Shifting Property for Riordan, Sheffer and Connection Constants Matrices

Emanuele Munarini^{∗} Politecnico di Milano Dipartimento di Matematica Piazza Leonardo da Vinci 32 20133 Milano Italy

              emanuele.munarini@polimi.it

Abstract

We study the shifting property of a matrix R = [r_{n,k}]_{n,k≥0} and a sequence (h_{n})_{n∈ℕ}, i.e., the identity

∑_{k=0}^{n} r_{n,k}h_{k+1} = ∑_{k=0}^{n} r_{n+1,k+1}h_{k} ,

when R is a Riordan matrix, a Sheffer matrix (exponential Riordan matrix), or a connection constants matrix (involving symmetric functions and continuants). Moreover, we consider the shifting identity for several sequences of combinatorial interest, such as the binomial coefficients, the polynomial coefficients, the Stirling numbers (and their q-analogues), the Lah numbers, the De Morgan numbers, the generalized Fibonacci numbers, the Bell numbers, the involutions numbers, the Chebyshev polynomials, the Stirling polynomials, the Hermite polynomials, the Gaussian coefficients, and the q-Fibonacci numbers.

∗The work is partially supported by MIUR (Ministero dell’Istruzione, dell’Università e della Ricerca).

1 Introduction

We say that a matrix R = [r_{n,k}]_{n,k≥0} and a sequence (h_{n})_{n∈ℕ} satisfy the shifting property when they satisfy the shifting identity

∑_{k=0}^{n} r_{n,k}h_{k+1} = ∑_{k=0}^{n} r_{n+1,k+1}h_{k}  (1)

for every n ∈ ℕ. This relation turns out to be satisfied by several combinatorial sequences, as already noted in the Abstract. In this paper, we will consider the following classes of matrices.

• Riordan matrices [12] (see also [13, 14, 15, 16]): R = [r_{n,k}]_{n,k≥0} = (g(t), f(t)) is an infinite lower triangular matrix whose columns have generating series

r_{k}(t) = ∑_{n≥k} r_{n,k} t^{n} = g(t)f(t)^{k} ,

where g(t) and f(t) are ordinary formal series with g_{0} = 1, f_{0} = 0 and f_{1} ≠ 0.

• Sheffer matrices [9, 10, 16] (or exponential Riordan matrices): R = [r_{n,k}]_{n,k≥0} = (g(t), f(t)) is an infinite lower triangular matrix whose columns have exponential generating series

r_{k}(t) = ∑_{n≥k} r_{n,k} \frac{t^{n}}{n!} = g(t)\frac{f(t)^{k}}{k!} ,

where g(t) and f(t) are exponential formal series with g_{0} = 1, f_{0} = 0 and f_{1} ≠ 0.

• Connection constants matrices [4, 3]: R = C^{(ρ,σ)} = [C_{n,k}^{(ρ,σ)}]_{n,k≥0} is an infinite lower

triangular matrix whose entries are the connection constants between the two persistent sequences of polynomials (p_{n}^{(ρ)}(x))_{n≥0} and (p_{n}^{(σ)}(x))_{n≥0}, i.e., are the coefficients for which

p_{n}^{(ρ)}(x) = ∑_{k=0}^{n} C_{n,k}^{(ρ,σ)}p_{k}^{(σ)}(x) ,  (2)

where ρ = (r_{1}, r_{2}, …) and σ = (s_{1}, s_{2}, …), and p_{n}^{(ρ)}(x) = (x − r_{1})(x − r_{2}) ⋯ (x − r_{n}).

Notice that, in the case of Riordan and Sheffer matrices, g(t) is invertible with respect to the product of series and f(t) is invertible with respect to the composition of series. The compositional inverse of the series f(t) is denoted by f̂(t). Moreover, Riordan and Sheffer matrices form a group with respect to the matrix product. In particular, we have

(g(t), f(t)) (G(t), F (t)) = (g(t)G(f(t)), F (f(t))) and (g(t), f(t))^{−1} = (\frac{1}{g(f̂(t))}, f̂(t)) .

For the connection constants matrices, we have the closure property

C^{(ρ,σ)}C^{(σ,τ)} = C^{(ρ,τ)} .

Moreover, all connection constants matrices are invertible, and the inverse matrices are given by

(C^{(ρ,σ)})^{−1} = C^{(σ,ρ)} .  (3)

2 General properties

Given a matrix R, we can have infinite sequences (h_{n})_{n∈ℕ} for which the shifting property is satisfied. However, for a normalized sequence (h_{n})_{n∈ℕ}, i.e., a sequence with h_{0} = 1, we have the following

Lemma 1. For every invertible infinite lower triangular matrix R = [r_{n,k}]_{n,k≥0}, there exists a unique normalized sequence (h_{n})_{n∈ℕ} for which the shifting property holds.

Proof. Since R is invertible, all its diagonal entries are non-zero. So, by identity (1), we obtain the recurrence

h_{n+1} = \frac{1}{r_{n,n}} (∑_{k=0}^{n} r_{n+1,k+1}h_{k} − ∑_{k=0}^{n−1} r_{n,k}h_{k+1}) .

This recurrence, with the initial value h_{0} = 1, defines an unique sequence.

In particular, for n = 0 in identity (1), we have r_{0,0}h_{1} = r_{1,1}h_{0}, So, if the sequence is normalized and r_{0,0} = r_{1,1}, then h_{1} = h_{0} = 1.

Lemma 2. Let R = [r_{n,k}]_{n,k≥0} be an invertible infinite lower triangular matrix and let (h_{n})_{n∈ℕ} be a normalized sequence satisfying the shifting property. If there exist a matrix S = [s_{n,k}]_{n,k≥0} such that

r_{n+1,k+1} = ∑_{i=k}^{n} r_{n,i}s_{i,k}  (4)

for every n, k ∈ ℕ, then the elements of the sequence (h_{n})_{n∈ℕ} satisfy the recurrence

h_{n+1} = ∑_{k=0}^{n} s_{n,k}h_{k} .  (5)

In particular, if R′ = [r_{n+1,k+1}]_{n,k≥0}, then

S = R^{−1}R′ .  (6)

Proof. Consider the normalized sequence defined by recurrence (5). This sequence and the matrix R satisfy the shifting property (1). Indeed, by recurrences (4) and (5), we have

∑_{k=0}^{n} r_{n+1,k+1}h_{k} = ∑_{k=0}^{n} (∑_{i=k}^{n} r_{n,i}s_{i,k}) h_{k} = ∑_{i=0}^{n} r_{n,i} (∑_{k=0}^{i} s_{i,k}h_{k}) = ∑_{i=0}^{n} r_{n,i}h_{i+1} .

Since by Lemma 1 there is only one normalized sequence satisfying the shifting property (1) for a given invertible matrix R, the first part of the lemma follows. To obtain the second part of the lemma, just note that identity (4) is equivalent to R′ = RS. Since R is invertible, we have identity (6).

3 Shifting property for Riordan matrices

Recall that the incremental ratio is the linear operator R : ℝ[[x]] → ℝ[[x]] defined by

u(t) = ∑_{n≥0} u_{n}t^{n}  ⟼  Ru(t) = \frac{u(t) − u_{0}}{t} = ∑_{n≥0} u_{n+1} t^{n} .

Moreover, the A-series of a Riordan matrix R = (g(t), f(t)) is the series

a(t) = ∑_{n≥0} a_{n} t^{n} = \frac{t}{f̂(t)} .

Theorem 3. A Riordan matrix R = (g(t), f(t)) and a normalized sequence (h_{n})_{n∈ℕ}, with ordinary generating series h(t), satisfy the shifting property, if and only if

h(t) = \frac{f̂(t)}{f̂(t) − t^{2}} ,  (7)

or, equivalently, if and only if

f̂(t) = \frac{th(t)}{Rh(t)} ,  (8)

or, equivalently, if and only if

h_{n+1} = ∑_{k=0}^{n} a_{k}h_{n−k}  (9)

where the numbers a_{k} are the coefficients of the A-series of the matrix R. If the shifting property holds, then the generating series for the shifting sums is

s(t) = g(t) Rh(f(t)) = g(t)\frac{f(t)}{t} h(f(t)) .  (10)

Proof. Let R = [r_{n,k}]_{n,k≥0} = (g(t), f(t)) be a Riordan matrix. Then, for every formal series u(t) = ∑_{n≥0} u_{n}t^{n}, we have the identity

g(t)u(f(t)) = ∑_{n≥0} [∑_{k=0}^{n} r_{n,k}u_{k}] t^{n}

and the Riordan matrix

R′ = [r_{n+1,k+1}]_{n,k≥0} = (g(t)\frac{f(t)}{t}, f(t)) .

Hence, identity (1) is equivalent to the identity

g(t)Rh(f(t)) = g(t)\frac{f(t)}{t} h(f(t))

(and this gives series (10)). Since g(t) is invertible with respect to multiplication, we have the equivalent identity

\frac{h(f(t)) − 1}{f(t)} = \frac{f(t)}{t} h(f(t)) .

Finally, since f(t) is invertible with respect to composition, we have the equivalent identity

\frac{h(t) − 1}{t} = \frac{t}{f̂(t)} h(t) or Rh(t) = a(t)h(t)

from which we obtain (7), (8) and (9).

Theorem 4. A Riordan matrix R = (g(t), f(t)) and a normalized sequence (h_{n})_{n∈ℕ}, with ordinary generating series h(t), satisfy the shifting property, if and only if they satisfy the absorbing identity

∑_{k=0}^{n} r_{n,k}h_{k} = ∑_{k=0}^{n} r_{n+k,2k} .  (11)

Proof. First, we have

∑_{n≥0} (∑_{k=0}^{n} r_{n+k,2k}) t^{n} = ∑_{n≥0} \frac{1}{t^{k}} (∑_{n≥k} r_{n,2k} t^{n}) = ∑_{n≥0} g(t)\frac{f(t)^{2k}}{t^{k}} = \frac{g(t)}{1 − f(t)^{2}/t} .

So, identity holds if and only if

g(t)h(f(t)) = \frac{tg(t)}{t − f(t)^{2}} .

Since f(t) is invertible with respect to composition and g(t) is invertible with respect to multiplication, this relation is equivalent to

h(t) = \frac{f̂(t)}{f̂(t) − t^{2}} ,

and, by Theorem (3), this condition holds if and only if the shifting property holds.

Next lemma will be useful to prove Theorem 6 and to obtain some examples.

Lemma 5. Let R = [r_{n,k}]_{n,k≥0} = (g(t), f(t)) be a Riordan matrix.

1. For every a ∈ ℕ, we have the Riordan matrix

R^{(a)} = (g(t) (\frac{f(t)}{t})^{a} , f(t)) = [r_{a+n,a+k}]_{n,k≥0}.  (12)

2. Let g(t) = 1. Then, for every a ∈ ℕ, we have the Riordan matrix

R^{[a]} = (f′(t) (\frac{f(t)}{t})^{a} , f(t)) = [\frac{a + n + 1}{a + k + 1} r_{a+n+1,a+k+1}]_{n,k≥0} .  (13)

Proof. Since

[t^{n}] g(t) (\frac{f(t)}{t})^{a} f(t)^{k} = [t^{a+n}] g(t)f(t)^{a+k} = r_{a+n,a+k} ,

we have identity (12). Moreover, if g(t) = 1, then

[t^{n}]f′(t) (\frac{f(t)}{t})^{a} f(t)^{k} = [t^{a+n}]f′(t)f(t)^{a+k} = [t^{a+n}] (\frac{f(t)^{a+k+1}}{a + k + 1})′
= \frac{a + n + 1}{a + k + 1} [t^{a+n+1}]f(t)^{a+k+1} = \frac{a + n + 1}{a + k + 1} r_{a+n+1,a+k+1} ,

and we also have identity (13).

By Theorem 3 (resp. Theorem 4), the shifting property (resp. absorbing property) for a Riordan matrix R = (g(t), f(t)) depends only on the series f(t), but not on the series g(t). So, for any sequence (h_{n})_{n∈ℕ}, there are infinite Riordan matrices satisfying the shifting property (resp. absorbing property). In particular, we have

Theorem 6. If a Riordan matrix R = [r_{n,k}]_{n,k≥0} = (g(t), f(t)) and a normalized sequence (h_{n})_{n∈ℕ} satisfy the shifting property, then

∑_{k=0}^{n} r_{a+n,a+k}h_{k+1} = ∑_{k=0}^{n} r_{a+n+1,a+k+1}h_{k}  (14)

∑_{k=0}^{n} r_{a+n,a+k}h_{k} = ∑_{k=0}^{n} r_{a+n+k,a+2k}  (15)

for every a ∈ ℕ. In this case, the generating series for the shifting sums is

s(t) = g(t) (\frac{f(t)}{t})^{a} Rh(f(t)) = g(t) (\frac{f(t)}{t})^{a+1} h(f(t))  (16)

and

s_{n} = ∑_{k=0}^{n} r_{a+n+k+1,a+2k+1} .  (17)

Moreover, if g(t) = 1, then

∑_{k=0}^{n} \frac{a + n + 1}{a + k + 1} r_{a+n+1,a+k+1}h_{k+1} = ∑_{k=0}^{n} \frac{a + n + 2}{a + k + 2} r_{a+n+2,a+k+2}h_{k}  (18)

∑_{k=0}^{n} \frac{a + n + 1}{a + k + 1} r_{a+n+1,a+k+1}h_{k} = ∑_{k=0}^{n} \frac{a + n + k + 1}{a + 2k + 1} r_{a+n+k+1,a+2k+1}  (19)

for every a ∈ ℕ. In this case, the generating series for the shifting sums is

s(t) = f′(t) (\frac{f(t)}{t})^{a} Rh(f(t)) = f′(t) (\frac{f(t)}{t})^{a+1} h(f(t))  (20)

and

s_{n} = ∑_{k=0}^{n} \frac{a + n + k + 2}{a + 2k + 2} r_{a+n+k+2,a+2k+2} .  (21)

Proof. In the Riordan matrices (12) and (13) the second series is always f(t), for all a ∈ ℕ. So, both matrices (12) and (13), with the sequence (h_{n})_{n∈ℕ}, satisfy the shifting property and the absorbing property, for all a ∈ ℕ. Moreover, series (16) and (20) derive from Riordan matrices (12) and (13) and identity (10). Finally, identities (17) and (21) are just a reformulation of the absorbing property.

Examples

1. If h_{n} is the generalized Fibonacci number^{1} f_{n}^{[m]}, with m ∈ ℕ, m ≥ 1, then, by Theorem 3, we have

h(t) = ∑_{n≥0} f_{n}^{[m]} t^{n} = \frac{1}{1 − t − t^{2} − ⋯ − t^{m}} and f̂(t) = \frac{t}{1 + t + t^{2} + ⋯ + t^{m−1}} .

In particular, for m = 2, we have the Fibonacci numbers f_{n}^{[2]} = f_{n} = F_{n+1} and

h(t) = \frac{1}{1 − t − t^{2}} ⟺ f̂(t) = \frac{t}{1 + t} ⟺ f(t) = \frac{t}{1 − t} .

f_{n}^{[8]} (\underline{A079262}), and so on. See [7].

1 For m = 2, 3, 4, 5, 6, 7, 8, we have the Fibonacci numbers f_{n}^{[2]} (\underline{A000045}), the Tribonacci numbers f_{n}^{[3]} (\underline{A000073}), the Tetranacci numbers f_{n}^{[4]} (\underline{A000078}), the Pentanacci numbers f_{n}^{[5]} (\underline{A001591}), the Hexanacci numbers f_{n}^{[6]} (\underline{A001592}), the Heptanacci numbers f_{n}^{[7]} (\underline{A122189},\underline{A066178}), the Octanacci numbers

Moreover, for m = 3, we have

h(t) = \frac{1}{1 − t − t^{2} − t^{3}} ⟺ f̂(t) = \frac{t}{1 + t + t^{2}}

⟺ f(t) = \frac{1 − t − \sqrt{1 − 2t − 3t^{2}}}{2t} .

Let us consider the Riordan matrix R = (1, f(t)). Since the polynomial coefficients^{2} [2] are defined by

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

by the Lagrange inversion formula we have

r_{n,k} = [t^{n}]f(t)^{k} = \frac{k}{n} [t^{n−k}] (\frac{t}{f̂(t)})^{n}

= \frac{k}{n} [t^{n−k}](1 + t + t^{2} + ⋯ + t^{m−1})^{n} = \binom{n; m}{n − k}\frac{k}{n}

for every n, k ∈ ℕ, n ≥ 1. So, identities (14) and (18) become

∑_{k=0}^{n} \binom{a + n; m}{n − k}\frac{a + k}{a + n} f_{k+1}^{[m]} = ∑_{k=0}^{n} \binom{a + n + 1; m}{n − k}\frac{a + k + 1}{a + n + 1} f_{k}^{[m]}  (22)

∑_{k=0}^{n} \binom{a + n + 1; m}{n − k}f_{k+1}^{[m]} = ∑_{k=0}^{n} \binom{a + n + 2; m}{n − k}f_{k}^{[m]} .  (23)

In particular, for m = 2, these two identities are equivalent and become

∑_{k=0}^{n} \binom{a + n}{a + k}f_{k+1} = ∑_{k=0}^{n} \binom{a + n + 1}{a + k + 1}f_{k} .  (24)

Moreover, and for a = 0, we have the identity

∑_{k=0}^{n} \binom{n}{k}f_{k+1} = ∑_{k=0}^{n} \binom{n + 1}{k + 1}f_{k} .  (25)

2For m = 2, 3, …, 10, we have the binomial coefficients \underline{A007318}, the trinomial coefficients \underline{A027907}, the quadrinomial coefficients \underline{A008287}, the pentanomial coefficients \underline{A035343}, the hexanomial or sextinomial coefficients \underline{A063260}, the heptanomial or septinomial coefficients \underline{A063265}, the octonomial coefficients \underline{A171890}, the 9-nomial coefficients \underline{A213652}, the 10-nomial coefficients \underline{A213651}.

Finally, by (17), we also have the identities

∑_{k=0}^{n} \binom{a + n; m}{n − k}\frac{a + k}{a + n} f_{k+1}^{[m]} = ∑_{k=0}^{n} \binom{a + n + k + 1; m}{n − k}\frac{a + 2k + 1}{a + n + k + 1}  (26)
∑_{k=0}^{n} \binom{a + n + 1; m}{n − k}\frac{a + k + 1}{a + n + 1} f_{k}^{[m]} = ∑_{k=0}^{n} \binom{a + n + k + 1; m}{n − k}\frac{a + 2k + 1}{a + n + k + 1} (27)

and by (21), we have the identities

∑_{k=0}^{n} \binom{a + n + 1; m}{n − k}f_{k+1}^{[m]} = ∑_{k=0}^{n} \binom{a + n + k + 2; m}{n − k}  (28)
∑_{k=0}^{n} \binom{a + n + 2; m}{n − k}f_{k}^{[m]} = ∑_{k=0}^{n} \binom{a + n + k + 2; m}{n − k} .  (29)

2. If h_{n} is the Catalan number C_{n} = \binom{2n}{n} \frac{1}{n+1} (\underline{A000108}), then, by Theorem 3, we have

h(t) = \frac{1 − \sqrt{1 − 4t}}{2t} and f̂(t) = t \frac{1 + \sqrt{1 − 4t}}{2} .

Notice that h(t) = t/f̂(t) = a(t). Consider the Riordan matrix R = (1, f(t)). By the Lagrange inversion formula, we have

r_{n,k} = [t^{n}]f(t)^{k} = \frac{k}{n} [t^{n−k}] (\frac{t}{f̂(t)})^{n} = \frac{k}{n} [t^{2n−k}] (\frac{1 − \sqrt{1 − 4t}}{2})^{n} = \binom{3n − 2k}{n − k}\frac{k}{3n − 2k}

for every n, k ∈ ℕ, n ≥ 1. So, replacing a with a + 1, identities (14) and (18) become

\begin{matrix}∑_{k=0}^{n} \binom{a + 3n − 2k + 1}{n − k}\frac{a + k + 1}{a + 3n − 2k + 1} C_{k+1} = \\ = ∑_{k=0}^{n} \binom{a + 3n − 2k + 2}{n − k}\frac{a + k + 2}{a + 3n − 2k + 2} C_{k} \end{matrix}  (30)

\begin{matrix}∑_{k=0}^{n} \binom{a + 3n − 2k + 1}{n − k}\frac{a + n + 1}{a + 3n − 2k + 1} C_{k+1} = \\ = ∑_{k=0}^{n} \binom{a + 3n − 2k + 2}{n − k}\frac{a + n + 2}{a + 3n − 2k + 2} C_{k} . \end{matrix}  (31)

Moreover, by (17), we have the identities

\begin{matrix}∑_{k=0}^{n} \binom{a + 3n − 2k + 1}{n − k}\frac{a + k + 1}{a + 3n − 2k + 1} C_{k+1} = \\ = ∑_{k=0}^{n} \binom{a + 3n − k + 2}{n − k}\frac{a + 2k + 2}{a + 3n − k + 2} \end{matrix}  (32)

\begin{matrix}∑_{k=0}^{n} \binom{a + 3n − 2k + 2}{n − k}\frac{a + k + 2}{a + 3n − 2k + 2} C_{k} = \\ = ∑_{k=0}^{n} \binom{a + 3n − k + 2}{n − k}\frac{a + 2k + 2}{a + 3n − k + 2} \end{matrix}  (33)

and by (21), we have the identities

\begin{matrix}∑_{k=0}^{n} \binom{a + 3n − 2k + 1}{n − k}\frac{a + n + 1}{a + 3n − 2k + 1} C_{k+1} = \\ = ∑_{k=0}^{n} \binom{a + 3n − k + 2}{n − k}\frac{a + n + k + 2}{a + 3n − k + 2} \end{matrix}  (34)

\begin{matrix}∑_{k=0}^{n} \binom{a + 3n − 2k + 2}{n − k}\frac{a + n + 2}{a + 3n − 2k + 2} C_{k} = \\ = ∑_{k=0}^{n} \binom{a + 3n − k + 2}{n − k}\frac{a + n + k + 2}{a + 3n − k + 2} . \end{matrix}  (35)

In this case, we obtain a closed form for the shifting identities. Indeed, since the generating series for the sums (30) is given by (14), we have

[t^{n}] (\frac{f(t)}{t})^{a+1} h(f(t)) = [t^{n}] (\frac{f(t)}{t})^{a+1} \frac{f(t)}{t} = [t^{n}] (\frac{f(t)}{t})^{a+2} =

= [t^{a+n+2}]f(t)^{a+2} = \binom{a + 3n + 2}{n}\frac{a + 2}{a + 3n + 2} = \binom{a + 3n + 1}{n}\frac{a + 2}{a + 2n + 2}.

So, we also have the identities

∑_{k=0}^{n} \binom{a + 3n − 2k + 1}{n − k}\frac{a + k + 1}{a + 3n − 2k + 1} C_{k+1} = \binom{a + 3n + 2}{n}\frac{a + 3}{a + 2n + 3} (36)

∑_{k=0}^{n} \binom{a + 3n − 2k + 2}{n − k}\frac{a + k + 2}{a + 3n − 2k + 2} C_{k} = \binom{a + 3n + 2}{n}\frac{a + 3}{a + 2n + 3}  (37)

∑_{k=0}^{n} \binom{a + 3n − k + 2}{n − k}\frac{a + 2k + 2}{a + 3n − k + 2} = \binom{a + 3n + 2}{n}\frac{a + 3}{a + 2n + 3} .  (38)

For a = 0, 1, …, 7 we have sequences \underline{A001764}, \underline{A006629}, \underline{A102893}, \underline{A006630}, \underline{A102594}, \underline{A006631}, \underline{A230547}, \underline{A233657}. Similarly, since the generating series for the sums (31) is given by (18), we have

[t^{n}]f′(t) (\frac{f(t)}{t})^{a+1} h(f(t)) = [t^{n}]f′(t) (\frac{f(t)}{t})^{a+1} \frac{f(t)}{t} = [t^{n}]f′(t) (\frac{f(t)}{t})^{a+2} =

= [t^{a+n+2}]f′(t)f(t)^{a+2} = [t^{a+n+2}] (\frac{f(t)^{a+3}}{a + 3} )′ = \frac{a + n + 3}{a + 3} [t^{a+n+3}]f(t)^{a+3}

= \binom{a + 3n + 3}{n}\frac{a + n + 3}{a + 3n + 3} = \binom{a + 3n + 2}{n}\frac{a + n + 3}{a + 2n + 3}.

So, we also have the identities

∑_{k=0}^{n} \binom{a + 3n − 2k + 1}{n − k}\frac{a + n + 1}{a + 3n − 2k + 1} C_{k+1} = \binom{a + 3n + 2}{n}\frac{a + n + 3}{a + 2n + 3}

∑_{k=0}^{n} \binom{a + 3n − 2k + 2}{n − k}\frac{a + n + 2}{a + 3n − 2k + 2} C_{k} = \binom{a + 3n + 2}{n}\frac{a + n + 3}{a + 2n + 3}

∑_{k=0}^{n} \binom{a + 3n − k + 2}{n − k}\frac{a + n + k + 2}{a + 3n − k + 2} = \binom{a + 3n + 2}{n}\frac{a + n + 3}{a + 2n + 3} .

3. If h_{n} is the central binomial coefficients \binom{2n}{n} (\underline{A000984}), then, by Theorem 3, we have

h(t) = \frac{1}{\sqrt{1 − 4t}} ,  f̂(t) = t\frac{1 + \sqrt{1 − 4t}}{4}  and  \frac{1}{\sqrt{1 − 4f(t)}} = \frac{f(t)}{4t − f(t)} .

So, the situation is very similar to the one we have already considered for the Catalan numbers. In particular, for the Riordan matrix R = (1, f(t)), we have

r_{n,k} = \binom{3n − 2k}{n − k}\frac{k}{3n − 2k} 2^{n} (n, k ∈ ℕ , n ≥ 1) .

So, after some simplifications of identities (14) and (18), we obtain

\begin{matrix}∑_{k=0}^{n} \binom{a + 3n − 2k}{n − k}\frac{a + k}{a + 3n − 2k} \binom{2k + 1}{k + 1} = \\ = ∑_{k=0}^{n} \binom{a + 3n − 2k + 1}{n − k}\frac{a + n + 1}{a + 3n − 2k + 1} \binom{2k}{k} \end{matrix}  (39)

\begin{matrix}∑_{k=0}^{n} \binom{a + 3n − 2k + 1}{n − k}\frac{a + n + 1}{a + 3n − 2k + 1} \binom{2k + 1}{k + 1} = \\ = ∑_{k=0}^{n} \binom{a + 3n − 2k + 2}{n − k}\frac{a + n + 2}{a + 3n − 2k + 2} \binom{2k}{k} . \end{matrix}  (40)

Again, by (17), we have the identities

\begin{matrix}∑_{k=0}^{n} \binom{a + 3n − 2k}{n − k}\frac{a + k}{a + 3n − 2k} \binom{2k + 1}{k + 1} = \\ = ∑_{k=0}^{n} \binom{a + 3n − k + 2}{n − k}\frac{a + 2k + 2}{a + 3n − k + 2} 2^{k} \end{matrix}  (41)

\begin{matrix}∑_{k=0}^{n} \binom{a + 3n − 2k + 1}{n − k}\frac{a + n + 1}{a + 3n − 2k + 1} \binom{2k}{k} = \\ = ∑_{k=0}^{n} \binom{a + 3n − k + 2}{n − k}\frac{a + 2k + 2}{a + 3n − k + 2} 2^{k} \end{matrix}  (42)

and, by (21), we have the identities

\begin{matrix}∑_{k=0}^{n} \binom{a + 3n − 2k + 1}{n − k}\frac{a + n + 1}{a + 3n − 2k + 1} \binom{2k + 1}{k + 1} = \\ = ∑_{k=0}^{n} \binom{a + 3n − k + 2}{n − k}\frac{a + n + k + 2}{a + 3n − k + 2} 2^{k} \end{matrix}  (43)

\begin{matrix}∑_{k=0}^{n} \binom{a + 3n − 2k + 2}{n − k}\frac{a + n + 2}{a + 3n − 2k + 2} \binom{2k}{k} = \\ = ∑_{k=0}^{n} \binom{a + 3n − k + 2}{n − k}\frac{a + n + k + 2}{a + 3n − k + 2} 2^{k} . \end{matrix}  (44)

These last identities admit a closed form. Indeed, by (20), we have

\begin{matrix}s(t) = f′(t) (\frac{f(t)}{t})^{a+1} h(f(t)) = f′(t) (\frac{f(t)}{t})^{a+1} \frac{1}{\sqrt{1 − 4f(t)}} \\ = f′(t) (\frac{f(t)}{t})^{a+1} \frac{f(t)}{4t − f(t)} . \end{matrix}

Then, using the Cauchy integral formula, we have

s_{n} = [t^{a+n+1}] \frac{f(t)^{a+2}}{4t − f(t)} f′(t) = \frac{1}{2πi} ∮ \frac{f(z)^{a+2}}{4z − f(z)} f′(z) \frac{dz}{z^{a+n+2}} .

Setting w = f(z), we have z = f̂(w), dw = f′(z) dz and

s_{n} = \frac{1}{2πi} ∮ (\frac{w}{f̂(w)})^{a+n+2} \frac{w}{4f̂(w) − w} \frac{dw}{w^{n+1}} = [t^{n}] (\frac{t}{f̂(t)})^{a+n+2} \frac{t}{4f̂(t) − t} =

= 2^{a+n+1} [t^{a+2n+1}]\frac{1}{\sqrt{1 − 4t}} (\frac{1 − \sqrt{1 − 4t}}{2})^{a+n+2} = \binom{a + 3n + 2}{n}2^{a+n+1} .

So, we have the identities

∑_{k=0}^{n} \binom{a + 3n − 2k + 1}{n − k}\frac{a + n + 1}{a + 3n − 2k + 1} \binom{2k + 1}{k + 1} = \binom{a + 3n + 2}{n}  (45)

∑_{k=0}^{n} \binom{a + 3n − 2k + 2}{n − k}\frac{a + n + 2}{a + 3n − 2k + 2} \binom{2k}{k} = \binom{a + 3n + 2}{n}  (46)

∑_{k=0}^{n} \binom{a + 3n − k + 2}{n − k}\frac{a + n + k + 2}{a + 3n − k + 2} 2^{k} = \binom{a + 3n + 2}{n} .  (47)

For a = 0, 1, 2, 3, 4 we have the sequences \underline{A025174}, \underline{A004319}, \underline{A236194}, \underline{A236194}, \underline{A236194}.

4. If h_{n} is the Chebyshev polynomial of the second kind U_{n}(x), then we have

h(t) = \frac{1}{1 − 2xt + t^{2}} ⟺ f(t) = \frac{2xt}{1 + t} .

So, if we consider the Riordan matrix

R = (1, \frac{2xt}{1 + t}) = [\binom{n − 1}{n − k}(−1)^{n−k}(2x)^{k}]_{n,k≥0} ,

then, replacing a with a + 1 in identities (14) and (18), we obtain

∑_{k=0}^{n} \binom{a + n}{n − k}(−1)^{n−k}(2x)^{k}U_{k+1}(x) = ∑_{k=0}^{n} \binom{a + n + 1}{n − k}(−1)^{n−k}(2x)^{k+1}U_{k}(x) (48)

\begin{matrix}∑_{k=0}^{n} \binom{a + n + 1}{n − k}\frac{a + n + 2}{a + k + 2} (−1)^{n−k}(2x)^{k}U_{k+1}(x) = \\ = ∑_{k=0}^{n} \binom{a + n + 2}{n − k}\frac{a + n + 3}{a + k + 3} (−1)^{n−k}(2x)^{k+1}U_{k}(x) . \end{matrix}  (49)

Moreover, by (17), we have the identities

∑_{k=0}^{n} \binom{a + n}{n − k}(−1)^{n−k}(2x)^{k}U_{k+1}(x) = ∑_{k=0}^{n} \binom{a + n + k + 1}{n − k}(−1)^{n−k}(2x)^{2k+1}  (50)

∑_{k=0}^{n} \binom{a + n + 1}{n − k}(−1)^{n−k}(2x)^{k+1}U_{k}(x) = ∑_{k=0}^{n} \binom{a + n + k + 1}{n − k}(−1)^{n−k}(2x)^{2k+1} (51)

and, by (21), we have the identities

\begin{matrix}∑_{k=0}^{n} \binom{a + n + 1}{n − k}\frac{a + n + 2}{a + k + 2} (−1)^{n−k}(2x)^{k}U_{k+1}(x) \\ = ∑_{k=0}^{n} \binom{a + n + k + 2}{n − k}\frac{a + n + k + 3}{a + 2k + 3} (−1)^{n−k}(2x)^{2k+1} \end{matrix}  (52)

\begin{matrix}∑_{k=0}^{n} \binom{a + n + 2}{n − k}\frac{a + n + 3}{a + k + 3} (−1)^{n−k}(2x)^{k+1}U_{k}(x) \\ = ∑_{k=0}^{n} \binom{a + n + k + 2}{n − k}\frac{a + n + k + 3}{a + 2k + 3} (−1)^{n−k}(2x)^{2k+1} . \end{matrix}  (53)

Similarly, if h_{n} is the Chebyshev polynomial of the first kind T_{n}(x), then we have

h(t) = \frac{1 − xt}{1 − 2xt + t^{2}} ⟺ f̂(t) = \frac{t − xt^{2}}{x − t}

⟺ f(t) = \frac{1 + t − \sqrt{1 + 2(1 − 2x)t + t^{2}}}{1 + t} .

Consider the Riordan matrix R = [R_{n,k}(x)]_{n,k≥0} = (f(t)/t, f(t)). Then, by the Lagrange inversion formula, we have

\begin{matrix}R_{n,k}(x) = [t^{n}]\frac{f(t)}{t} f(t)^{k} = [t^{n+1}]f(t)^{k+1} = \\ = \frac{k + 1}{n + 1} [t^{n−k}] (\frac{t}{f̂(t)})^{n+1} = \frac{k + 1}{n + 1} [t^{n−k}] (\frac{x − t}{1 − xt})^{n+1} \end{matrix}

and so

R_{n,k}(x) = ∑_{i=0}^{n−k} \binom{n}{i}\binom{2n − k − i}{n}\frac{k + 1}{n − i + 1} (−1)^{i}x^{2n−2i−k+1} .

For these polynomials, we have the shifting identities

∑_{k=0}^{n} R_{a+n,a+k}(x)T_{k+1}(x) = ∑_{k=0}^{n} R_{a+n+1,a+k+1}(x)T_{k}(x)  (54)

∑_{k=0}^{n} \frac{a + n + 1}{a + k + 1} R_{a+n,a+k}(x)T_{k+1}(x) = ∑_{k=0}^{n} \frac{a + n + 2}{a + k + 2} R_{a+n+1,a+k+1}(x)T_{k}(x) . (55)

Finally, as we did for (17) and (21), we can also obtain the identities

∑_{k=0}^{n} R_{a+n,a+k}(x)T_{k+1}(x) = ∑_{k=0}^{n} R_{a+n+k+1,a+k+1}(x)  (56)

∑_{k=0}^{n} R_{a+n+1,a+k+1}(x)T_{k}(x) = ∑_{k=0}^{n} R_{a+n+k+1,a+k+1}(x)  (57)

and

∑_{k=0}^{n} \frac{a + n + 1}{a + k + 1} R_{a+n,a+k}(x)T_{k+1}(x) = ∑_{k=0}^{n} \frac{a + n + k + 2}{a + 2k + 2} R_{a+n+k+1,a+k+1}(x)  (58)

∑_{k=0}^{n} \frac{a + n + 2}{a + k + 2} R_{a+n+1,a+k+1}(x)T_{k}(x) = ∑_{k=0}^{n} \frac{a + n + k + 2}{a + 2k + 2} R_{a+n+k+1,a+k+1}(x) . (59)

4 Shifting property for Sheffer matrices

First, we consider the particular case of Sheffer matrices with g(t) = 1, and then we consider the general case which is more complex.

Theorem 7. A Sheffer matrix R = (1, f(t)) and a normalized sequence (h_{n})_{n∈ℕ}, with exponential generating series h(t), satisfy the shifting property, if and only if

h(t) = e^{∫_{0}^{t} f′(f̂(u)) du} ,  (60)

or, equivalently, if and only if

f̂(t) = ∫_{0}^{t} \frac{h(u)}{h′(u)} du .  (61)

In particular, the numbers h_{n} satisfy the recurrence

h_{n+1} = ∑_{k=0}^{n} s_{n,k}h_{k}  (62)

where

s_{n,k} = \frac{n!}{k!} [t^{n−k}]\frac{1}{f̂′(t)} .  (63)

If the shifting property holds, then the exponential generating series for the shifting sums is

s(t) = f′(t)e^{∫_{0}^{t} f′(u)^{2}du} .  (64)

Proof. Let R = [r_{n,k}]_{n,k≥0} = (g(t), f(t)) be a Sheffer matrix. Then, for every exponential formal series u(t) = ∑_{n≥0} u_{n} \frac{t^{n}}{n!}, we have the identity

g(t) u(f(t)) = ∑_{n≥0} [∑_{k=0}^{n} r_{n,k}u_{k}] \frac{t^{n}}{n!}

and (for g(t) = 1) the Sheffer matrix

R′ = [r_{n+1,k+1}]_{n,k≥0} = (f′(t), f(t)) .

Hence, if g(t) = 1, then identity (1) turns out to be equivalent to the identity

h′(f(t)) = f′(t) h(f(t)) .  (65)

Since f(t) is invertible with respect to composition, we have the equivalent identity

h′(t) = f′(f̂(t)) h(t)

from which we obtain (60). Since f′(f̂(t)) = 1/f̂′(t), we also have the identity

f̂′(t) = \frac{h(t)}{h′(t)}

from which we have (61). Finally, recurrence (62) derives from recurrence (5), where, by identity (6), the Sheffer matrix S = [s_{n,k}]_{n,k≥0} is defined by

S = R^{−1}R′ = (1, f̂(t))(f′(t), f(t)) = (f′(f̂(t)), t) = (\frac{1}{f̂′(t)}, t) .

This also implies (63). Finally, by (65) and (60), we obtain the exponential generating series for the shifting sums:

s(t) = f′(t) h(f(t)) = f′(t) e^{∫_{0}^{f(t)} f′(f̂(v)) dv} .

Setting u = f̂(v), we have v = f(u), dv = f′(u) du and s(t) assumes the form in (64).

Examples

1. By Theorem 7, we have

f(t) = ln \frac{1}{1 − t} ⟺ f̂(t) = 1 − e^{−t} ⟺ h(t) = e^{e^{t}−1} .

In this case, the numbers h_{n} are the Bell numbers b_{n} (\underline{A000110}). Since f̂(t) = 1 − e^{−t} and f̂′(t) = e^{−t}, we have s_{n,k} = \frac{n!}{k!}[t^{n−k}]e^{t} = \binom{n}{k}, and recurrence (62) becomes the usual recurrence for the Bell numbers:

b_{n+1} = ∑_{k=0}^{n} \binom{n}{k}b_{k} .

In particular, for the Sheffer matrix of the Stirling numbers of the first kind [5] (\underline{A132393})

R = [\left[{n \atop k}\right]]_{n,k≥0} = (1, ln \frac{t}{1 − t}) ,

we have the shifting identity

∑_{k=0}^{n} \left[{n \atop k}\right]b_{k+1} = ∑_{k=0}^{n} \left[{n + 1 \atop k + 1}\right]b_{k} .  (66)

By identity (64), we obtain the series

s(t) = \frac{1}{1 − t} e^{\frac{t}{1−t}}

(which is the exponential generating series for sequence \underline{A002720}). Since

e^{\frac{t}{1−t}} = ∑_{n≥0} ℓ_{n} \frac{t^{n}}{n!}

is the exponential generating series of the cumulative Lah numbers (\underline{A000262}), we also have the identities

∑_{k=0}^{n} \left[{n \atop k}\right]b_{k+1} = ∑_{k=0}^{n} \left[{n + 1 \atop k + 1}\right]b_{k} = ∑_{k=0}^{n} \binom{n}{k}(n − k)!ℓ_{k} .  (67)

2. By Theorem 7, we have

f(t) = e^{t} − 1 ⟺ f̂(t) = ln(1 + t) ⟺ h(t) = e^{t+t^{2}/2} .

In this case, the numbers h_{n} are the involution numbers i_{n} (\underline{A000085}). Since f̂(t) = ln(1 + t) and f̂′(t) = \frac{1}{1+t}, we have

s_{n,k} = \frac{n!}{k!}[t^{n−k}](1 + t) = \begin{cases}1, & if k = n; \\ n, & if k = n − 1; \\ 0, & otherwise , \end{cases}

and recurrence (62) becomes the usual recurrence for the involution numbers: i_{n+1} = i_{n} + ni_{n−1} (for n ≥ 1).

In particular, for the Sheffer matrix of the Stirling numbers of the second kind [5] (\underline{A008277})

R = [\left\{{n \atop k}\right\}]_{n,k≥0} = (1, e^{t} − 1) ,

we have the shifting identity

∑_{k=0}^{n} \left\{{n \atop k}\right\}i_{k+1} = ∑_{k=0}^{n} \left\{{n + 1 \atop k + 1}\right\}i_{k}  (68)

Finally, by identity (64), we obtain the series

s(t) = e^{t+e^{t} sinh t}

which is the exponential generating series for the Dowling numbers (\underline{A007405}).

3. By Theorem 7, we have

f(t) = \frac{t}{1 − t} ⟺ f̂(t) = \frac{t}{1 + t} ⟺ h(t) = e^{t+t^{2}+t^{3}/3} .

In this case, the numbers h_{n} form sequence \underline{A049425}. Since f̂(t) = \frac{t}{1+t} and f̂′(t) = \frac{1}{(1+t)^{2}} , we have

s_{n,k} = \frac{n!}{k!}[t^{n−k}](1 + t)^{2} = \begin{cases}1, & if k = n; \\ 2n, & if k = n − 1; \\ n(n − 1), & if k = n − 2; \\ 0, & otherwise , \end{cases}

and recurrence (62) becomes h_{n+1} = h_{n} + 2nh_{n−1} + n(n − 1)h_{n−2} (for n ≥ 2).

In particular, for the Sheffer matrix of the Lah numbers (\underline{A105287})

R = [|{n \atop k}|]_{n,k≥0} = (1, \frac{t}{1 − t}) ,

we have the shifting identity

∑_{k=0}^{n} |{n \atop k}|h_{k+1} = ∑_{k=0}^{n} |{n + 1 \atop k + 1}|h_{k} .  (69)

Finally, by identity (64), we obtain the series

s(t) = \frac{1}{(1 − t)^{2}} e^{\frac{3t−3t^{2}+t^{3}}{3(1−t)^{3}}} .

4. If h_{n} is the Stirling polynomial S_{n}(x) = ∑_{k=0}^{n} \left\{{n \atop k}\right\}x^{k}, then, by identity (61), we have

h(t) = e^{x(e^{t}−1)} ⟺ f̂(t) = \frac{1 − e^{−t}}{x} ⟺ f(t) = ln \frac{1}{1 − xt} .

So, if we consider the Sheffer matrix

R = (1, ln \frac{1}{1 − xt}) = [\left[{n \atop k}\right]x^{n}]_{n,k≥0} ,

then the shifting identity simplifies in

∑_{k=0}^{n} \left[{n \atop k}\right]S_{k}(x) = x ∑_{k=0}^{n} \left[{n + 1 \atop k + 1}\right]S_{k+1}(x) .  (70)

By identity (64), we obtain the series

s(t) = \frac{x}{1 − xt} e^{\frac{x^{2}t}{1−xt}} .

Since

e^{\frac{x^{2}t}{1−xt}} = ∑_{n≥0} x^{n}L_{n}(x) \frac{t^{n}}{n!}

where the coefficients

L_{n}(x) = ∑_{k=0}^{n} |{n \atop k}|x^{k}

are the Lah polynomials, we also have the identities

∑_{k=0}^{n} \left[{n \atop k}\right]S_{k}(x) = x ∑_{k=0}^{n} \left[{n + 1 \atop k + 1}\right]S_{k+1}(x) = ∑_{k=0}^{n} \binom{n}{k}(n − k)!L_{k}(x) .  (71)

5. If h_{n} is the Hermite polynomial H_{n}(x) [9], then, by identity (61), we have

h(t) = e^{2xt−t^{2}} ⟺ f̂(t) = −\frac{1}{2} ln (1 − \frac{x}{t}) ⟺ f(t) = −x(e^{−2t} − 1) .

So, if we consider the Sheffer matrix

R = (1, −x(e^{−2t} − 1)) = [\left\{{n \atop k}\right\}(−1)^{n−k}2^{n}x^{k}]_{n,k≥0} ,

then the shifting identity simplifies in

∑_{k=0}^{n} \left\{{n \atop k}\right\}(−1)^{n−k}x^{k}H_{k}(x) = 2 ∑_{k=0}^{n} \left\{{n + 1 \atop k + 1}\right\}(−1)^{n−k}x^{k+1}H_{k+1}(x) .  (72)

By identity (64), we obtain the series s(t) = 2xe^{−2t+x^{2}(1−e^{−4t})} .

More in general, we have

Theorem 8. A Sheffer matrix R = (g(t), f(t)) and a normalized sequence (h_{n})_{n∈ℕ}, with exponential generating series h(t), satisfy the shifting property, if and only if

h′(t) = f′(f̂(t))h(t) + \frac{g′(f̂(t))}{g(f̂(t))} ∫_{0}^{t} h(u) du ,  (73)

Proof. First, denoting with D the derivative with respect to t, we have

∑_{n≥k} r_{n+1,k+1} \frac{t^{n}}{n!} = D ∑_{n≥k} r_{n,k+1} \frac{t^{n}}{n!} = D [g(t)\frac{f(t)^{k+1}}{(k + 1)!}] = g′(t)\frac{f(t)^{k+1}}{(k + 1)!} + g(t)f′(t)\frac{f(t)^{k}}{k!} .

So, we have

∑_{n≥0} [∑_{k=0}^{n} r_{n+1,k+1}h_{n}] \frac{t^{n}}{n!} = ∑_{k≥0} h_{n} [∑_{n≥k}^{n} r_{n+1,k+1} \frac{t^{n}}{n!}]

= g′(t) ∑_{k≥0} h_{n} \frac{f(t)^{k+1}}{(k + 1)!} + g(t)f′(t) ∑_{k≥0} h_{n} \frac{f(t)^{k}}{k!}

= g′(t) ∑_{k≥1} h_{n−1} \frac{f(t)^{k}}{k!} + g(t)f′(t)h(f(t))

= g′(t) ∫_{0}^{f(t)} h(u) du + g(t)f′(t)h(f(t)) .

Consequently, identity (1) becomes

g(t)h′(f(t)) = g′(t) ∫_{0}^{f(t)} h(u) du + g(t)f′(t)h(f(t)) .

Since g(t) is invertible with respect to multiplication and f(t) is invertible with respect to composition, this identity is equivalent to identity (73).

Theorem 9. Let R = [r_{n,k}]_{n,k≥0} = (1, f(t)) be a Sheffer matrix and let (h_{n})_{n∈ℕ} be a normalized sequence with exponential generating series h(t). If R and (h_{n})_{n∈ℕ} satisfy the shifting property, then the Sheffer matrix R′ = [r_{n+1,k+1}]_{n,k≥0} = (f′(t), f(t)) satisfies the shifting property with respect to the normalized sequence (H_{n})_{n∈ℕ} whose exponential generating series H(t) is given by

H(t) = 1 + h′(t) ∫_{0}^{t} \frac{du}{h(u)} ,  (74)

or, equivalently, by

H′(t) = \frac{h′′(t)}{h′(t)} H(t) − \frac{h(t)h′′(t) − h′(t)^{2}}{h(t)h′(t)} .  (75)

Proof. The series H(t) is defined by identity (73), where g(t) = f′(t). By identity (61), we have

f̂′(t) = \frac{h(t)}{h′(t)} and f̂′′(t) = \frac{h′(t)^{2} − h(t)h′′(t)}{h′(t)^{2}} .

Since

f′(f̂(t)) = \frac{1}{f̂′(t)} and f′′(f̂(t)) = −\frac{f̂′′(t)}{f̂′(t)^{3}} ,

we have

\frac{g′(f̂(t))}{g(f̂(t))} = \frac{f′′(f̂(t))}{f′(f̂(t))} = −\frac{f̂′′(t))}{f̂′(t)^{2}} = −\frac{h′(t)^{2} − h(t)h′′(t)}{h(t)^{2}} .

So, the series H(t) is defined by the equation

H′(t) = \frac{h′(t)}{h(t)} H(t) − \frac{h′(t)^{2} − h(t)h′′(t)}{h(t)^{2}} ∫_{0}^{t} H(u) du .

Now, let w(t) = ∫_{0}^{t} H(u) du. Then w′(t) = H(t), w′′(t) = H′(t), and the above equation becomes

w′′(t) − \frac{h′(t)}{h(t)} w′(t) + \frac{h′(t)^{2} − h(t)h′′(t)}{h(t)^{2}} w(t) = 0 .

It is easy to see that w(t) = h(t) is a particular solution of this equation. So, we can set w(t) = h(t)z(t). Then, we have

w′(t) = h′(t)z(t) + h(t)z′(t)
w′′(t) = h′′(t)z(t) + 2h′(t)z′(t) + h(t)z′′(t)

and the previous equation becomes

h(t)z′′(t) + h′(t)z′(t) = 0

or, equivalently, (h(t)z′(t))′ = 0. Hence, we have h(t)z′(t) = K, with K constant. Since h_{0} = 1 and z_{0} = 0, we have z_{1} = 1. So K = 1, and h(t)z′(t) = 1, that is

z′(t) = \frac{1}{h(t)} .

Since z_{0} = 0, by integrating, we have

z(t) = ∫_{0}^{t} \frac{1}{h(u)} du ,

that is

w(t) = h(t) ∫_{0}^{t} \frac{1}{h(u)} du or ∫_{0}^{t} H(u) du = h(t) ∫_{0}^{t} \frac{1}{h(u)} du ,

By differentiating this last equation, we obtain identity (74). Finally, by differentiating (74), we obtain identity (75).

Examples 1. By (73), for the Sheffer matrix

R = [\left[{n + 1 \atop k + 1}\right]]_{n,k≥0} = (\frac{1}{1 − t}, ln \frac{t}{1 − t}) ,

we have that the series h(t) satisfies the integro-differential equation

h′(t) = e^{t} ∫_{0}^{t} h(u) du + e^{t} h(t) .

By differentiating, we obtain the differential equation

h′′(t) − (1 + e^{t})h′(t) − e^{t}h(t) = 0

from which we have that the numbers h_{n} satisfy the recurrence

h_{n+2} = h_{n+1} + ∑_{k=0}^{n} \binom{n}{k}h_{k+1} + ∑_{k=0}^{n} \binom{n}{k}h_{k} .

Let b(t) = e^{e^{t}−1} be the exponential generating series for the Bell numbers. Then, by equation (74) and Example 1 on page 16, we have

h(t) = 1 + b′(t) ∫_{0}^{t} \frac{du}{b(u)} = 1 + e^{e^{t}+t−1} ∫_{0}^{t} e^{−(e^{u}−1)} du .

Since

e^{x(e^{t}−1)} = ∑_{n≥0} S_{n}(x) \frac{t^{n}}{n!}

is the generating series for the Stirling polynomials, we have the explicit expression

h_{n} = δ_{n,0} + ∑_{k=1}^{n} \binom{n}{k}S_{k}(−1)b_{n−k+1} .

These numbers form sequence \underline{A040027}. By equation (75), we also have

h′(t) = (1 + e^{t})h(t) − 1 ,

from which we obtain the following other recurrence

h_{n+1} = h_{n} + ∑_{k=0}^{n} \binom{n}{k}h_{k} − δ_{n,0} .

Finally, we have the shifting property

∑_{k=0}^{n} \left[{n + 1 \atop k + 1}\right]h_{k+1} = ∑_{k=0}^{n} \left[{n + 2 \atop k + 2}\right]h_{k} .  (76)

These sums form sequence \underline{A002793}.

2. By (73), for the Sheffer matrix

R = [\left\{{n + 1 \atop k + 1}\right\}]_{n,k≥0} = (e^{t}, e^{t} − 1) ,

we have that the series h(t) satisfies the integro-differential equation

h′(t) = ∫_{0}^{t} h(u) du + (1 + t)h(t) .

By differentiating, we obtain the differential equation

h′′(t) − (1 + t)h′(t) − 2h(t) = 0

from which we have that the numbers h_{n} satisfy the recurrence

h_{n+2} = h_{n+1} + (n + 2)h_{n}

with the initial values h_{0} = h_{1} = 1. These numbers form sequence \underline{A000932}. Moreover, by equation (74) and Example 2 on page 17, we have

h(t) = 1 + (1 + t)e^{t+t^{2}/2} ∫_{0}^{t} e^{−(u+u^{2}/2)} du .

Finally, we have the shifting identity

∑_{k=0}^{n} \left\{{n + 1 \atop k + 1}\right\}h_{k+1} = ∑_{k=0}^{n} \left\{{n + 2 \atop k + 2}\right\}h_{k} .  (77)

3. By (73), for the Sheffer matrix

R = [|{n + 1 \atop k + 1}|]_{n,k≥0} = (\frac{1}{(1 − t)^{2}} , \frac{1}{1 − t}) ,

we have that the series h(t) satisfies the integro-differential equation

h′(t) = 2(1 + t) ∫_{0}^{t} h(u) du + (1 + t)^{2}h(t) .

By differentiating, we obtain the differential equation

(1 + t)h′′(t) − (2 + 3t + 3t^{2} + t^{3})h′(t) − 3(1 + t)^{2}h(t) = 0 .

From this equation, we have that the numbers h_{n} satisfy the recurrence

h_{n+4} + nh_{n+3} − 3(n + 3)h_{n+2} − 3(n + 3)(n + 2)h_{n+1} − (n + 3)(n + 2)(n + 1)h_{n} = 0 with the initial values h_{0} = h_{1} = 1, h_{2} = 5, h_{3} = 17. The first few values of this sequence are: 1, 1, 5, 17, 69, 339, 1677, 9321, 55137, 343659, 2285289, 15910857, 116120781. This sequence does not appear in [11]. Moreover, by equation (74) and Example 3 on page 18, we have

h(t) = 1 + (1 + t)^{2}e^{t+t^{2}+t^{3}/3} ∫_{0}^{t} e^{−(u+u^{2}+u^{3}/3)} du .

Finally, we have the shifting identity

∑_{k=0}^{n} |{n + 1 \atop k + 1}|h_{k+1} = ∑_{k=0}^{n} |{n + 2 \atop k + 2}|h_{k} .  (78)

4. By (73), for the Sheffer matrix

R = [\binom{n}{k}\frac{n!}{k!}]_{n,k≥0} = (\frac{1}{1 − t}, \frac{1}{1 − t}) ,

we have that the series h(t) satisfies the integro-differential equation

h′(t) = (1 + t)^{2}h(t) + (1 + t) ∫_{0}^{t} h(u) du .

By differentiating, we obtain the differential equation

(1 + t)h′′(t) − (2 + 3t + 3t^{2} + t^{3})h′(t) − 2(1 + t)^{2}h(t) = 0 .

From this equation, we have that the numbers h_{n} satisfy the recurrence

h_{n+4} + nh_{n+3} − (3n + 8)h_{n+2} − (3n + 7)(n + 2)h_{n+1} − (n + 2)^{2}(n + 1)h_{n} = 0

with the initial conditions h_{0} = h_{1} = 1, h_{2} = 4, h_{3} = 13. The first few values are: 1, 1, 4, 13, 50, 231, 1106, 5909, 33818, 205055, 1326226, 9014181, 64329034, 480660103. This sequence does not appear in [11]. In this case we have the shifting identity

∑_{k=0}^{n} \binom{n}{k}\frac{n!}{k!} h_{k+1} = ∑_{k=0}^{n} \binom{n + 1}{k + 1}\frac{(n + 1)!}{(k + 1)!} h_{k} .  (79)

5 Shifting property for connection constants matrices

The elementary and the homogeneous symmetric functions [6, 4] are respectively defined by

\binom{x_{1}, x_{2}, …, x_{n}}{k} = e_{k}(x_{1}, x_{2}, …, x_{n}) = ∑_{1≤i_{1}<···<i_{k}≤n} x_{i_{1}} ⋯ x_{i_{k}}

\begin{aligned}( \\ ( & {x_{1}, x_{2}, …, x_{n} \atop k} & ) \\ & & ) \end{aligned} = h_{k}(x_{1}, x_{2}, …, x_{n}) = ∑_{{i_{1},...,in≥0 \atop i_{1}+···+in=k}} x_{1}^{i_{1}} ⋯ x_{n}^{i_{n}}

and have generating series

∑_{k≥0} \binom{x_{1}, x_{2}, …, x_{n}}{k}t^{k} = (1 + x_{1}t)(1 + x_{2}t) ⋯ (1 + x_{n}t)

∑_{k≥0} \begin{aligned}( \\ ( & {x_{1}, x_{2}, …, x_{n} \atop k} & ) \\ & & ) \end{aligned}t^{k} = \frac{1}{(1 − x_{1}t)(1 − x_{2}t) ⋯ (1 − x_{n}t)} .

We proved [4] that every connection constant can be expressed in terms of these symmetric functions. More precisely, given the zero sequence 0 = (0, 0, …), we proved that

C_{n,k}^{(ρ,0)} = \binom{r_{1}, …, r_{n}}{n − k}(−1)^{n−k}  (80)
C_{n,k}^{(0,σ)} = \begin{aligned}( \\ ( & {s_{1}, …, s_{k+1} \atop n − k} & ) \\ & & ) \end{aligned}  (81)

and that

C_{n,k}^{(ρ,σ)} = ∑_{j=k}^{n} \binom{r_{1}, …, r_{n}}{n − j}\begin{matrix}( \\ ( & {s_{1}, …, s_{k+1} \atop j − k} & ) \\ & & ) \end{matrix}(−1)^{n−j} .  (82)

Let us consider, now, the following generalization of the connection constants C_{n,k}^{(ρ,0)} and C_{n,k}^{(0,σ)}. Given two sequences σ = (s_{1}, s_{2}, …) and τ = (t_{1}, t_{2}, …), we define the coefficients

N_{n,k}^{(σ,τ)} = \binom{s_{1}, …, s_{n}}{n − k}\frac{1}{t_{1} ⋯ t_{n}}  (t_{k} ≠ 0 , for every k ∈ ℕ)  (83)
M_{n,k}^{(σ,τ)} = \begin{aligned}( \\ ( & {s_{1}, …, s_{k+1} \atop n − k} & ) \\ & & ) \end{aligned}t_{1}t_{2} ⋯ t_{k} .  (84)

The M_{n,k}^{(σ,τ)} are the generalized De Morgan numbers^{3}. Since the homogeneous symmetric functions satisfy the recurrence

\begin{aligned}( \\ ( & {x_{1}, x_{2}, …, x_{n+1} \atop k + 1} & ) \\ & & ) \end{aligned} = \begin{aligned}( \\ ( & {x_{1}, x_{2}, …, x_{n} \atop k + 1} & ) \\ & & ) \end{aligned} + \begin{aligned}( \\ ( & {x_{1}, x_{2}, …, x_{n+1} \atop k} & ) \\ & & ) \end{aligned}x_{n+1} ,  (85)

the generalized De Morgan numbers satisfy the recurrence

M_{n+1,k+1}^{(σ,τ)} = t_{k+1}M_{n,k}^{(σ,τ)} + s_{k+2}M_{n,k+1}^{(σ,τ)} .  (86)

Finally, we define the generalized continuants K_{n}^{(σ,τ)} by the recurrence

K_{n+2}^{(σ,τ)} = t_{n+2}K_{n+1}^{(σ,τ)} + s_{n+2}K_{n}^{(σ,τ)}  (87)

3Notice that this is a slight generalization of the De Morgan numbers considered in [4].

with the initial conditions K_{0}^{(σ,τ)} = 1 and K_{1}^{(σ,τ)} = t_{1}, [5]. For convenience, we also define K_{−1}^{(σ,τ)} = 0.

Then, we have

Theorem 10. The matrix R = [M_{n,k}^{(σ,τ)}]_{n,k≥0} and the sequence (K_{n}^{(σ,τ)})_{n≥0} satisfy the shifting

property, that is

∑_{k=0}^{n} M_{n,k}^{(σ,τ)}K_{k+1}^{(σ,τ)} = ∑_{k=0}^{n} M_{n+1,k+1}^{(σ,τ)}K_{k}^{(σ,τ)} .  (88)

Proof. By the recurrences (86) and (87), we have

∑_{k=0}^{n} M_{n+1,k+1}^{(σ,τ)}K_{k}^{(σ,τ)} = ∑_{k=0}^{n} M_{n,k}^{(σ,τ)}t_{k+1}K_{k}^{(σ,τ)} + ∑_{k=0}^{n−1} M_{n,k+1}^{(σ,τ)}s_{k+2}K_{k}^{(σ,τ)}

= ∑_{k=0}^{n} M_{n,k}^{(σ,τ)}t_{k+1}K_{k}^{(σ,τ)} + ∑_{k=1}^{n} M_{n,k}^{(σ,τ)}s_{k+1}K_{k−1}^{(σ,τ)}

= ∑_{k=0}^{n} M_{n,k}^{(σ,τ)}(t_{k+1}K_{k}^{(σ,τ)} + s_{k+1}K_{k−1}^{(σ,τ)})

= ∑_{k=0}^{n} M_{n,k}^{(σ,τ)}K_{k+1}^{(σ,τ)} .

Using recurrences (86) and (87), we obtain the following particular instances of the sifting identity (88). In particular, we obtain the q-analogues of some identities obtained in the previous sections. Recall that the q-numbers are defined as [n] := 1 + q + q^{2} + ⋯ + q^{n−1}, and that the q-factorials are defined as [n]! = [n][n − 1] ⋯ [2][1].

Examples

1. For s_{n} = n−1 and t_{n} = n, we obtain the ordinary De Morgan numbers M_{n,k} (\underline{A131689}) defined by the recurrence

M_{n+1,k+1} = (k + 1)M_{n,k} + (k + 1)M_{n,k+1} .

Moreover, the numbers K_{n}^{(σ,τ)} = h_{n} satisfy the recurrence

h_{n+2} = (n + 2)h_{n+1} + (n + 1)h_{n}

with the initial conditions h_{0} = h_{1} = 1. They have exponential generating series

h(t) = \frac{e^{−t}}{(1 − t)^{2}}

and form sequence \underline{A000255}. Finally, we have the shifting identity

∑_{k=0}^{n} M_{n,k}h_{k+1} = ∑_{k=0}^{n} M_{n+1,k+1}h_{k} .  (89)

2. For s_{n} = q^{n−1} and t_{n} = 1, we obtain the Gaussian coefficients

\begin{aligned}( \\ ( & {1, q, q^{2}, …, q^{k} \atop n − k} & ) \\ & & ) \end{aligned} = \binom{n}{k}_{q} = \frac{[n]!}{[k]![n − k]!}

satisfying the recurrence

\binom{n + 1}{k + 1}_{q} = \binom{n}{k}_{q} + q^{k+1}\binom{n}{k + 1}_{q} .

Moreover, we have the q-Fibonacci numbers K_{n}^{(σ,τ)} = f_{n}(q), defined by the recurrence f_{n+2}(q) = f_{n+1}(q) + q^{n+1}f_{n}(q) with the initial conditions f_{0}(q) = f_{1}(q) = 1, [1, 8]. So, we have the shifting identity

∑_{k=0}^{n} \binom{n}{k}_{q}f_{k+1}(q) = ∑_{k=0}^{n} \binom{n + 1}{k + 1}_{q}f_{k}(q) .  (90)

Clearly, for q = 1, we reobtain identity (25).

3. For s_{n} = [n − 1] and t_{n} = 1, we have the q-Stirling numbers of the second kind [4]:

\begin{aligned}( \\ ( & {[0], [1], [2], …, [k] \atop n − k} & ) \\ & & ) \end{aligned} = \begin{aligned}( \\ ( & {[1], [2], …, [k] \atop n − k} & ) \\ & & ) \end{aligned} = \left\{{n \atop k}\right\}_{q}

satisfying the recurrence

\left\{{n + 1 \atop k + 1}\right\}_{q} = \left\{{n \atop k}\right\}_{q} + [k + 1]\left\{{n \atop k + 1}\right\}_{q} .

Moreover, we have the q-involution numbers K_{n}^{(σ,τ)} = i_{n}(q) satisfying the recurrence i_{n+2}(q) = i_{n+1}(q) + [n + 1]i_{n}(q) with the initial conditions i_{0}(q) = i_{1}(q) = 1. So, we have the shifting identity

∑_{k=0}^{n} \left\{{n \atop k}\right\}_{q}i_{k+1}(q) = ∑_{k=0}^{n} \left\{{n + 1 \atop k + 1}\right\}_{q}i_{k}(q) .  (91)

Clearly, for s_{n} = n − 1 (i.e., for q = 1), we reobtain identity (68).

4. For s_{n} = [n − 1] and t_{n} = [n], we obtain the q-De Morgan numbers M_{n,k}(q) = \left\{{n \atop k}\right\}_{q}[k]! satisfying the recurrence

M_{n+1,k+1}(q) = [k + 1]M_{n,k}(q) + [k + 1]M_{n,k+1}(q) .

Moreover, the numbers K_{n}^{(σ,τ)} = h_{n}(q) satisfy the recurrence

h_{n+2}(q) = [n + 2]h_{n+1}(q) + [n + 1]h_{n}(q)

with the initial conditions h_{0}(q) = h_{1}(q) = 1. Finally, we have the shifting identity

∑_{k=0}^{n} M_{n,k}(q)h_{k+1}(q) = ∑_{k=0}^{n} M_{n+1,k+1}(q)h_{k}(q)  (92)

which is a q-analogue of identity (89).

Lemma 11. If t_{k} ≠ 0 for every k ∈ ℕ, then the matrices

M^{(σ,τ)} = [M_{n,k}^{(σ,τ)}]_{n,k≥0}  and  N^{(σ,τ)} = [N_{n,k}^{(σ,τ)}]_{n,k≥0}

are invertible, and their inverses are

[M_{n,k}^{(σ,τ)}]_{n,k≥0}^{−1} = [N_{n,k}^{(σ,τ)}(−1)^{n−k}]_{n,k≥0}
[N_{n,k}^{(σ,τ)}]_{n,k≥0}^{−1} = [M_{n,k}^{(σ,τ)}(−1)^{n−k}]_{n,k≥0} .

Proof. Both M^{(σ,τ)} and N^{(σ,τ)} are infinite lower triangular matrices with non-zero diagonal entries. So, they are invertible. Moreover, by relations (80) and (81) and by property (3), we have at once the identities

[\begin{matrix}( \\ ( & {s_{1}, …, s_{k+1} \atop n − k} & ) \\ & & ) \end{matrix}]_{n,k≥0}^{−1} = [\binom{r_{1}, …, r_{n}}{n − k}(−1)^{n−k}]_{n,k≥0}

[\binom{r_{1}, …, r_{n}}{n − k}]_{n,k≥0}^{−1} = [\begin{matrix}( \\ ( & {s_{1}, …, s_{k+1} \atop n − k} & ) \\ & & ) \end{matrix}(−1)^{n−k}]_{n,k≥0}

which imply the assertion.

Theorem 12. The matrix R = [N_{n,k}^{(σ,τ)}]_{n,k≥0} and the sequence (h_{n}^{(σ,τ)})_{n,k≥0} defined by the recurrence

h_{n+1}^{(σ,τ)} = ∑_{k=0}^{n} s_{n,k}^{(σ,τ)}h_{k}^{(σ,τ)}

with h_{0}^{(σ,τ)} = 1, where

s_{n,k}^{(σ,τ)} = ∑_{i=k}^{n} \begin{aligned}( \\ ( & {s_{1}, …, s_{i+1} \atop n − i} & ) \\ & & ) \end{aligned}\binom{s_{1}, …, s_{i+1}}{i − k}\frac{(−1)^{n−i}}{t_{i+1}} ,  (93)

satisfy the shifting property.

Proof. For the matrix R = [N_{n,k}^{(σ,τ)}]_{n,k≥0}, we have R′ = [N_{n+1,k+1}^{(σ,τ)}]_{n,k≥0} and R^{−1} = [M_{n,k}^{(σ,τ)}(−1)^{n−k}]_{n,k≥0} .

Consequently, the entries of the matrix S = R^{−1}R′ are

s_{n,k}^{(σ,τ)} = ∑_{i=k}^{n} M_{n,i}^{(σ,τ)}(−1)^{n−i}N_{i+1,k+1}^{(σ,τ)} ,

which simplify in (93). Now, the theorem follows from Lemma 2.

Examples 1. By Theorem 12, for s_{n} = q^{n−1} and t_{n} = 1, we have the q-coefficients

\begin{aligned}( \\ ( & {1, q, …, q^{k} \atop n − k} & ) \\ & & ) \end{aligned} = \binom{n}{k}_{q} and \binom{1, q, …, q^{n−1}}{n − k} = \binom{n}{k}_{q}q^{\binom{n−k}{2}},

and

s_{n,k}(q) = ∑_{i=k}^{n} \binom{n}{i}_{q}\binom{i + 1}{k + 1}_{q}q^{\binom{i−k}{2}}(−1)^{n−i} .

For the q-Fibonacci numbers defined by the recurrence

F_{n+1}(q) = ∑_{k=0}^{n} s_{n,k}(q)F_{k}(q)

with the initial condition F_{0}(q) = 1, we have the shifting property

∑_{k=0}^{n} \binom{n}{k}_{q}q^{\binom{n−k}{2}}F_{k+1}(q) = ∑_{k=0}^{n} \binom{n + 1}{k + 1}_{q}q^{\binom{n−k}{2}}F_{k}(q)  (94)

which is another q-analogue of identity (25).

2. By Theorem 12, for s_{n} = [n − 1] and t_{n} = 1, we have the q-coefficients

\begin{aligned}( \\ ( & {[1], …, [k] \atop n − k} & ) \\ & & ) \end{aligned} = \left\{{n \atop k}\right\}_{q} and \binom{[1], …, [n]}{n − k} = \left[{n \atop k}\right]_{q},

and

s_{n,k}(q) = ∑_{i=k}^{n} \left\{{n \atop i}\right\}_{q}\left[{i + 1 \atop k + 1}\right]_{q}(−1)^{n−i} .

For the q-Bell numbers defined by the recurrence

B_{n+1}(q) = ∑_{k=0}^{n} s_{n,k}(q)B_{k}(q)

with the initial condition B_{0}(q) = 1, we have the shifting property

∑_{k=0}^{n} \left[{n \atop k}\right]_{q}B_{k+1}(q) = ∑_{k=0}^{n} \left[{n + 1 \atop k + 1}\right]_{q}B_{k}(q)  (95)

which is a q-analogue of identity (66). Clearly, identity (66) can also be reobtained for s_{n} = n − 1 and t_{n} = 1.

References

[1] L. Carlitz, Fibonacci notes–3: q-Fibonacci numbers, Fibonacci Quart. 12 (1974), 317– 322.

[2] L. Comtet, Advanced Combinatorics, Reidel, 1974.

[3] E. Damiani, O. D’Antona, and G. Naldi, On the connection constants, Studies in Appl. Math. 85 (1991), 289–302.

[4] O. M. D’Antona and E. Munarini, A combinatorial interpretation of the connection constants for persistent sequences of polynomials, European J. Combin. 26 (2005), 1105– 1118.

[5] R. L. Graham, D. E. Knuth, and O. Patashnik, Concrete Mathematics, Addison-Wesley, 1989.

[6] I. G. Macdonald, Symmetric Functions and Hall Polynomials, second ed., Oxford University Press, 1995.

[7] E. Munarini, A combinatorial interpretation of the generalized Fibonacci numbers, Adv. in Appl. Math. 19 (1998), 306–318.

[8] E. Munarini, Generalized q-Fibonacci numbers, Fibonacci Quart. 43 (2005), 234–242.

[9] S. Roman, The Umbral Calculus, Pure and Applied Mathematics Series, Vol. 111, Academic Press, 1984.

[10] S. M. Roman and G.-C. Rota, The umbral calculus, Adv. in Math. 27 (1978), 95–188.

[11] N. J. A. Sloane, The On-Line Encyclopedia of Integer Sequences, http://oeis.org.

[12] L. W. Shapiro, S. Getu, W. J. Woan, and L. C. Woodson, The Riordan group, Discrete Appl. Math. 34 (1991), 229–239.

[13] L. W. Shapiro, A survey of the Riordan group, Talk at a meeting of the American Mathematical Society, Richmond, Virginia, 1994.

[14] L. W. Shapiro, Bijections and the Riordan group, Theoret. Comput. Sci. 307 (2003), 403–413.

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[16] W. Wang and T. Wang, Generalized Riordan arrays, Discrete Math. 308 (2008), 6466– 6500.

2000 Mathematics Subject Classification: Primary: 05A19, Secondary: 05A15, 05A10, 11C08. Keywords: combinatorial sum, Riordan matrix, Sheffer matrix, connection constant, generating series, differential equation.

(Concerned with sequences \underline{A000045}, \underline{A000073}, \underline{A000078}, \underline{A000085}, \underline{A000108}, \underline{A000110}, \underline{A000255}, \underline{A000262}, \underline{A000932}, \underline{A000984}, \underline{A001591}, \underline{A001592}, \underline{A001764}, \underline{A002720}, \underline{A002793}, \underline{A004319}, \underline{A006629}, \underline{A006630}, \underline{A006631}, \underline{A007318}, \underline{A007405}, \underline{A008277}, \underline{A008287}, \underline{A025174}, \underline{A027907}, \underline{A035343}, \underline{A040027}, \underline{A049425}, \underline{A063260}, \underline{A063265}, \underline{A066178}, \underline{A079262}, \underline{A102594}, \underline{A102893}, \underline{A105287}, \underline{A122189}, \underline{A131689}, \underline{A132393}, \underline{A171890}, \underline{A213651}, \underline{A213652}, \underline{A230547}, \underline{A233657}, and \underline{A236194}.)

Received September 28 2016; revised versions received October 3 2016; July 6 2017; August 24 2017; August 25 2017. Published in Journal of Integer Sequences, August 31 2017.

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