B_{n+1,k+1}(c) = (2(k + 1) + c) B_{n,k+1}(c) + (2(n − k) + c) B_{n,k}(c),  (3.4)

with B_{0,0}(c) = 1 and we define B_{n,k}(c) ≜ 0, for all k ∉ ℕ_{n}. Proof. The point t = 0 is an ordinary point of B^{c}(x, y, t), so B^{c}(x, y, t) has a Maclaurin power series, converging absolutely and uniformly for |t| < |\frac{ln x−ln y}{x−y} |.

Define the partial differential operator

D(x, y, t; c) ≜ (1 − \frac{x + y}{2} t) \frac{∂}{∂t} + \frac{x − y}{2} (x\frac{∂}{∂x} − y \frac{∂}{∂y} ) − c\frac{x + y}{2} .  (3.5)

A direct calculation shows that

D(x, y, t; c)B^{c}(x, y, t) = 0.  (3.6)

Substituting in Eq. (3.6) for B^{c}(x, y, t) the uniformly convergent series (3.2) gives

∑_{n=0}^{+∞} 2^{−n} (D_{n}(x, y; c)B_{n}(x, y; c)) \frac{t^{n}}{n!} = 0,

wherein

D_{n}(x, y; c) ≜ \frac{1}{2}T_{1} + \frac{x − y}{2} (x\frac{∂}{∂x} − y \frac{∂}{∂y} ) − (n + c)\frac{x + y}{2}  (3.7)

and T_{p} is the difference shift operator such that T_{p}B_{n}(x, y; c) = B_{n+p}(x, y; c). This holds for any t, so we have

D_{n}(x, y; c)B_{n}(x, y; c) = 0.  (3.8)

Substituting in Eq. (3.8) for B_{n}(x, y; c) the bivariate homogeneous polynomial (3.3) gives

∑_{k=0}^{n} (D_{n,k}(x, y; c)B_{n,k}(c)) x^{n−k}y^{k} = 0,

wherein

D_{n,k}(x, y; c) ≜ \frac{1}{2}T_{1,0} − ((n − k + 1) + c/2) T_{0,−1} − (k + c/2)  (3.9)

and T_{p,q} is the bivariate difference shift operator such that T_{p,q}B_{n,k}(c) = B_{n+p,k+q}(c). This holds for any x and y, so we have

D_{n,k}(x, y; c)B_{n,k}(c) = 0,  (3.10)

which is just Eq. (3.4).

¿From the fact that B^{c}(x, y, 0) = 1, we obtain B_{0}(x, y; c) = B_{0,0}(c) = 1.

We have that B(x, y, t) = B(y, x, t) for all x, y, t ∈ ℂ, hence B_{n}(x, y; c) = B_{n}(y, x; c) for all n ∈ ℕ, and B_{n,k}(c) = B_{n,n−k}(c) for all c ∈ ℂ.

3.1 Special cases

(i) For x = 0 or y = 0, we get

B^{c}(0, z, t) = B^{c}(z, 0, t) = e^{\frac{c}{2} zt}.  (3.11)

This implies that

B_{n}(0, z; c) = B_{n}(z, 0; c) = (cz)^{n} ,  (3.12)

and this yields in turn that

B_{n,k}(c) = c^{n}δ_{n=k}.  (3.13)

(ii) For y = ±x, we get

B^{c}(x, x, t) = \frac{1}{(1 − xt)^{c}} ,  (3.14)
B^{c}(x, −x, t) = sech^{c}(xt).  (3.15)

This gives

B_{n}(x, x; c) = 2^{n}c^{(n)}x^{n},  (3.16)
B_{n}(x, −x; c) = 2^{n} (lim_{t→0} D_{t}^{n} sech^{c} t) x^{n},  (3.17)

and this yields in turn

∑_{k=0}^{n} B_{n,k}(c) = 2^{n}c^{(n)},  (3.18)

∑_{k=0}^{n}(−1)^{k}B_{n,k}(c) = 2^{n} (lim_{t→0} D_{t}^{n} sech^{c} t) .  (3.19)

n\ k  1  2  3  4  5  6
1  1
2  1  1
3  1  6  1
4  1  23  23  1
5  1  76  230  76  1
6  1  237  1682  1682  237  1

Table 1: The number triangle B_{n,k}

3.2 The numbers B_{n,k}(1) and B_{n,k}(2)

Putting c = 0 in Eq. (3.2) shows that B_{n}(x, y; 0) = δ_{n=0}, so B_{n,k}(0) = δ_{n=0}.

(i) For c = 1, Eq. (3.4) becomes

B_{n+1,k+1}(1) = (2(k + 1) + 1) B_{n,k+1}(1) + (2(n − k) + 1) B_{n,k}(1),

so

B_{n,k}(1) = B_{n+1,k+1},  (3.20)

with B_{n,k} the numbers derived by MacMahon [5, p. 331], (Sloane’s \underline{A060187}), cf. Table 1. In this case, Eqs. (3.18) and (3.19) become

∑_{k=0}^{n} B_{n+1,k+1} = 2^{n}n!,  (3.21)

∑_{k=0}^{n}(−1)^{k}B_{n+1,k+1} = 2^{n}E_{n},  (3.22)

with E_{n} the Euler (or secant) numbers [1, p. 804, 23.1.2], (|E_{2n}| are Sloane’s \underline{A000364}). The numbers B_{n,k} are thus (also) generated by (for |t| < \frac{1}{2} |\frac{ln y}{1−y} | and y ≠ 1)

\frac{1 − y}{e^{−(1−y)t} − ye^{+(1−y)t}} = ∑_{n=0}^{+∞} ∑_{k=0}^{n} B_{n+1,k+1}y^{k} \frac{t^{n}}{n!}.  (3.23)

We can also obtain from Eq. (3.23) the following more standard generating function for the B_{n,k}, (i.e., on the same footing as Eq. (3.29) below), (for |t| < |\frac{ln y}{1−y} | and y ≠ 1)

\frac{1}{2} ln \frac{\frac{e^{− \frac{1−y^{2}}{2} t}+ye^{+ \frac{1−y^{2}}{2} t}}{1+y}}{\frac{e^{− \frac{1−y^{2}}{2} t}−ye^{+ \frac{1−y^{2}}{2} t}}{1−y}} = ∑_{n=1}^{+∞} ∑_{k=1}^{n} B_{n,k}y^{2k−1} \frac{t^{n}}{n!}.  (3.24)

Eqs. (3.23) and (3.24) appear to be new.

(ii) For c = 2, Eq. (3.4) becomes

B_{n+1,k+1}(2) = (k + 2) 2B_{n,k+1}(2) + (n − k + 1) 2B_{n,k}(2),

n\ k  1  2  3  4  5  6
1  1
2  1  1
3  1  4  1
4  1  11  11  1
5  1  26  66  26  1
6  1  57  302  302  57  1

Table 2: The number triangle A_{n,k}

so

B_{n,k}(2) = 2^{n}A_{n+1,k+1},  (3.25)

with A_{n,k} the Eulerian numbers [2], (Sloane’s \underline{A008292}), cf. Table 2. Another notation for the Eulerian numbers is \left⟨{n \atop k}\right⟩ = A_{n,k+1}.

In this case, Eqs. (3.18) and (3.19) become

∑_{k=0}^{n} A_{n+1,k+1} = (n + 1)!,  (3.26)

∑_{k=0}^{n}(−1)^{k}A_{n+1,k+1} = 2^{n+2} (2^{n+2} − 1) \frac{B_{n+2}}{n + 2},  (3.27)

with B_{n} the Bernoulli numbers [1, p. 804, 23.1.2], (|B_{n}| are Sloane’s \underline{A027641} and \underline{A027642}). In Eq. (3.27) we used D_{t}^{n} sech^{2} t = D_{t}^{n+1} tanh t. The Eulerian numbers A_{n,k} are thus (also) generated by

\frac{1}{(\frac{e^{− \frac{1−y}{2} t}−ye^{+ \frac{1−y}{2} t}}{1−y} )2} = ∑_{n=0}^{+∞} ∑_{k=0}^{n} A_{n+1,k+1}y^{k} \frac{t^{n}}{n!}.  (3.28)

The well-known standard generating function for the Eulerian numbers is

\frac{1 − y}{1 − ye^{(1−y)t}} = 1 + ∑_{n=1}^{+∞} ∑_{k=1}^{n} A_{n,k}y^{k} \frac{t^{n}}{n!}.  (3.29)

For further convenience we define A_{n,k} ≜ 0 and B_{n,k} ≜ 0, for all k ∉ ℤ_{+,n}.

3.3 Examples of some B_{n}(x, y; c)

The first six B_{n}(x, y; c) are:

B_{0}(x, y; c) = 1,
B_{1}(x, y; c) = cx + cy,
B_{2}(x, y; c) = c^{2}x^{2} + 2c (2 + c) xy + c^{2}y^{2},
B_{3}(x, y; c) = c^{3}x^{3} + c (3c^{2} + 12c + 8) x^{2}y + c (3c^{2} + 12c + 8) xy^{2} + c^{3}y^{3},

B_{4}(x, y; c) = \begin{matrix}c^{4}x^{4} \\ +4c (c^{3} + 6c^{2} + 8c + 4) x^{3}y \\ +2c (3c^{3} + 24c^{2} + 56c + 32) x^{2}y^{2} \\ +4c (c^{3} + 6c^{2} + 8c + 4) xy^{3} \\ +c^{4}y^{4} \end{matrix} ,

B_{5}(x, y; c) = \begin{matrix}c^{5}x^{5} \\ +c (5c^{4} + 40c^{3} + 80c^{2} + 80c + 32) x^{4}y \\ +c (10c^{4} + 120c^{3} + 480c^{2} + 720c + 352) x^{3}y^{2} \\ +c (10c^{4} + 120c^{3} + 480c^{2} + 720c + 352) x^{2}y^{3} \\ +c (5c^{4} + 40c^{3} + 80c^{2} + 80c + 32) xy^{4} \\ +c^{5}y^{5} \end{matrix} .

4 Properties of the B_{n}(x, y; c)

4.1 Additive property with respect to the parameter c

Obviously, for all a, b ∈ ℂ,

B^{a+b}(x, y, t) = B^{a}(x, y, t)B^{b}(x, y, t),  (4.1)

and from this follows, for all n ∈ ℕ,

B_{n}(x, y; a + b) = ∑_{k=0}^{n} \binom{n}{k}B_{n−k}(x, y; a)B_{k}(x, y; b).  (4.2)

Substituting Eq. (3.3) in Eq. (4.2) gives

B_{n,k}(a + b) = ∑_{p=0}^{n} \binom{n}{p} ∑_{q=0}^{k} B_{n−p,k−q}(a)B_{p,q}(b).  (4.3)

For instance, by letting a = b = 1, Eq. (4.3) yields the following quadratic expansion of Eulerian numbers A_{n,k} in the MacMahon numbers B_{n,k},

A_{n+1,k+1} = \frac{1}{2^{n}} ∑_{p=0}^{n} \binom{n}{p} ∑_{q=0}^{k} B_{n−p+1,k−q+1}B_{p+1,q+1}.  (4.4)

4.2 Infinite series

Proposition 4.1. For all c ∈ ℂ and for all n ∈ ℕ,

B_{n}(x, y; c) = \{ \begin{matrix}\frac{(x−y)^{n+c}}{x^{c}} ∑_{k=0}^{+∞} c^{(k)} (2k + c)^{n} \frac{(y/x)^{k}}{k!} & if & |y| < |x|; \\ \frac{(y−x)^{n+c}}{y^{c}} ∑_{k=0}^{+∞} c^{(k)} (2k + c)^{n} \frac{(x/y)^{k}}{k!} & if & |x| < |y|, \end{matrix}  (4.5)

where x, y ∈ ℂ and the series converges absolutely.

Proof. (i) Applying Eq. (2.10) to B^{c}(x, y, t) gives, for all (x, y) ∈ D_{|y|<|x|} ≜ {(x, y) ∈ ℂ^{2} : |y| < |x|} and for all t ∈ Λ_{t} (x, y) ≜ {t ∈ ℂ : Re((1 − y/x)t) < (ln |x| − ln |y|)}, the absolutely convergent series

B^{c}(x, y, t) = \frac{(x − y)^{c}}{x^{c}} ∑_{k=0}^{+∞} c^{(k)} \frac{(y/x)^{k}}{k!} e^{+(k+c/2)(x−y)t}.

Expanding herein e^{+(k+c/2)(x−y)t} in Maclaurin series gives

B^{c}(x, y, t) = \frac{(x − y)^{c}}{x^{c}} ∑_{k=0}^{+∞} c^{(k)} \frac{(y/x)^{k}}{k!} ∑_{n=0}^{+∞} (k + c/2)^{n} (x − y)^{n} \frac{t^{n}}{n!}.

Both series are absolutely convergent, so we may interchange the order of summation [10, p. 175, Theorem 8.3], yielding

B^{c}(x, y, t) = ∑_{n=0}^{+∞} (\frac{(x − y)^{n+c}}{x^{c}} ∑_{k=0}^{+∞} c^{(k)} (k + c/2)^{n} \frac{(y/x)^{k}}{k!} ) \frac{t^{n}}{n!}.

On the other hand holds by Proposition 3.1, for all t ∈ Ω_{t} (x, y) ≜ \{t ∈ ℂ : |t| < |\frac{ln x−ln y}{x−y} |\}, that

B^{c}(x, y, t) = ∑_{n=0}^{+∞} 2^{−n}B_{n}(x, y; c)\frac{t^{n}}{n!}.

For (x, y) ∈ D_{|y|<|x|}, Λ_{t} (x, y) ∩ Ω_{t} (x, y) ≠ ∅. Then for all t ∈ Λ_{t} (x, y) ∩ Ω_{t} (x, y) holds

∑_{n=0}^{+∞} 2^{−n}B_{n}(x, y; c)\frac{t^{n}}{n!} = ∑_{n=0}^{+∞} (\frac{(x − y)^{n+c}}{x^{c}} ∑_{k=0}^{+∞} c^{(k)} (k + c/2)^{n} \frac{(y/x)^{k}}{k!} ) \frac{t^{n}}{n!},

and the first part of Eq. (4.5) follows.

(ii) Similar.

In particular, Eq. (4.5) becomes, for c = m ∈ ℤ_{+},

B_{n}(x, y; m) = (x − y)^{n+m} ∑_{k=0}^{+∞} \binom{m−1+k}{k} (2k + m)^{n} x^{−(k+m)}y^{k},  (4.6)

and for c = −m ∈ ℤ_{−},

B_{n}(x, y; −m) = (x − y)^{n−m} ∑_{k=0}^{m}(−1)^{k}\binom{m}{k} (2k − m)^{n} x^{m−k}y^{k}.  (4.7)

Moreover, Eq. (4.5) reduces to the following special form, for all c ∈ ℂ,

∑_{k=0}^{+∞} c^{(k)} (2k + c)^{n} \frac{z^{k}}{k!} = \frac{B_{n}(1, z; c)}{(1 − z)^{n+c}}, |z| < 1,  (4.8)

∑_{k=0}^{+∞} c^{(k)} (2k + c)^{n} \frac{z^{−k}}{k!} = z^{c} \frac{B_{n}(z, 1; c)}{(z − 1)^{n+c}}, |z| > 1.  (4.9)

In particular, Eqs. (4.8) and (4.9) become,

(i) for c = 1, for all n ∈ ℕ,

∑_{k=0}^{+∞} (2k + 1)^{n} z^{k} = \frac{B_{n}(1, z)}{(1 − z)^{n+1}} , |z| < 1,  (4.10)

∑_{k=0}^{+∞} (2k + 1)^{n} z^{−k} = z \frac{B_{n}(z, 1)}{(z − 1)^{n+1}} , |z| > 1.  (4.11)

Herein is B_{n}(x, y) the MacMahon homogeneous bivariate polynomial,

B_{n}(x, y; 1) ≜ B_{n}(x, y) = ∑_{k=0}^{n} B_{n+1,k+1}x^{n−k}y^{k}.  (4.12)

(ii) for c = 2, for all n ∈ ℤ_{+},

∑_{l=1}^{+∞} l^{n}z^{l} = \frac{z}{(1 − z)^{n+1}} A_{n−1}(1, z), |z| < 1,  (4.13)

∑_{l=1}^{+∞} l^{n}z^{−l} = \frac{z}{(z − 1)^{n+1}} A_{n−1}(z, 1), |z| > 1.  (4.14)

Herein is A_{n}(x, y) the Eulerian homogeneous bivariate polynomial,

2^{−n}B_{n}(x, y; 2) ≜ A_{n}(x, y) = ∑_{k=0}^{n} A_{n+1,k+1}x^{n−k}y^{k}.  (4.15)

Notice that the left hand side of Eq. (4.13) is by definition the polylogarithm of negative integer order, Li_{−n}(z) [6]. Further, combining Eq. (4.13) with [13, Eq. (14)], we get the interesting identity, for all n ∈ ℕ,

∑_{p=1}^{n}(−1)^{n−p}p!S(n, p)z^{p−1} = A_{n−1}(z, z − 1).  (4.16)

Taking in Eq. (4.7) the lim_{y→−x} and using Eq. (3.15) we obtain, for all m, n ∈ ℕ,

lim_{t→0} D_{t}^{n} cosh^{m} t = \frac{1}{2^{m}} ∑_{k=0}^{m} \binom{m}{k} (2k − m)^{n} .  (4.17)

Taking in Eq. (4.7) the lim_{y→x} and using Eq. (3.16) yields, for all m ∈ ℕ,

\frac{1}{2^{m}} ∑_{k=0}^{m}(−1)^{k}\binom{m}{k} (2k − m)^{m} = (−1)^{m}m!.  (4.18)

4.3 Generating expression

Proposition 4.2. For all n ∈ ℕ and for all c, z ∈ ℂ,

B_{n}(1, z; c) = (1 − z)^{n+c} (c + 2zD_{z})^{n} (1 − z)^{−c} .  (4.19)

Proof. It is easy to show from Eq. (2.11) that, for all n ∈ ℕ, for all b, x ∈ ℂ and for all z ∈ {z ∈ ℂ : |z| < 1}, the following identity holds

(x + 2zD_{z})^{n} (1 + z)^{b} = ∑_{k=0}^{+∞} (2k + x)^{n} b_{(k)} \frac{z^{k}}{k!} .

Then

(1 + z)^{a} (x + 2zD_{z})^{n} (1 + z)^{b} = ∑_{k=0}^{+∞} (∑_{l=0}^{k} \binom{k}{l}a_{(k−l)}b_{(l)} (2l + x)^{n}) \frac{z^{k}}{k!} .

Putting x = c, a = n + c, b = −c and substituting z → −z, we get

(1 − z)^{n+c} (c + 2zD_{z})^{n} (1 − z)^{−c}
= ∑_{k=0}^{+∞} ((−1)^{k} ∑_{l=0}^{k} \binom{k}{l} (n + c)_{(k−l)} (−c)_{(l)} (2l + c)^{n}) \frac{z^{k}}{k!} .

Due to the fact that (1 − z)^{n+c} (c + 2zD_{z})^{n} (1 − z)^{−c} is a polynomial of degree n in z, we must have that

(−1)^{k} ∑_{l=0}^{k} \binom{k}{l} (n + c)_{(k−l)} (−c)_{(l)} (2l + c)^{n} = 0,

for all k ∉ ℤ_{+,n}. Hence using Eq. (5.5) below, Eq. (3.3) and the fact that B_{n,k}(c) ≜ 0, for all k ∉ ℤ_{+,n}, Eq. (4.19) follows.

In particular, for c = 1, we obtain

(1 − z)^{n+1} (1 + 2zD_{z})^{n} (1 − z)^{−1} = ∑_{k=0}^{n} B_{n+1,k+1}z^{k},  (4.20)

and for c = 2, we obtain

(1 − z)^{n+2} (1 + zD_{z})^{n} (1 − z)^{−2} = ∑_{k=0}^{n} A_{n+1,k+1}z^{k}.  (4.21)

These appear to be new generating expressions for the MacMahon and Eulerian numbers.

5 Properties of the B_{n,k}(c)

Proposition 5.1. For all m, n ∈ ℕ and for all c ∈ ℂ,

∑_{k=0}^{m} \binom{m}{k} (n + c)^{(m−k)} (k!B_{n,k}(c)) = c^{(m)} (2m + c)^{n} .  (5.1)

Proof. For all c ∈ ℂ and for all z ∈ ℂ such that |z| < 1 we have the absolutely convergent series (4.8). As |z| < 1, we can apply Eq. (2.10) and get

∑_{m=0}^{+∞} c^{(m)} (2m + c)^{n} \frac{z^{m}}{m!} = ∑_{k=0}^{n} k!B_{n,k}(c)\frac{z^{k}}{k!} ∑_{l=0}^{+∞} (n + c)^{(l)} \frac{z^{l}}{l!} .

Interchanging the summation order gives

∑_{m=0}^{+∞} c^{(m)} (2m + c)^{n} \frac{z^{m}}{m!} = ∑_{l=0}^{+∞} ∑_{k=0}^{n} \binom{k+l}{k} (n + c)^{(l)} k!B_{n,k}(c)\frac{z^{k+l}}{(k + l)!}.

With the definition B_{n,k}(c) ≜ 0, for all k ∉ ℤ_{+,n}, we can write this as

∑_{m=0}^{+∞} c^{(m)} (2m + c)^{n} \frac{z^{m}}{m!} = ∑_{l=0}^{+∞} ∑_{k=0}^{+∞} \binom{k+l}{k} (n + c)^{(l)} k!B_{n,k}(c)\frac{z^{k+l}}{(k + l)!}.

This is equivalent to

∑_{m=0}^{+∞} c^{(m)} (2m + c)^{n} \frac{z^{m}}{m!} = ∑_{m=0}^{+∞} ∑_{k=0}^{m} \binom{m}{k} (n + c)^{(m−k)} k!B_{n,k}(c)\frac{z^{m}}{m!},

and since z is arbitrary, Eq. (5.1) follows.

In particular, for c = 1, we obtain

∑_{k=0}^{m} \binom{n+m−k}{n}B_{n+1,k+1} = (2m + 1)^{n} ,  (5.2)

and for c = 2, we obtain

∑_{k=0}^{m} \binom{n+1+m−k}{n+1}A_{n+1,k+1} = (m + 1)^{n+1} .  (5.3)

These are well-known partial sums of the MacMahon and Eulerian number triangles [5, p. 328 and p. 331].

5.1 Expressions

Proposition 5.2. For all n ∈ ℕ, for all k ∈ ℕ_{n} and for all c ∈ ℂ,

k!B_{n,k}(c) = ∑_{l=0}^{k} \binom{k}{l} (− (n + c))^{(k−l)} c^{(l)} (2l + c)^{n} ,  (5.4)

= (−1)^{k} ∑_{l=0}^{k} \binom{k}{l} (n + c)_{(k−l)} (−c)_{(l)} (2l + c)^{n} .  (5.5)

Proof. (i) We will show that Eq. (5.4) is a solution of Eq. (5.1). Substitute Eq. (5.4) in Eq. (5.1) and get

∑_{k=0}^{m} \binom{m}{k} (n + c)^{(m−k)} ∑_{l=0}^{k} \binom{k}{l} (− (n + c))^{(k−l)} c^{(l)} (2l + c)^{n} = c^{(m)} (2m + c)^{n} .

Interchanging the summation order gives

∑_{l=0}^{m} \binom{m}{l} (∑_{q=0}^{m−l} \binom{m−l}{q} (n + c)^{(m−l−q)} (− (n + c))^{(q)}) c^{(l)} (2l + c)^{n} = c^{(m)} (2m + c)^{n} .

Due to the orthogonality relation (2.14) this simplifies to

∑_{l=0}^{m} \binom{m}{l}δ_{l=m}c^{(l)} (2l + c)^{n} = c^{(m)} (2m + c)^{n} ,

and this is an identity.

(ii) Use z_{(n)} = (−1)^{n} (−z)^{(n)}.

In particular, for c = −m ∈ ℤ_{−},

(i) for n ≥ m

B_{n,k}(−m) = (−1)^{k} ∑_{l=max(0,k+m−n)}^{min(k,m)} \binom{n−m}{k−l}\binom{m}{l} (2l − m)^{n} ,  (5.6)

(ii) for n < m

B_{n,k}(−m) = ∑_{l=0}^{min(k,m)}(−1)^{l}\binom{m−n−1+(k−l)}{k−l}\binom{m}{l} (2l − m)^{n} .  (5.7)

An equivalent form of Eqs. (5.4) and (5.5) is

k!B_{n,k}(c) = ∑_{l=0}^{k}(−1)^{k−l}\binom{k}{l} \frac{Γ (n + c + 1) Γ (c + l)}{Γ (n + c + 1 − (k − l)) Γ (c)} (2l + c)^{n} .  (5.8)

In particular, for c = 1, we obtain

B_{n+1,k+1} = ∑_{l=0}^{k}(−1)^{k−l}\binom{n+1}{k−l} (2l + 1)^{n} .  (5.9)

Expression (5.9) coincides with that given by MacMahon [5, p. 331]. For c = 2 and using Eq. (3.25), we get the familiar result

A_{n+1,k+1} = ∑_{l=0}^{k}(−1)^{k−l}\binom{n+2}{k−l} (l + 1)^{n+1} ,  (5.10)

or equivalently, for all n − 1 ∈ ℤ_{+} and for all k ∈ ℤ_{+,n−1},

\left⟨{n−1 \atop k−1}\right⟩ = A_{n−1,k} = ∑_{l=0}^{k}(−1)^{l}\binom{n}{l} (k − l)^{n−1} .  (5.11)

Let S (j, i) denote the Stirling numbers of the second kind (Sloane’s \underline{A008277}).

Proposition 5.3. For all n ∈ ℕ, for all k ∈ ℕ_{n} and for all c ∈ ℂ,

B_{n,k}(c) = (−1)^{k} ∑_{j=0}^{n} \binom{n}{j}2^{j}c^{n−j} ∑_{i=0}^{min(k,j)}(−1)^{i}\binom{n−i}{k−i}S (j, i) c^{(i)}.  (5.12)

Proof. Using Eqs. (5.15) and (5.16) from Proposition 5.4 below, we have

B_{n,k}(c) = ∑_{l=0}^{n} b_{n,k,l}c^{l},

= (−1)^{k} ∑_{l=0}^{n}(−1)^{l} ∑_{p=0}^{l}(−1)^{p}\binom{n}{p}2^{n−p} ∑_{i=l−p}^{min(k,n−p)} \binom{n−i}{k−i}S (n − p, i) s (i, l − p) c^{l}.

Interchanging the order of the first two summation gives

B_{n,k}(c) = (−1)^{k} ∑_{p=0}^{n} ∑_{l=p}^{n}(−1)^{l−p}\binom{n}{p}2^{n−p} ∑_{i=l−p}^{min(k,n−p)} \binom{n−i}{k−i}S (n − p, i) s (i, l − p) c^{l},

= (−1)^{k} ∑_{p=0}^{n} ∑_{l−p=0}^{n−p}(−1)^{l−p}\binom{n}{n−p}2^{n−p} ∑_{i=l−p}^{min(k,n−p)} \binom{n−i}{k−i}S (n − p, i) s (i, l − p) c^{l−p+p},

= (−1)^{k} ∑_{p=0}^{n} ∑_{j=0}^{n−p}(−1)^{j}\binom{n}{n−p}2^{n−p} ∑_{i=j}^{min(k,n−p)} \binom{n−i}{k−i}S (n − p, i) s (i, j) c^{j+p},

= (−1)^{k} ∑_{q=0}^{n} \binom{n}{q}2^{q}c^{n−q} ∑_{j=0}^{q}(−c)^{j} ∑_{i=j}^{min(k,q)} \binom{n−i}{k−i}S (q, i) s (i, j) .

Taking into account that S (n, k) = s (n, k) ≜ 0, for all k ∉ ℕ_{n}, we can write this as

B_{n,k}(c) = (−1)^{k} ∑_{q=0}^{n} \binom{n}{q}2^{q}c^{n−q} ∑_{j=0}^{q}(−c)^{j} ∑_{i=0}^{k} \binom{n−i}{k−i}S (q, i) s (i, j) .

Interchanging the two last summations yields

B_{n,k}(c) = (−1)^{k} ∑_{q=0}^{n} \binom{n}{q}2^{q}c^{n−q} ∑_{i=0}^{k} \binom{n−i}{k−i}S (q, i) ∑_{j=0}^{i} s (i, j) (−c)^{j}.

Using the fundamental property of the Stirling numbers of the first kind [1, p. 824, 24.1.3, I, B, 1],

(−c)_{(i)} = ∑_{j=0}^{i} s (i, j) (−c)^{j},

we obtain

B_{n,k}(c) = (−1)^{k} ∑_{q=0}^{n} \binom{n}{q}2^{q}c^{n−q} ∑_{i=0}^{k} \binom{n−i}{k−i}S (q, i) (−c)_{(i)}.

By using the identity (−c)_{(i)} = (−1)^{i}c^{(i)}, writing j for q and replacing the upper limit in the second sum with min(k, j), Eq. (5.12) follows.

In particular, for c = 1, we get

B_{n+1,k+1} = (−1)^{k} ∑_{j=0}^{n} \binom{n}{j}2^{j} ∑_{i=0}^{min(k,j)}(−1)^{i}\binom{n−i}{k−i}i!S (j, i) ,  (5.13)

and for c = 2, we get

A_{n+1,k+1} = (−1)^{k} ∑_{j=0}^{n} \binom{n}{j} ∑_{i=0}^{min(k,j)}(−1)^{i}\binom{n−i}{k−i}(i + 1)!S (j, i) .  (5.14)

These appear to be new expressions for the MacMahon and Eulerian numbers, in terms of Stirling numbers of the second kind.

5.2 Polynomial expression

Proposition 5.4. For all n ∈ ℕ and for all z ∈ ℂ,

B_{n,k}(c) = ∑_{l=0}^{n} b_{n,k,l}c^{l},  (5.15)

where, for all k, l ∈ ℕ_{n},

b_{n,k,l} = (−1)^{k+l} ∑_{p=max(0,l−k)}^{l}(−1)^{p}\binom{n}{p}2^{n−p} ∑_{i=l−p}^{min(k,n−p)} \binom{n−i}{k−i}s (i, l − p) S (n − p, i) .  (5.16)

Proof. Using

(c + 2zD_{z})^{n} = ∑_{l=0}^{n} \binom{n}{l}c^{n−l}2^{l} (zD_{z})^{l} ,

we get

(1 − z)^{n+c} (c + 2zD_{z})^{n} (1 − z)^{−c} = (1 − z)^{n+c} ∑_{l=0}^{n} \binom{n}{l}c^{n−l}2^{l} (zD_{z})^{l} (1 − z)^{−c} .

Using herein the formula [9, p. 144],

(zD_{z})^{l} = ∑_{p=0}^{l} S (l, p) z^{p}D_{z}^{p},

we obtain

(1 − z)^{n+c} (c + 2zD_{z})^{n} (1 − z)^{−c}
= (1 − z)^{n+c} ∑_{l=0}^{n} \binom{n}{l}c^{n−l}2^{l} ∑_{p=0}^{l} S (l, p) z^{p}D_{z}^{p} (1 − z)^{−c} ,

= ∑_{l=0}^{n} \binom{n}{l}c^{n−l}2^{l} ∑_{p=0}^{l} S (l, p) z^{p}c^{(p)} (1 − z)^{n−p} ,

= ∑_{l=0}^{n} \binom{n}{l}c^{n−l}2^{l} ∑_{p=0}^{l} S (l, p) (−c)_{(p)} (1 − z)^{n−p} (−z)^{p} ,

= ∑_{p=0}^{n} ∑_{l=p}^{n} \binom{n}{l}c^{n−l}2^{l}S (l, p) (−c)_{(p)} (1 − z)^{n−p} (−z)^{p} .

Substituting herein the binomial expansion for (1 − z)^{n−p} gives

(1 − z)^{n+c} (c + 2zD_{z})^{n} (1 − z)^{−c}
= ∑_{p=0}^{n} ∑_{l=p}^{n} \binom{n}{l}c^{n−l}2^{l}S (l, p) (−c)_{(p)} ∑_{q=0}^{n−p} \binom{n−p}{q} (−z)^{q} (−z)^{p} ,

= ∑_{p=0}^{n} ∑_{l=p}^{n} ∑_{m=p}^{n} \binom{n}{l}\binom{n−p}{m−p}c^{n−l}2^{l} (−c)_{(p)} S (l, p) (−z)^{m} ,

= ∑_{p=0}^{n} ∑_{m=p}^{n} ∑_{l=p}^{n} \binom{n}{l}\binom{n−p}{m−p}c^{n−l}2^{l} (−c)_{(p)} S (l, p) (−z)^{m} ,

= ∑_{m=0}^{n} ∑_{p=0}^{m} \binom{n−p}{m−p} (−c)_{(p)} ∑_{l=p}^{n} \binom{n}{l}c^{n−l}2^{l}S (l, p) (−z)^{m} .

Making use of the fundamental relation of the Stirling numbers of the first kind,

(−c)_{(p)} = ∑_{k=0}^{p} s (p, k) (−c)^{k} ,

we get

(1 − z)^{n+c} (c + 2zD_{z})^{n} (1 − z)^{−c}
= ∑_{m=0}^{n} ∑_{p=0}^{m} \binom{n−p}{m−p} ∑_{k=0}^{p}(−1)^{k}s (p, k) ∑_{q=p}^{n} \binom{n}{q}2^{q}S (q, p) c^{n−q+k} (−z)^{m} ,

= ∑_{m=0}^{n} ∑_{p=0}^{m} \binom{n−p}{m−p} ∑_{k=0}^{p}(−1)^{k}s (p, k) ∑_{l=0}^{n−p} \binom{n}{n−l}2^{n−l}S (n − l, p) c^{l+k} (−z)^{m} .

Summing over diagonals in the two last summations and putting S (n, k) = s (n, k) ≜ 0, for all k ∉ ℕ_{n}, this becomes

(1 − z)^{n+c} (c + 2zD_{z})^{n} (1 − z)^{−c}
= ∑_{m=0}^{n} ∑_{p=0}^{m} \binom{n−p}{m−p} ∑_{q=0}^{n} ∑_{l=0}^{n−p}(−1)^{q−l}\binom{n}{l}2^{n−l}S (n − l, p) s (p, q − l) c^{q} (−z)^{m} ,

= ∑_{m=0}^{n} ∑_{q=0}^{n} ∑_{p=0}^{m} ∑_{l=0}^{n−p} \binom{n−p}{m−p}(−1)^{q−l}\binom{n}{l}2^{n−l}S (n − l, p) s (p, q − l) c^{q} (−z)^{m} ,

= ∑_{m=0}^{n} ∑_{q=0}^{n} ∑_{p=0}^{m} ∑_{l=0}^{n} \binom{n−p}{m−p}(−1)^{q−l}\binom{n}{l}2^{n−l}S (n − l, p) s (p, q − l) c^{q} (−z)^{m} ,

= ∑_{m=0}^{n} ∑_{q=0}^{n} ∑_{l=0}^{n} ∑_{p=0}^{m} \binom{n−p}{m−p}(−1)^{q−l}\binom{n}{l}2^{n−l}S (n − l, p) s (p, q − l) c^{q} (−z)^{m} ,

= ∑_{m=0}^{n} ∑_{q=0}^{n} ∑_{l=0}^{n}(−1)^{q−l}\binom{n}{l}2^{n−l} ∑_{p=max(0,q−l)}^{min(m,n−l)} \binom{n−p}{m−p}S (n − l, p) s (p, q − l) c^{q} (−z)^{m} .

Renaming indexes gives

(1 − z)^{n+c} (c + 2zD_{z})^{n} (1 − z)^{−c}
= ∑_{k=0}^{n}(−1)^{k} ∑_{l=0}^{n}(−1)^{l} ∑_{j=0}^{n} \binom{n}{j}(−1)^{j}

(2^{n−j} ∑_{i=max(0,l−n+n−j)}^{min(k,n−j)} \binom{n−i}{k−i}S (n − j, i) s (i, l − n + n − j)) c^{l}z^{k},

= ∑_{k=0}^{n}(−1)^{k} ∑_{l=0}^{n}(−1)^{l} ∑_{j=0}^{n} \binom{n}{j}(−1)^{n−j}

(2^{j} ∑_{i=max(0,j+l−n)}^{min(k,j)} \binom{n−i}{k−i}S (j, i) s (i, j + l − n)) c^{l}z^{k}.

Applying Proposition 4.2 and identification with

B_{n}(1, z; c) = ∑_{k=0}^{n} (∑_{l=0}^{n} b_{n,k,l}c^{l}) z^{k},

yields, for all k, l ∈ ℕ_{n},

b_{n,k,l} = (−1)^{n+k+l} ∑_{j=n−l}^{n}(−1)^{j}\binom{n}{j} (2^{j} ∑_{i=j−(n−l)}^{min(k,j)} \binom{n−i}{k−i}S (j, i) s (i, j − (n − l))) .

This can be rewritten as

b_{n,k,l} = (−1)^{n+l+k} ∑_{j=n−l}^{n}(−1)^{j}\binom{n}{j}2^{j} ∑_{i=j−(n−l)}^{min(k,j)} \binom{n−i}{k−i}S (j, i) s (i, j − (n − l)) ,

= (−1)^{k} ∑_{j−(n−l)=0}^{l}(−1)^{j−(n−l)}\binom{n}{j−(n−l)+(n−l)}

2^{j−(n−l)+(n−l)} ∑_{i=j−(n−l)}^{min(k,j−(n−l)+(n−l))} \binom{n−i}{k−i}S (j − (n − l) + (n − l), i) s (i, j − (n − l)) ,

= (−1)^{k} ∑_{q=0}^{l}(−1)^{q}\binom{n}{q+(n−l)}

2^{q+(n−l)} ∑_{i=q}^{min(k,q+(n−l))} \binom{n−i}{k−i}S (q + (n − l), i) s (i, q) ,

= (−1)^{k} ∑_{q=0}^{l}(−1)^{q}\binom{n}{l−q}2^{n−(l−q)} ∑_{i=q}^{min(k,n−(l−q))} \binom{n−i}{k−i}S (n − (l − q), i) s (i, q) ,

= (−1)^{k} ∑_{p=max(0,l−k)}^{l}(−1)^{l−p}\binom{n}{p}2^{n−p} ∑_{i=l−p}^{min(k,n−p)} \binom{n−i}{k−i}S (n − p, i) s (i, l − p) .

Using basic properties of the Stirling numbers, it is easy to derive the following special values for the b_{n,k,l},

b_{n,n,l} = b_{n,0,l} = δ_{n=l},  (5.17)
b_{n,k,n} = \binom{n}{k},  (5.18)
b_{n,k,0} = δ_{k=0}δ_{n=0}.  (5.19)

5.3 Symmetry

Recall Eq. (5.5) and a variant of it obtained by replacing k with n − k,

k!B_{n,k}(c) = (−1)^{k} ∑_{l=0}^{k} \binom{k}{l} (n + c)_{(k−l)} (−c)_{(l)} (2l + c)^{n} ,
(n − k)!B_{n,n−k}(c) = (−1)^{n−k} ∑_{l=0}^{n−k} \binom{n−k}{l} (n + c)_{(n−k−l)} (−c)_{(l)} (2l + c)^{n} .

Due to the symmetry B_{n,k}(c) = B_{n,n−k}(c), there must hold, for all n ∈ ℕ, for all k ∈ ℕ_{n} and for all c ∈ ℂ, that

\frac{(−1)^{k}}{k!} ∑_{l=0}^{k} \binom{k}{l} (n + c)_{(k−l)} (−c)_{(l)} (2l + c)^{n}
= \frac{(−1)^{n−k}}{(n − k)!} ∑_{l=0}^{n−k} \binom{n−k}{l} (n + c)_{(n−k−l)} (−c)_{(l)} (2l + c)^{n} .  (5.20)

In particular, for k = 0, Eq. (5.20) yields

\frac{(−1)^{n}}{n!} ∑_{l=0}^{n}(−1)^{l}\binom{n}{l} \frac{c + n}{c + l} (c + 2l)^{n} = \frac{c^{n}}{c^{(n)}} ,  (5.21)

and for k = 1,

\frac{(−1)^{n}}{n!} ∑_{l=0}^{n}(−1)^{l}\binom{n}{l} \frac{(c + n + 1) (c + n)}{(c + l + 1)(c + l)} (c + 2l)^{n+1}

= \frac{c (c + 2)^{n+1} − (c + n + 1) c^{n+1}}{c^{(n)}} .  (5.22)

5.4 A result of Ruiz

Summing Eq. (5.8) over k from 0 to n and using result (3.18), we get, for all n ∈ ℕ and for all c ∈ ℂ,

c^{(n)} = 2^{−n} ∑_{k=0}^{n} B_{n,k}(c),
= ∑_{k=0}^{n} ∑_{l=0}^{k}(−1)^{k−l}\binom{n+c}{k−l}\binom{c−1+l}{l} (l + c/2)^{n} ,

or

∑_{l=0}^{n} (∑_{q=0}^{n−l}(−1)^{q}\binom{n+c}{q}) \binom{c−1+l}{l} (l + c/2)^{n} = c^{(n)}.

Using the binomial identity

∑_{q=0}^{n−l}(−1)^{q}\binom{n+c}{q} = (−1)^{n−l} \binom{n+c−1}{n−l},

we get

∑_{l=0}^{n} (−1)^{n−l} \binom{n+c−1}{n−l}\binom{c−1+l}{l} (l + c/2)^{n} = c^{(n)},

or

\binom{n+c−1}{n} ∑_{l=0}^{n} (−1)^{n−l} \binom{n}{l} (l + c/2)^{n} = c^{(n)},

or

∑_{l=0}^{n} (−1)^{n−l} \binom{n}{l} (l + c/2)^{n} = n!,

or finally

\frac{1}{n!} ∑_{l=0}^{n} (−1)^{l} \binom{n}{l} ((−c/2) − l)^{n} = 1.  (5.23)

Identity (5.23) is a result of Ruiz [8]. Written in the form

\frac{1}{n!} ∑_{l=0}^{n} (−1)^{n−l} \binom{n}{l} (c + 2l)^{n} = 2^{n},  (5.24)

it can be regarded as the first identity in a series of identities of which Eqs. (5.21) and (5.22) are the next two successors. Ruiz’s result however is special because the sum in Eq. (5.24) is independent of c.

In addition, applying D_{c}^{m} to Eq. (5.24) we obtain the following derived identities, for all n ∈ ℕ, for all m ∈ ℕ_{n} and for all c ∈ ℂ,

∑_{l=0}^{n} (−1)^{l} \binom{n}{l} (c + 2l)^{n−m} = (−1)^{n} 2^{n}n!δ_{m=0}.  (5.25)

6 Properties of the b_{n,k,l}

Due to the symmetry relation B_{n,k}(c) = B_{n,n−k}(c), we have that b_{n,n−k,l} = b_{n,k,l}, for all n ∈ ℕ and for all k, l ∈ ℕ_{n}.

6.1 Recursion relation for the b_{n,k,l}

Proposition 6.1. For all n ∈ ℕ and for all k, l ∈ ℕ_{n},

b_{n+1,k+1,l+1} = 2(k + 1)b_{n,k+1,l+1} + b_{n,k+1,l} + 2(n − k)b_{n,k,l+1} + b_{n,k,l},  (6.1)

with b_{0,0,0} = 1 and b_{n,k,l} ≜ 0 if k, l ∉ ℤ_{+,n}.

Proof. ¿From the fact that B_{n,k}(0) = δ_{n=0} we find that b_{n,k,0} = δ_{k=0}δ_{n=0} and hence b_{0,0,0} = 1. Substituting Eq. (5.15) in Eq. (3.4), we get

∑_{l=0}^{n+1} b_{n+1,k+1,l}c^{l} = (2(k + 1) + c) ∑_{l=0}^{n} b_{n,k+1,l}c^{l} + (2(n − k) + c) ∑_{l=0}^{n} b_{n,k,l}c^{l}.

This is equivalent to

∑_{l=0}^{n+1} b_{n+1,k+1,l}c^{l} = ∑_{l=0}^{n} b_{n,k+1,l}c^{l+1} + ∑_{l=0}^{n} b_{n,k,l}c^{l+1}

+2(k + 1) ∑_{l=0}^{n} b_{n,k+1,l}c^{l} + 2(n − k) ∑_{l=0}^{n} b_{n,k,l}c^{l},

or

∑_{l=0}^{n+1} b_{n+1,k+1,l}c^{l} = ∑_{q=1}^{n+1} b_{n,k+1,q−1}c^{q} + ∑_{q=1}^{n+1} b_{n,k,q−1}c^{q}

+2(k + 1) ∑_{l=0}^{n} b_{n,k+1,l}c^{l} + 2(n − k) ∑_{l=0}^{n} b_{n,k,l}c^{l},

or, because b_{n,k,l} ≜ 0 if k, l ∉ ℤ_{+,n}, we get

∑_{l=0}^{n+1} b_{n+1,k+1,l}c^{l} = ∑_{q=0}^{n+1} b_{n,k+1,q−1}c^{q} + ∑_{q=0}^{n+1} b_{n,k,q−1}c^{q}

+2(k + 1) ∑_{l=0}^{n+1} b_{n,k+1,l}c^{l} + 2(n − k) ∑_{l=0}^{n+1} b_{n,k,l}c^{l}.

This holds for all c, so we obtain

b_{n+1,k+1,l} = 2(k + 1)b_{n,k+1,l} + b_{n,k+1,l−1} + 2(n − k)b_{n,k,l} + b_{n,k,l−1}.

6.2 Partial sums of the b_{n,k,l}

The following result shows that various partial sums over the number pyramid b_{n,k,l} are related to several important number triangles.

Proposition 6.2. For all n ∈ ℕ and for all k, l ∈ ℕ_{n},

∑_{l=0}^{n} b_{n,k,l} = B_{n+1,k+1},  (6.2)

∑_{l=0}^{n}(−1)^{l}b_{n,k,l} = (−1)^{n−k} (\binom{n}{k} − 2o_{n}δ_{k>0}\binom{n−1}{k−1}) ,  (6.3)

∑_{k=0}^{n} b_{n,k,l} = (−1)^{n−l}2^{n}s(n, l),  (6.4)

∑_{k=0}^{n}(−1)^{k}b_{n,k,l} = e_{n}(−1)^{n/2}2^{n}δ_{l≤n/2}T_{n/2,l}.  (6.5)

Proof. (i) This immediately follows from Eq. (3.20). (ii) We have, for all z, t ∈ ℂ,

B^{−1}(1, z, t)
= \frac{1}{1 − z} e^{− \frac{1−z}{2} t} − \frac{z}{1 − z} e^{+ \frac{1−z}{2} t},
= ∑_{n=0}^{+∞} 2^{−n}(−1)^{n} (1 − z)^{n−1} \frac{t^{n}}{n!} − z ∑_{n=0}^{+∞} 2^{−n} (1 − z)^{n−1} \frac{t^{n}}{n!},

= ∑_{n=0}^{+∞} 2^{−n} ((−1)^{n} − z) (1 − z)^{n−1} \frac{t^{n}}{n!},

= ∑_{n=0}^{+∞} 2^{−n} (δ_{n=0} + δ_{n>0} ((−1)^{n} − z) ∑_{k=0}^{n−1} \binom{n−1}{k}(−z)^{k}) \frac{t^{n}}{n!},

= ∑_{n=0}^{+∞} 2^{−n} (δ_{n=0} + δ_{n>0} ((−1)^{n} ∑_{k=0}^{n−1} \binom{n−1}{k}(−z)^{k} + ∑_{k=0}^{n−1} \binom{n−1}{k}(−z)^{k+1})) \frac{t^{n}}{n!},

= ∑_{n=0}^{+∞} 2^{−n} (δ_{n=0} + δ_{n>0} ((−1)^{n} ∑_{k=0}^{n−1} \binom{n−1}{k}(−z)^{k} + ∑_{l=1}^{n} \binom{n−1}{l−1}(−z)^{l})) \frac{t^{n}}{n!},

or

B^{−1}(1, z, t)
= ∑_{n=0}^{+∞} 2^{−n} (δ_{n=0} + δ_{n>0} ( \begin{matrix}(−1)^{n} \\ + ∑_{k=1}^{n−1} ((−1)^{n}\binom{n−1}{k} + \binom{n−1}{k−1}) (−z)^{k} \\ +(−z)^{n} \end{matrix} )) \frac{t^{n}}{n!}.

With the identity

(−1)^{n}\binom{n−1}{k} + \binom{n−1}{k−1} = ((−1)^{n} + (1 − (−1)^{n}) \frac{k}{n}) \binom{n}{k},

the expression inside the first pair of parentheses becomes

δ_{n=0} + δ_{n>0} ((−1)^{n} + ∑_{k=1}^{n−1} ((−1)^{n} + (1 − (−1)^{n}) \frac{k}{n}) \binom{n}{k}(−z)^{k} + (−z)^{n})

= ∑_{k=0}^{n} (δ_{n=0} + δ_{n>0} ((−1)^{n} + (1 − (−1)^{n}) \frac{k}{n})) \binom{n}{k}(−z)^{k},

= ∑_{k=0}^{n}(−1)^{n−k}\binom{n}{k} (1 − δ_{n>0} (1 − (−1)^{n}) \frac{k}{n}) z^{k},

and we get for the full expression

B^{−1}(1, z, t) = ∑_{n=0}^{+∞} 2^{−n} ∑_{k=0}^{n}(−1)^{n−k}\binom{n}{k} (1 − δ_{n>0} (1 − (−1)^{n}) \frac{k}{n}) z^{k} \frac{t^{n}}{n!},

or

B^{−1}(1, z, t) = ∑_{n=0}^{+∞} 2^{−n} ∑_{k=0}^{n}(−1)^{n−k} (\binom{n}{k} − 2o_{n}δ_{k>0}\binom{n−1}{k−1}) z^{k} \frac{t^{n}}{n!}.

Identifying this expression with

B^{−1}(1, z, t) = ∑_{n=0}^{+∞} ∑_{k=0}^{n} (2^{−n} ∑_{l=0}^{n} b_{n,k,l}(−1)^{l}) z^{k} \frac{t^{n}}{n!}

yields

∑_{l=0}^{n}(−1)^{l}b_{n,k,l} = (−1)^{n−k} (\binom{n}{k} − 2o_{n}δ_{k>0}\binom{n−1}{k−1}) .

(iii) Applying Eq. (3.18) to the following double sum we obtain, for all c ∈ ℂ,

∑_{k=0}^{n} ∑_{l=0}^{n} b_{n,k,l}c^{l} = ∑_{k=0}^{n} B_{n,k}(c) = 2^{n}c^{(n)},

or

∑_{l=0}^{n} ((−1)^{n−l}2^{−n} ∑_{k=0}^{n} b_{n,k,l}) c^{l} = c_{(n)}.

Identifying this with the definition of the Stirling numbers of the first kind gives

(−1)^{n−l}2^{−n} ∑_{k=0}^{n} b_{n,k,l} = s(n, l).

(iv) This result is related to the Maclaurin series of sech^{c} t, considered in the next section. There it is shown that (i)

∑_{k=0}^{2m+1}(−1)^{k}b_{2m+1,k,l} = 0,

and (ii) (see Proposition 7.1 below)

∑_{k=0}^{n}(−1)^{k}b_{n,k,l} = (−1)^{n/2}2^{n}e_{n}δ_{l≤n/2}T_{n/2,l},

where the number triangle T_{m,l} is Sloane’s sequence \underline{A088874}.

We can add to the sums given by Proposition 6.2, Eq. (3.25), which expressed in terms of the b_{n,k,l} reads

∑_{l=0}^{n} b_{n,k,l}2^{l} = 2^{n}A_{n+1,k+1}.  (6.6)

6.3 The numbers T_{n,k}

We now give an explicit expression for the numbers \underline{A088874}.

Proposition 6.3. For all n ∈ ℕ and for all l ∈ ℕ_{n},

e_{n}δ_{l≤n/2}(−1)^{n/2−l}T_{n/2,l} = ∑_{p=0}^{l}(−1)^{p}\binom{n}{p}w_{n−p,l−p},  (6.7)

wherein

w_{n,m} ≜ 2^{n} ∑_{k=m}^{n} S (n, k) s (k, m) (1/2)^{k} .  (6.8)

Proof. Summing over k from 0 to n in the expression for b_{n,k,l} given by Proposition 5.4, we get

∑_{k=0}^{n}(−1)^{k}b_{n,k,l} = ∑_{p=0}^{l}(−1)^{l−p}\binom{n}{p}2^{n−p} ∑_{k=0}^{n} ∑_{i=l−p}^{min(k,n−p)} \binom{n−i}{k−i}S (n − p, i) s (i, l − p) .

Using S (n, k) = s (n, k) ≜ 0, ∀k ∉ ℕ_{n}, we can write this as

∑_{k=0}^{n}(−1)^{k}b_{n,k,l} = ∑_{p=0}^{l}(−1)^{l−p}\binom{n}{p}2^{n−p} ∑_{k=0}^{n} ∑_{i=0}^{k} \binom{n−i}{k−i}S (n − p, i) s (i, l − p) .

Interchanging the last two summations gives

∑_{k=0}^{n}(−1)^{k}b_{n,k,l} = ∑_{p=0}^{l}(−1)^{l−p}\binom{n}{p}2^{n−p} ∑_{i=0}^{n} (∑_{k=i}^{n} \binom{n−i}{k−i}) S (n − p, i) s (i, l − p) ,

= ∑_{p=0}^{l}(−1)^{l−p}\binom{n}{p}2^{n−p} ∑_{i=0}^{n} 2^{n−i}S (n − p, i) s (i, l − p) ,

or

2^{−n} ∑_{k=0}^{n}(−1)^{k}b_{n,k,l} = ∑_{p=0}^{l}(−1)^{l−p}\binom{n}{p} ∑_{i=l−p}^{n−p} 2^{n−p−i}S (n − p, i) s (i, l − p) .

Using herein definition (6.8) and the result (7.8) from Proposition 7.1 below, Eq. (6.7) follows.

n\ k  0  1  2  3  4  5  6  7  8
0  1
1  0  1
2  0  2  3
3  0  16  30  15
4  0  272  588  420  105
5  0  7 936  18 960  16 380  6 300  945
6  0  353 792  911 328  893 640  429 660  103 950  10 395
7  0  22 368 256  61 152 000  65 825 760  36 636 600  11 351 340  1 891 890  135 135
8  0  1 903 757 312  5 464 904 448  6 327 135 360  3 918 554 640  1 427 025 600  310 269 960  37 837 800  2 027 025

Table 3: The number triangle T_{n,k}

We give the onset of the number triangle T_{n,k} in Table 3. In particular, T_{n,0} = δ_{n=0}, T_{n,1} are the tangent numbers (Sloane’s \underline{A000182}), and T_{n,n} = 1.3.5...(2n − 1) are the double factorial numbers (Sloane’s \underline{A001147}).

7 The Maclaurin series of sech^{c} t

Substituting x = −y = 1 in Eqs. (3.2) and (3.3), and using Eq. (3.15), shows that the Maclaurin series of the c-th power of sech t is given by, for all c ∈ ℂ,

sech^{c} t = ∑_{n=0}^{+∞} p_{n}(c)\frac{t^{n}}{n!},  (7.1)

where

p_{n}(c) = 2^{−n} ∑_{k=0}^{n}(−1)^{k}B_{n,k}(c),  (7.2)

= ∑_{l=0}^{n} (2^{−n} ∑_{k=0}^{n}(−1)^{k}b_{n,k,l}) c^{l}.  (7.3)

The polynomials p_{n}(c) have the following properties.

(i) p_{2k+1}(c) = 0, for all k ∈ ℕ and for all c ∈ ℂ, because sech^{c} t is even in t.

(ii) p_{n}(c) has degree n/2. This can be seen as follows. We have, for all m ∈ ℕ, lim_{c→0} D_{c}^{m} sech^{c} t = (ln sech t)^{m} and (ln sech t)^{m} = O (t^{2m}). Hence p_{n}(c) must have degree n/2.

(iii) p_{n}(0) = δ_{n=0}.

Then due to (i) we have, for all m ∈ ℕ,

∑_{k=0}^{2m+1}(−1)^{k}b_{2m+1,k,l} = 0,  (7.4)

and we define, in accordance with (ii),

∑_{k=0}^{2m}(−1)^{k}b_{2m,k,l} ≜ (−1)^{m}2^{2m}δ_{l≤m}T_{m,l}.  (7.5)

Hence

p_{2m}(c) = lim_{t→0} D_{t}^{2m} sech^{c} t = (−1)^{m} ∑_{l=0}^{m} T_{m,l}c^{l},  (7.6)

so

sech^{c} t = ∑_{m=0}^{+∞} ((−1)^{m} ∑_{l=0}^{m} T_{m,l}c^{l}) \frac{t^{2m}}{(2m)!}.  (7.7)

Due to (iii), T_{n,0} = δ_{n=0}, for all n ∈ ℕ.

We now clarify the nature of the numbers T_{m,l} defined in Eq. (7.5).

Proposition 7.1. The number triangle T_{n,l}, for all n ∈ ℕ and for all l ∈ ℕ_{n}, satisfies the recursion relation

T_{n+1,l+1} = ∑_{p=0}^{n−l} 2^{p} (\binom{p+l+1}{p+1} + δ_{l>0}\binom{p+l}{p+1}) T_{n,l+p},  (7.8)

with T_{n,0} = δ_{n=0}.

Proof. Starting from the identity

D_{t}^{2} sech^{c} t = c^{2} sech^{c} t − c(c + 1) sech^{c+2} t,

and substituting herein the series expansion for sech^{c} t, Eq. (7.7), we get

∑_{n=1}^{+∞}(−1)^{n} ∑_{l=0}^{n} T_{n,l}c^{l} \frac{t^{2(n−1)}}{(2(n − 1))!}

= c^{2} ∑_{n=0}^{+∞}(−1)^{n} ∑_{l=0}^{n} T_{n,l}c^{l} \frac{t^{2n}}{(2n)!} − c(c + 1) ∑_{n=0}^{+∞}(−1)^{n} ∑_{l=0}^{n} T_{n,l}(c + 2)^{l} \frac{t^{2n}}{(2n)!}.

As this holds for all t, we must have that

− ∑_{l=0}^{n+1} T_{n+1,l}c^{l} = c^{2} ∑_{l=0}^{n} T_{n,l}c^{l} − c(c + 1) ∑_{l=0}^{n} T_{n,l}(c + 2)^{l}.

This can be rearranged in the form

∑_{l=0}^{n} T_{n+1,l+1}c^{l} = (c + 1) ∑_{l=0}^{n} T_{n,l}(c + 2)^{l} − ∑_{l=0}^{n} T_{n,l}c^{l+1}.

By the binomial theorem,

∑_{l=0}^{n} T_{n+1,l+1}c^{l} = ∑_{l=0}^{n} T_{n,l} ∑_{k=0}^{l} \binom{l}{k}2^{l−k}c^{k+1}

+ ∑_{l=0}^{n} T_{n,l} ∑_{k=0}^{l} \binom{l}{k}2^{l−k}c^{k} − ∑_{l=0}^{n} T_{n,l}c^{l+1}.

Interchanging the order of summation in the double sum terms gives

∑_{l=0}^{n} T_{n+1,l+1}c^{l} = ∑_{k=0}^{n} ∑_{q=0}^{n−k} 2^{q}\binom{q+k}{k}T_{n,q+k}c^{k+1}

+ ∑_{k=0}^{n} ∑_{q=0}^{n−k} 2^{q}\binom{q+k}{k}T_{n,q+k}c^{k} − ∑_{k=1}^{n+1} T_{n,k−1}c^{k}.

We can rearrange this further into

∑_{l=0}^{n} T_{n+1,l+1}c^{l} = ∑_{l=1}^{n+1} ∑_{q=0}^{n−(l−1)} 2^{q}\binom{q+l−1}{l−1}T_{n,q+l−1}c^{l}

+ ∑_{l=0}^{n} ∑_{q=0}^{n−l} 2^{q}\binom{q+l}{l}T_{n,q+l}c^{l} − ∑_{l=1}^{n+1} T_{n,l−1}c^{l}.

As this holds for all c, we must have that

T_{n+1,1} = ∑_{q=0}^{n} 2^{q}T_{n,q};

∑_{l=1}^{n} T_{n+1,l+1}c^{l} = ∑_{l=1}^{n} ( \begin{matrix}∑_{q=0}^{n−(l−1)} 2^{q}\binom{q+l−1}{l−1}T_{n,q+l−1} \\ + ∑_{q=0}^{n−l} 2^{q}\binom{q+l}{l}T_{n,q+l} − T_{n,l−1} \end{matrix} ) c^{l}
+T_{n,n}c^{n+1} − T_{n,n}c^{n+1},

or

T_{n+1,1} = ∑_{q=0}^{n} 2^{q}T_{n,q};

T_{n+1,l+1} = ∑_{q=0}^{n−(l−1)} 2^{q}\binom{q+l−1}{l−1}T_{n,q+l−1} + ∑_{q=0}^{n−l} 2^{q}\binom{q+l}{l}T_{n,q+l} − T_{n,l−1}, l ∈ ℤ_{+,n},

or

T_{n+1,1} = ∑_{q=0}^{n} 2^{q}T_{n,q};

T_{n+1,l+1} = ∑_{q=1}^{n−l+1} 2^{q}\binom{q+l−1}{l−1}T_{n,q+l−1} + ∑_{q=0}^{n−l} 2^{q}\binom{q+l}{l}T_{n,q+l}, l ∈ ℤ_{+,n},

or

T_{n+1,1} = ∑_{q=0}^{n} 2^{q}T_{n,q};

T_{n+1,l+1} = ∑_{p=0}^{n−l} 2^{p+1}\binom{p+l}{l−1}T_{n,p+l} + ∑_{p=0}^{n−l} 2^{p}\binom{p+l}{l}T_{n,p+l}, l ∈ ℤ_{+,n},

or

T_{n+1,1} = ∑_{p=0}^{n} 2^{p}T_{n,p}, l = 0;

T_{n+1,l+1} = ∑_{p=0}^{n−l} 2^{p} (2\binom{p+l}{l−1} + \binom{p+l}{l}) T_{n,p+l}, l ∈ ℤ_{+,n},

or

T_{n+1,1} = ∑_{p=0}^{n} 2^{p}T_{n,p}, l = 0;

T_{n+1,l+1} = ∑_{p=0}^{n−l} 2^{p} (2\binom{p+l}{p+1} + \binom{p+l}{p}) T_{n,p+l}, l ∈ ℤ_{+,n}.

With the basic additive (Pascal’s first) binomial identity

\binom{p+l}{p+1} + \binom{p+l}{p} = \binom{p+l+1}{p+1},

we can write this also as

T_{n+1,1} = ∑_{p=0}^{n} 2^{p}T_{n,p};

T_{n+1,l+1} = ∑_{p=0}^{n−l} 2^{p} (\binom{p+l+1}{p+1} + \binom{p+l}{p+1}) T_{n,l+p}, l ∈ ℤ_{+,n}.

Both these equations can be combined into the following single equation, for all l ∈ ℕ_{n},

T_{n+1,l+1} = ∑_{p=0}^{n−l} 2^{p} (\binom{p+l+1}{p+1} + δ_{l>0}\binom{p+l}{p+1}) T_{n,l+p}.

Finally, from the fact that sech^{0} t = 1 we conclude that T_{n,0} = δ_{n=0}.

Identifying our results with those given in Sloane [11, \underline{A085734} and \underline{A088874}] we see that the T_{n,k}, satisfying recursion relation (7.8) and boundary condition T_{n,0} = δ_{n=0}, are indeed Sloane’s sequence \underline{A088874} (there constructed by Deleham using a delta operator). Thus Eq. (7.7) shows that the number triangle T_{n,k} is as fundamental for the Maclaurin series of sech^{c} t as the Euler numbers E_{n} are for the Maclaurin series of sech t.

Eq. (7.7) also immediately leads to the orthogonality relation, for all n, k ∈ ℕ,

∑_{p=0}^{n} \binom{2n}{2p} ∑_{l=max(0,k+p−n)}^{min(k,p)} (−1)^{l} T_{n−p,k−l}T_{p,l} = δ_{n=0}δ_{k=0}.  (7.9)

The series (7.7) shows that the numbers E_{2m}^{(c)} ≜ (−1)^{m} ∑_{l=0}^{m} T_{m,l}c^{l}, for all m ∈ ℕ and for all c ∈ ℂ, are a set of generalized Euler numbers (although they are polynomials in c) in the sense of Luo, et al. [7]. It seems more natural however, to regard the integer numbers T_{m,l} as a more fundamental set, because they are independent of c.

7.1 Multinomial Euler numbers E_{n}^{m}

For the important special case that c = m ∈ ℕ, it might be convenient to introduce a generalization of the Euler numbers that are also integers.

By identifying Eq. (7.7) with the Maclaurin series of sech^{m} t, written in the form

sech^{m} t = ∑_{n=0}^{+∞} E_{n}^{m} \frac{t^{n}}{n!},  (7.10)

we get, for all p ∈ ℕ, E_{2p+1}^{m} = 0 and

E_{2p}^{m} = (−1)^{p} ∑_{l=0}^{p} T_{p,l}m^{l}.  (7.11)

On the other hand, we also have that sech^{m} t = (∑_{n=0}^{+∞} E_{n} \frac{t^{n}}{n!})^{m}, so we obtain,

E_{n}^{m} = δ_{m=0}δ_{n=0} + δ_{m>0} ∑_{K:|K|=n} \binom{n}{K} ∏_{i=1}^{#(K)} E_{k_{i}}.  (7.12)

Expression (7.12) suggests that the E_{n}^{m} be called multinomial Euler numbers. Equating the right hand sides of Eq. (7.11) and Eq. (7.12) gives

(−1)^{n} ∑_{l=0}^{n} T_{n,l}m^{l} = δ_{m=0}δ_{n=0} + δ_{m>0} ∑_{K:|K|=n} \binom{n}{K} ∏_{i=1}^{#(K)} E_{k_{i}}.  (7.13)

This reduces, for m = 1, to

(−1)^{n} ∑_{l=0}^{n} T_{n,l} = E_{2n}.  (7.14)

Some particular multinomial Euler numbers are mentioned in Sloane [11]. Eq. (7.11) reduces to the following particular cases: E_{0}^{m} = 1, E_{2}^{m} = −m, E_{4}^{m} = m (3m + 2) (rhombic matchstick numbers, Sloane’s \underline{A045944}) and E_{6}^{m} = −m (15m^{2} + 30m + 16) (not in Sloane). Also, E_{n}^{0} = δ_{n=0}, E_{n}^{1} = E_{n} (Euler or sech numbers, |E_{n}| is Sloane’s \underline{A000364}), E_{n−1}^{2} are the tanh numbers (due to D_{t}^{n} sech^{2} t = D_{t}^{n+1} tanh t, with |E_{n−1}^{2}| being Sloane’s \underline{A000182}).

We give the onset of the number square E_{n}^{m}, for even n, in Table 4.

n\ m  1  2  3  4  5  6
0  1  1  1  1  1  1
2  -1  -2  -3  -4  -5  -6
4  5  16  33  56  85  120
6  -61  -272  -723  -1 504  -2 705  -4 416
8  1 385  7 936  25 953  64 256  134 185  249 600
10  -50 521  -353 792  -1 376 643  -3 963 904  -9 451 805  -19 781 376

Table 4: The number square E_{n}^{m}

7.2 Even multinomial parity numbers e_{n}^{m}

By identifying Eq. (7.7) with the Maclaurin series of cosh^{m} t, written in the form

cosh^{m} t = ∑_{n=0}^{+∞} e_{n}^{m} \frac{t^{n}}{n!},  (7.15)

we get, for all p ∈ ℕ, e_{2p+1}^{m} = 0 and

e_{2p}^{m} = (−1)^{p} ∑_{l=0}^{p} T_{p,l}(−m)^{l}.  (7.16)

On the other hand, we also have that cosh^{m} t = (∑_{n=0}^{+∞} e_{n} \frac{t^{n}}{n!})^{m}, so we obtain,

e_{n}^{m} = δ_{m=0}δ_{n=0} + δ_{m>0} ∑_{K:|K|=n} \binom{n}{K} ∏_{i=1}^{#(K)} e_{k_{i}}.  (7.17)

Expression (7.17) suggests that the e_{n}^{m} be called even multinomial parity numbers. Eq. (4.17) gives us the following explicit expression for the e_{n}^{m} numbers,

e_{n}^{m} = \frac{1}{2^{m}} ∑_{j=0}^{m} \binom{m}{j}(m − 2j)^{n}.  (7.18)

Equating the right hand sides of Eq. (7.16) and Eq. (7.18) gives

(−1)^{n} ∑_{l=0}^{n} T_{n,l}(−m)^{l} = \frac{1}{2^{m}} ∑_{j=0}^{m} \binom{m}{j}(m − 2j)^{2n}.  (7.19)

In particular, for m = 1, Eq. (7.19) reduces to

(−1)^{n} ∑_{l=0}^{n}(−1)^{l}T_{n,l} = 1,  (7.20)

and, for m = 2, to

(−1)^{n} ∑_{l=0}^{n} T_{n,l} (−2)^{l} = 2^{2n−1}.  (7.21)

n\ m  1  2  3  4  5  6
0  1  1  1  1  1  1
2  1  2  3  4  5  6
4  1  8  21  40  65  96
6  1  32  183  544  1 205  2 256
8  1  128  1 641  8 320  26 465  64 896
10  1  512  14 763  131 584  628 805  2 086 656

Table 5: The number square e_{n}^{m}

Various particular even multinomial parity numbers are mentioned in Sloane [11]. For instance, e_{0}^{m} = 1, e_{2}^{m} = m, e_{4}^{m} = m (3m − 2) (octagonal numbers, Sloane’s \underline{A000567}) and e_{6}^{m} = m (15m^{2} − 30m + 16) (not in Sloane). Also, e_{n}^{0} = δ_{n=0}, e_{n}^{1} = e_{n}, e_{n}^{2} = δ_{n=0}+δ_{n>0}e_{n}2^{n−1} (e_{2p}^{2} is Sloane’s \underline{A009117}), e_{n}^{3} = e_{n}(3^{n} + 3)/4 (e_{2p}^{3} is Sloane’s \underline{A054879}), e_{n}^{4} = e_{n} (4^{n} + 4.2^{n}) /8 (e_{2p}^{4} is Sloane’s \underline{A092812}) and e_{n}^{5} = e_{n} (10 + 5.3^{n} + 5^{n}) /16 (not in Sloane). In general, expression (7.18) shows that e_{2p}^{m} equals the number of closed walks, based at a vertex, of length 2p along the edges of an m-dimensional cube [12].

We give the onset of the number square e_{n}^{m}, for even n, in Table 5.

7.3 Relations between the E_{n}^{m} and the e_{n}^{m} numbers

(i) Evidently, due to the fact that cosh^{m} t sech^{m} t = 1, for all m ∈ ℕ, holds the following orthogonality relation, for all n, m ∈ ℕ,

∑_{i=0}^{n} \binom{n}{i}e_{n−i}^{m}E_{i}^{m} = δ_{n=0}.  (7.22)

Combining Eqs. (7.22) and (7.18), and using the fact that e_{0}^{m} = 1, for all m ∈ ℕ, we obtain the following recursion relation for the E_{n}^{m},

E_{n}^{m} = δ_{n=0} − δ_{n>0} ∑_{i=0}^{n−1} \binom{n}{i} \frac{1}{2^{m}} ∑_{j=0}^{m} \binom{m}{j}(m − 2j)^{n−i}E_{i}^{m}.  (7.23)

In particular, for m = 1, Eq. (7.23) yields

E_{2p} = δ_{p=0} − δ_{p>0} ∑_{j=0}^{p−1} \binom{2p}{2j}E_{2j}.  (7.24)

Due to the symmetry of the binomial expression in Eq. (7.22) and because E_{0}^{m} = 1, for all m ∈ ℕ, we get equivalently a recursion relation for the e_{n}^{m} in terms of the E_{n}^{m} numbers,

e_{n}^{m} = δ_{n=0} − δ_{n>0} ∑_{i=0}^{n−1} \binom{n}{i}E_{n−i}^{m}e_{i}^{m}.  (7.25)

In particular, for m = 1, by using e_{n}^{1} = e_{n} and e_{0} = 1 = E_{0}, we get

E_{0} = δ_{p=0} − δ_{p>0} ∑_{j=0}^{p−1} \binom{2p}{2j}E_{2(p−j)}.  (7.26)

(ii) Comparing Eq. (7.11) with Eq. (7.16) we see that the numbers E_{2n}^{m} are the extension to negative integers m of the numbers e_{2n}^{m}, as expected from their generating functions given in Eqs. (7.10) and (7.15).

(iii) Recall Faa di Bruno’s formula for the n-th derivative of a composition of two functions [1, p. 823, 24.1.2, II, C.], for all n ∈ ℤ_{+},

D_{t}^{n}f (g (t)) = ∑_{k=1}^{n} (D_{g}^{k}f (g)) (t) ∑_{P (n):|K|=k} n! ∏_{i=1}^{n} \frac{(D_{t}^{i}g (t))^{k_{i}}}{(i!)^{k_{i}} k_{i}!} ,

where P (n) ≜ {K ≜ {k_{1}, k_{2}, ..., k_{n} ∈ ℕ} : 1k_{1} + 2k_{2} + ... + nk_{n} = n}. An element K ∈ P (n) represents a partition of a set of cardinality n into k_{1} classes of cardinality 1, k_{2} classes of cardinality 2, up to k_{n} classes of cardinality n.

Applied to f ∘ g, with g (t) = cosh^{m} t and f(g) = 1/g, we get

E_{n}^{m} = lim_{t→0} D_{t}^{n}f (g (t)) ,
= ∑_{k=1}^{n}(−1)^{k}k! ∑_{P (n):|K|=k} n! ∏_{i=1}^{n} \frac{(e_{i}^{m})^{k_{i}}}{(i!)^{k_{i}} k_{i}!}.

Define, for all n ∈ ℕ, S_{e}^{m}(n, 0) ≜ δ_{n=0} and if n > 0, for all k ∈ ℤ_{+,n},

S_{e}^{m}(n, k) ≜ ∑_{P (n):|K|=k} n! ∏_{i=1}^{n} \frac{(e_{i}^{m})^{k_{i}}}{(i!)^{k_{i}} k_{i}!}.  (7.27)

Then

E_{n}^{m} = ∑_{k=1}^{n}(−1)^{k}k!S_{e}^{m}(n, k).  (7.28)

Eq. (7.28) expresses the multinomial Euler numbers in terms of the even multinomial parity numbers through the intermediate numbers S_{e}^{m}(n, k).

The even parity symbol e_{i} in the product in Eq. (7.27) makes that all k_{i} with odd index i must be taken zero, so S_{e}^{m}(n, k) and hence E_{n}^{m} are both zero for odd n. With n = 2p, we get, for all p ∈ ℤ_{+},

S_{e}^{m}(2p, k) = ∑_{P_{e}(2p):|K_{e}|=k}(2p)! ∏_{j=1}^{p} \frac{(e_{2j}^{m})^{k_{2j}}}{((2j)!)^{k_{2j}} k_{2j}!},

where P_{e}(2p) ≜ {K_{e} ≜ {k_{2}, k_{4}, ..., k_{n} ∈ ℕ} : k_{2} + 2k_{4} + ... + pk_{2p} = p}. An element K_{e} ∈ P_{e}(2p) represents a partition of a set of 2p elements into 0 classes of cardinality 1, k_{2} classes of cardinality 2, 0 classes of cardinality 3, k_{4} classes of cardinality 4, up to k_{2p} classes of cardinality 2p.

In particular, for m = 1, we get from Eq. (7.28), for all p ∈ ℤ_{+},

S_{e}(2p, k) ≜ S_{e}^{1}(2p, k) = ∑_{P_{e}(2p):|K_{e}|=k}(2p)! ∏_{j=1}^{p} \frac{1}{((2j)!)^{k_{2j}} k_{2j}!},

i.e., the number of ways of partitioning a set of 2p elements into k non-empty subsets, each of even cardinality, and

E_{2p} = ∑_{k=1}^{2p}(−1)^{k}k!S_{e}(2p, k).  (7.29)

This seems to be a new expression for the (even) Euler numbers. Here the sum involves partitions into subsets of even cardinality. A similar sum, involving partitions into subsets of any cardinality, is the well-known result for the Stirling numbers of the second kind,

1 = ∑_{k=0}^{2p}(−1)^{k}k!S(2p, k).

It thus turns out that the numbers S_{e}^{m}(n, k), (which by comparing Eq. (7.27) with Eq. (7.30) might be called “even multinomial Stirling numbers of the second kind”), are more natural to the E_{n}^{m} than the S(n, k). This can be seen by applying Faa di Bruno’s formula to f ∘ g, with g (t) = e^{t} and f(g) = (\frac{1}{2} (g + 1/g))^{−m}, and using [1, p. 823, 24.1.2, II B],

S(n, k) = ∑_{P (n):|K|=k} \frac{n!}{∏_{i=1}^{n} (i!)^{k_{i}} k_{i}!}.  (7.30)

We get

E_{n}^{m} = ∑_{k=1}^{n} (lim_{t→0} D_{t}^{k} sech^{m} (ln (1 + t))) S(n, k),  (7.31)

an expression more complicated than E_{n}^{m} = lim_{t→0} D_{t}^{n} sech^{m} (t). For m = 1, the numbers defined by the expression inside the parentheses in (7.31) are Sloane’s \underline{A009014}.

We can derive another expression for the E_{n}^{m} in terms of the S(n, k), directly from the generating function sech^{m} (x), as was done in Luo, et al. [7], but it turns out to involve a double sum. In the particular case m = 1 however, we can obtain this other expression from our results by combining Eqs. (3.22) and (5.13), and then it reads

E_{n} = ∑_{k=0}^{n} \binom{n}{k} ∑_{l=0}^{k}(−1)^{l}l!2^{k−l}S (k, l) .  (7.32)

References

[1] M. Abramowitz and I. A. Stegun, Handbook of Mathematical Functions, 9th ed., Dover, 1970.

[2] L. Comtet, Advanced Combinatorics: The Art of Finite and Infinite Expansions, rev. enl. ed., Reidel, 1974.

[3] G. R. Franssens, Functions with derivatives given by polynomials in the function itself or a related function, 2005, to appear.

[4] I. S. Gradshteyn and I. M. Ryzhik, Table of Integrals, Series and Products, Academic Press, 1980.

[5] P. A. MacMahon, The divisors of numbers, Proc. London Math. Soc. 19 (1920), 305–340.

[6] L. Lewin, Polylogarithms and Associated Functions, North-Holland, 1981.

[7] Q. M. Luo, F. Qi and L. Debnath, Generalizations of Euler numbers and polynomials, Int. J. Math. Sci. (2003), 3893–3901.

[8] S. M. Ruiz, An algebraic identity leading to Wilson’s Theorem, Math. Gaz. 80 (1996), 579–582.

[9] S. Roman, The Umbral Calculus, Academic Press, 1984.

[10] W. Rudin, Principles of Mathematical Analysis, McGraw-Hill, 1976.

[11] N. J. A. Sloane, On-Line Encyclopedia of Integer Sequences, available at

    http://www.research.att.com/~njas/sequences.

[12] R. P. Stanley, Enumerative Combinatorics, Wadsworth and Brooks/Cole, 1986.

[13] E. W. Weisstein, Stirling Number of the Second Kind, available from

    http://mathworld.wolfram.com/StirlingNumberoftheSecondKind.html

2000 Mathematics Subject Classification: Primary 11B83, 05A15; Secondary 11Y55, 33B10, 30B10.

Keywords: Generating function, binomial coefficients, Deleham numbers, Eulerian numbers, MacMahon numbers, Stirling numbers, Euler numbers.

(Concerned with sequences \underline{A000182}, \underline{A000364}, \underline{A000567}, \underline{A001147}, \underline{A008277}, \underline{A008292}, \underline{A009014}, \underline{A009117}, \underline{A027641}, \underline{A027642}, \underline{A045944}, \underline{A054879}, \underline{A060187}, \underline{A085734}, \underline{A088874}, and \underline{A092812}.)

Received June 17 2005; revised version received August 8 2006. Published in Journal of Integer Sequences, August 21 2006.

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