Incomplete Tribonacci Numbers and Polynomials

José L. Ramírez^{1}

Instituto de Matemáticas y sus Aplicaciones

Calle 74 No. 14 - 14 Bogotá Colombia

josel.ramirez@ima.usergioarboleda.edu.co

Víctor F. Sirvent^{2} Universidad Simón Bolívar

Departamento de Matemáticas Apartado 89000 Caracas 1086-A

Venezuela

vsirvent@usb.ve

Abstract

We define the incomplete tribonacci sequence of numbers and polynomials. We study recurrence relations, some properties of these numbers and polynomials, and the generating function of the incomplete tribonacci numbers.

1 Introduction

Fibonacci numbers and their generalizations have many interesting properties and applications in many fields of science and art (cf. [10]). The Fibonacci numbers F_{n} are defined by the recurrence relation

1The first author was partially supported by Universidad Sergio Arboleda.

2The second author was partially supported by FWF project Nr. P23990.

F_{0} = 0, F_{1} = 1, F_{n+1} = F_{n} + F_{n−1}, n ⩾ 1.

The first few terms are 0,1,1,2,3,5,8,13,… (sequence \underline{A000045})^{3}. Another important sequence is the Lucas sequence. These numbers are defined by the recurrence relation

L_{0} = 2, L_{1} = 1, L_{n+1} = L_{n} + L_{n−1}, n ⩾ 1.

The first few terms are 2,1,3,7,11,18,29,37,… (sequence \underline{A000032}). The Fibonacci and Lucas numbers have been studied extensively. In particular, there is a beautiful combinatorial identity for Fibonacci numbers and Lucas numbers (cf. [10]):

F_{n} = \begin{matrix}⌊ & \frac{n−1}{2} & ⌋ \\ ∑ \\ i=0 \end{matrix} \binom{n − i − 1}{i},  L_{n} = \begin{matrix}⌊ & \frac{n}{2} & ⌋ \\ ∑ \\ i=0 \end{matrix} \frac{n}{n − j} \binom{n − j}{j}.  (1)

In analogy with (1), Filipponi [6] introduced the incomplete Fibonacci numbers F_{n}(s) and the incomplete Lucas numbers L_{n}(s). They are defined by

F_{n}(s) = ∑_{j=0}^{s} \binom{n − 1 − j}{j}, (n = 1, 2, 3, …; 0 ≤ s ≤ ⌊\frac{n − 1}{2}⌋);

and

L_{n}(s) = ∑_{j=0}^{s} \frac{n}{n − j} \binom{n − j}{j}, (n = 1, 2, 3, …; 0 ≤ s ≤ ⌊\frac{n}{2}⌋) .

Generating functions of the incomplete Fibonacci and Lucas numbers were determined by Pintér and Srivastava [15]. Djordjević [3] defined and studied incomplete generalized Fibonacci and Lucas numbers. Djordjević and Srivastava [4] defined incomplete generalized Jacobsthal and Jacobsthal-Lucas numbers. Tasci and Cetin Firengiz [21] defined the incomplete Fibonacci and Lucas p-numbers. Tasci et al. [22] defined the incomplete bivariate Fibonacci and Lucas p-polynomials. Ramírez [18] introduced the incomplete k-Fibonacci and k-Lucas numbers, the incomplete h(x)-Fibonacci and h(x)-Lucas polynomials [17], and the bi-periodic incomplete Fibonacci sequences [16].

The tribonacci numbers are defined by the recurrence relation:

t_{0} = 0, t_{1} = t_{2} = 1, t_{n+2} = t_{n+1} + t_{n} + t_{n−1} for n ⩾ 1.  (2)

The first few terms of the tribonacci numbers are 0, 1, 1, 2, 4, 7, 13, 24, 44, 81,…, (sequence \underline{A000073}). The tribonacci numbers have been studied in different contexts; see [1, 5, 8, 9, 11, 12, 13, 14, 19]. Alladi and Hoggatt [1] defined the tribonacci triangle; see

0 1 2 3 4 5 6 7 ⋯
0 1
1 1 1
2 1 3 1
3 1 5 5 1
4 1 7 13 7 1
5 1 9 25 25 9 1
6 1 11 41 63 41 11 1
7 1 13 61 129 129 61 13 1
⋮  ⋮

Table 1: Tribonacci triangle

Table 1. It was used to derive the expansion of the tribonacci numbers and each element is defined in similar way as in the Pascal Triangle (cf. [10]).

Let B(n, i) be the element in the n-th row and i-th column of the tribonacci triangle. By the definition of the triangle, we have

B(n + 1, i) = B(n, i) + B(n, i − 1) + B(n − 1, i − 1),  (3)

where B(n, 0) = B(n, n) = 1. The sum of elements on the rising diagonal lines in the tribonacci triangle is the tribonacci number t_{n} (cf. [1]), i.e.,

t_{n} = ∑_{i=0}^{⌊ \frac{n−1}{2} ⌋} B(n − 1 − i, i).

Moreover, Barry [2] showed that these coefficients satisfy the relation

B(n, i) = ∑_{j=0}^{i} \binom{i}{j}\binom{n − j}{i}.

Therefore,

t_{n} = ∑_{i=0}^{⌊ \frac{n−1}{2} ⌋} B(n − 1 − i, i) = ∑_{i=0}^{⌊ \frac{n−1}{2} ⌋} ∑_{j=0}^{i} \binom{i}{j}\binom{n − 1 − i − j}{i}.  (4)

A large class of polynomials can also be defined by Fibonacci-like recurrences and tribonaccilike recurrences, such that yield Fibonacci numbers and tribonacci numbers. Such polynomials are called Fibonacci polynomials [10] and tribonacci polynomials [7], respectively.

3Many integer sequences and their properties are given in the On-Line Encyclopedia of Integer Sequences [20].

In 1883, Catalan and Jacobsthal introduced Fibonacci polynomials (cf. [10]). The polynomials F_{n}(x), studied by Catalan, are defined by the recurrence relation

F_{0}(x) = 0, F_{1}(x) = 1, F_{n+1}(x) = xF_{n}(x) + F_{n−1}(x), n ⩾ 1.

The Fibonacci polynomials, studied by Jacobsthal, are defined by

J_{0}(x) = 1, J_{1}(x) = 1, J_{n+1}(x) = J_{n}(x) + xJ_{n−1}(x), n ⩾ 1.

The Lucas polynomials L_{n}(x), originally studied in 1970 by Bicknell, are defined by

L_{0}(x) = 2, L_{1}(x) = x, L_{n+1}(x) = xL_{n}(x) + L_{n−1}(x), n ⩾ 1.

Hoggatt and Bicknell [7] introduced tribonacci polynomials. The tribonacci polynomials T_{n}(x) are defined by the recurrence relation

T_{0}(x) = 0, T_{1}(x) = 1, T_{2}(x) = x^{2}, T_{n+2}(x) = x^{2}T_{n+1}(x) + xT_{n}(x) + T_{n−1}(x), n ⩾ 2.

Note that T_{n}(1) = t_{n} for all integer positive n. The first few tribonacci polynomials are

\begin{matrix}T_{1}(x) = 1, & T_{5}(x) = x^{8} + 3x^{5} + 3x^{2}, \\ T_{2}(x) = x^{2}, & T_{6}(x) = x^{10} + 4x^{7} + 6x^{4} + 2x, \\ T_{3}(x) = x^{4} + x, & T_{7}(x) = x^{12} + 5x^{9} + 10x^{6} + 7x^{3} + 1, \\ T_{4}(x) = x^{6} + 2x^{3} + 1, & T_{8}(x) = x^{14} + 6x^{11} + 15x^{8} + 16x^{5} + 6x^{2}. \end{matrix}

In analogy with the tribonacci triangle, we define the tribonacci polynomial triangle; see Table 2.

0  1  2  3  4  5 ⋯
0 1
1 x^{2}  x
2 x^{4} 2x^{3} + 1  x^{2}
3 x^{6} 3x^{5} + 2x^{2}  3x^{4} + 2x  x^{3}
4 x^{8} 4x^{7} + 3x^{4}  6x^{6} + 6x^{3} + 1  4x^{5} + 3x^{2}  x^{4}
5 x^{10} 5x^{9} + 4x^{6} 10x^{8} + 12x^{5} + 3x^{2} 10x^{7} + 12x^{4} + 3x 5x^{6} + 4x^{3} x^{5}
⋮  ⋮

Table 2: Tribonacci polynomial triangle

Let B(n, i)(x) be the element in the n-th row and i-th column of the tribonacci polynomial triangle. Then

B(n + 1, i)(x) = x^{2}B(n, i)(x) + xB(n, i − 1)(x) + B(n − 1, i − 1)(x),

where B(n, 0)(x) = x^{2n} and B(n, n) = x^{n}.

It can be proved by induction on n, that the sum of elements on the rising diagonal lines in the tribonacci polynomial triangle is the tribonacci polynomial T_{n}(x), i.e.,

T_{n}(x) = ∑_{i=0}^{⌊ \frac{n−1}{2} ⌋} B(n − 1 − i, i)(x).

Moreover,

B(n, i)(x) = ∑_{j=0}^{i} \binom{i}{j}\binom{n − j}{i}x^{2n−i−3j}.

Therefore,

T_{n}(x) = ∑_{i=0}^{⌊ \frac{n−1}{2} ⌋} B(n − 1 − i, i)(x) = ∑_{i=0}^{⌊ \frac{n−1}{2} ⌋} ∑_{j=0}^{i} \binom{i}{j}\binom{n − 1 − i − j}{i}x^{2n−3(i+j)−2}.  (5)

From Equations (4) and (5), we introduce the incomplete tribonacci polynomials and incomplete tribonacci numbers. In this way we obtain new recurrence relations, new identities and the generating function of the incomplete tribonacci numbers.

2 Incomplete tribonacci polynomials and incomplete tribonacci numbers

Definition 1. For n ⩾ 1, incomplete tribonacci polynomials are defined as

T_{n}^{(s)}(x) = ∑_{i=0}^{s} B(n − 1 − i, i)(x)  (6)

= ∑_{i=0}^{s} ∑_{j=0}^{i} \binom{i}{j}\binom{n − i − j − 1}{i}x^{2n−3(i+j)−2}, 0 ⩽ s ⩽ ⌊\frac{n − 1}{2}⌋ .  (7)

We define the incomplete tribonacci numbers, t_{n}(s) as the value of T_{n}^{(s)}(x) at x = 1, i.e., t_{n}(s) = T_{n}^{(s)}(1).

In Tables 3 and 4, some values of incomplete tribonacci polynomials and incomplete tribonacci numbers are provided.

From the definitions follow

T (⌊\frac{n−1}{2}⌋)_{n}(x) = T_{n}(x) and t_{n} (⌊\frac{n − 1}{2}⌋) = t_{n}.

n/s 0  1  2  3
1  1
2 x^{2}
3 x^{4}  x^{4} + x
4 x^{6}  x^{6} + 2x^{3} + 1
5 x^{8} x^{8} + 3x^{5} + 2x^{2}  x^{8} + 3x^{5} + 3x^{2}
6 x^{10} x^{10} + 4x^{7} + 3x^{4}  x^{10} + 4x^{7} + 6x^{4} + 2x
7 x^{12} x^{12} + 5x^{9} + 4x^{6} x^{12} + 5x^{9} + 10x^{6} + 6x^{3} + 1 x^{12} + 5x^{9} + 10x^{6} + 7x^{3} + 1
8 x^{14} x^{14} + 6x^{11} + 5x^{8} x^{14} + 6x^{11} + 15x^{8} + 12x^{5} + 3x^{2} x^{14} + 6x^{11} + 15x^{8} + 16x^{5} + 6x^{2}
Table 3: Polynomials T_{n}^{(s)}(x), for 1 ⩽ n ⩽ 8 and 0 ⩽ s ⩽ 3.

Some special cases of (7) are

T_{n}^{(0)}(x) = x^{2n−2}, (n ≥ 1);  (8)
T_{n}^{(1)}(x) = x^{2n−2} + (n − 2)x^{2n−5} + (n − 3)x^{2n−8}, (n ≥ 3);  (9)
T (⌊\frac{n−1}{2}⌋)_{n}(x) = T_{n}(x), (n ≥ 1);  (10)
T (⌊\frac{n−3}{2}⌋)_{n}(x) = \begin{cases}T_{n}(x) − (\frac{n}{2}x^{\frac{n+2}{2}} + \frac{n−2}{2}x^{\frac{n−4}{2}} ) , & if n ≥ 3 and even; \\ T_{n}(x) − x^{\frac{n−1}{2}} , & if n ≥ 3 and odd. \end{cases}  (11)

Throughout the rest of this section, we describe some recurrence properties of the polynomials T_{n}^{(s)}(x) and numbers t_{n}(s).

Proposition 2. The non-linear recurrence relation of the incomplete tribonacci polynomials T_{n}^{(s)}(x) is

T_{n+3}^{(s+1)}(x) = x^{2}T_{n+2}^{(s+1)}(x) + xT_{n+1}^{(s)}(x) + T_{n}^{(s)}(x), 0 ⩽ s ⩽ ⌊\frac{n − 1}{2}⌋ .  (12)

The relation (12) can be transformed into the non-homogeneous recurrence relation

T_{n+3}^{(s)}(x) = x^{2}T_{n+2}^{(s)}(x) + xT_{n+1}^{(s)}(x) + T_{n}^{(s)}(x) − (xB(n − s, s)(x) + B(n − 1 − s, s)(x)) . (13)

Proof. We use the Definition 1 to rewrite the right-hand side of (12) as

x^{2} ∑_{i=0}^{s+1} B(n + 1 − i, i)(x) + x ∑_{i=0}^{s} B(n − i, i)(x) + ∑_{i=0}^{s} B(n − 1 − i, i)(x).

n/s 0 1 2  3  4  5  6  7
1 1
2 1
3 1 2
4 1 4
5 1 6 7
6 1 8 13
7 1 10 23 24
8 1 12 37 44
9 1 14 55 80  81
10 1 16 77 140 149
11 1 18 103 232 273 274
12 1 20 133 364 493 504
13 1 22 167 544 865 926 927
14 1 24 205 780 1461 1692 1705
15 1 26 247 1080 2369 3050 3135 3136
16 1 28 293 1452 3693 5376 5753 5768

Table 4: Numbers t_{n}(s), for 1 ⩽ n ⩽ 16 and 0 ⩽ s ⩽ 7.

Therefore,

x^{2}T_{n+2}^{(s+1)}(x) + xT_{n+1}^{(s)}(x) + T_{n}^{(s)}(x) =

= x^{2} ∑_{i=0}^{s+1} B(n + 1 − i, i)(x) + x ∑_{i=1}^{s+1} B(n − i + 1, i − 1)(x) + ∑_{i=1}^{s+1} B(n − i, i − 1)(x)

= ∑_{i=0}^{s+1} (x^{2}B(n + 1 − i, i)(x) + xB(n − i + 1, i − 1)(x) + B(n − i, i − 1)(x))

− xB(n + 1, −1)(x) − B(n, −1)(x)

= ∑_{i=0}^{s+1} B(n + 2 − i, i)(x) = T_{n+3}^{(s+1)}(x).

Corollary 3. The non-linear recurrence relation of the incomplete tribonacci numbers t_{n}(s) is

t_{n+3}(s + 1) = t_{n+2}(s + 1) + t_{n+1}(s) + t_{n}(s), 0 ⩽ s ⩽ ⌊\frac{n − 1}{2}⌋ .  (14)

The relation (14) can be transformed into the non-homogeneous recurrence relation

t_{n+3}(s) = t_{n+2}(s) + t_{n+1}(s) + t_{n}(s) − (B(n − s, s) + B(n − 1 − s, s)) .  (15)

Corollary 4. For n ≥ 2s + 2,

∑_{i=0}^{h−1} t_{n+i}(s) = \frac{1}{2} (t_{n+h+2}(s + 1) − t_{n+2}(s + 1) + t_{n}(s) − t_{n+h}(s)) .  (16)

Proof. We proceed by induction on h. The sum (16) clearly holds for h = 1; see (14). Now suppose that the result is true for all i < h. We prove it for h:

∑_{i=0}^{h} t_{n+i}(s) = ∑_{i=0}^{h−1} t_{n+i}(s) + t_{n+h}(s)

= \frac{1}{2} (t_{n+h+2}(s + 1) − t_{n+2}(s + 1) + t_{n}(s) − t_{n+h}(s)) + t_{n+h}(s)

= \frac{1}{2} (t_{n+h+2}(s + 1) − t_{n+2}(s + 1) + t_{n}(s) + t_{n+h}(s))

= \frac{1}{2} (t_{n+h+3}(s + 1) − t_{n+2}(s + 1) + t_{n}(s) − t_{n+h+1}(s)) .

The following proposition shows the sum of the n-th row of the Table 3. Proposition 5. The following equality holds:

∑_{s=0}^{l} T_{n}^{(s)}(x) = (l + 1)T_{n}(x) − ∑_{i=0}^{l} ∑_{j=0}^{i} i\binom{i}{j}\binom{n − i − j − 1}{i}x^{2n−3(i+j)−2},  (17)

where l = ⌊\frac{n−1}{2}⌋. Proof. We know

T_{n}^{(s)}(x) = ∑_{i=0}^{s} B(n − 1 − i, i)(x).

Then, we have

∑_{s=0}^{l} T_{n}^{(s)}(x) = B(n − 1 − 0, 0)(x) + (B(n − 1 − 0, 0)(x) + B(n − 1 − 1, 1)(x)) + ⋯

+ (B(n − 1 − 0, 0)(x) + B(n − 1 − 1, 1)(x) + ⋯ + B(n − 1 − l, l)(x))

= (l + 1)B(n − 1 − 0, 0)(x) + lB(n − 1 − 1, 1)(x) + ⋯ + B(n − 1 − l, l)(x).

Hence

∑_{s=0}^{l} T_{n}^{(s)}(x) = ∑_{i=0}^{l}(l + 1 − i)B(n − i − 1, i)(x)

= ∑_{i=0}^{l}(l + 1)B(n − i − 1, i)(x) − ∑_{i=0}^{l} iB(n − i − 1, i)(x)

= (l + 1)T_{n}(x) − ∑_{i=0}^{l} iB(n − i − 1, i)(x).

The following corollary shows the sum of the n-th row of the Table 4. It is obtained from (17) with x = 1.

Corollary 6. The following equality holds:

∑_{s=0}^{l} t_{n}(s) = (l + 1)t_{n} − ∑_{i=0}^{l} ∑_{j=0}^{i} i\binom{i}{j}\binom{n − i − j − 1}{i},

where l = ⌊\frac{n−1}{2}⌋.

Let

a_{n} = \begin{matrix}⌊ & \frac{n−1}{2} & ⌋ \\ ∑ \\ i=0 \end{matrix} ∑_{j=0}^{i} i\binom{i}{j}\binom{n − i − j − 1}{i}.

We conjecture that the generating function of (a_{n})_{n=0}^{∞} is

\frac{z^{3} + z^{4}}{(1 − z − z^{2} − z^{3})^{2}} = (z^{3} + z^{4})T^{2}(z),

where T (z) is the generating function of the tribonacci numbers and T^{2}(z) is the generating function of the convolution of tribonacci sequence; see sequence \underline{A073778}. If this conjecture is true, then

\begin{matrix}⌊ & \frac{n−1}{2} & ⌋ \\ ∑ \\ s=0 \end{matrix} t_{n}(s) = (l + 1)t_{n} − ∑_{j=0}^{n}(t_{j−2} + t_{j−3})t_{n−j+1}.

3 Generating function of the incomplete tribonacci numbers

In this section, we give the generating functions of the incomplete tribonacci numbers.

Lemma 7. Let (s_{n})_{n=0}^{∞} be a complex sequence satisfying the following non-homogeneous and non-linear recurrence relation

s_{n} = s_{n−1} + s_{n−2} + s_{n−3} + r_{n}, (n > 2),  (18)

where (r_{n})_{n=0}^{∞} is a given complex sequence. Then the generating function U(t) of the sequence (s_{n})_{n=0}^{∞} is

U(t) = \frac{G(t) + s_{0} − r_{0} + (s_{1} − s_{0} − r_{1})t + (s_{2} − s_{1} − s_{0} − r_{2})t^{2}}{1 − t − t^{2} − t^{3}}  (19)

where G(t) denotes the generating function of (r_{n})_{n=0}^{∞}.

Proof. Since U(t) and G(t) are the generating functions of (s_{n})_{n=0}^{∞} and (r_{n})_{n=0}^{∞}, respectively. Their power series representations are

U(t) = s_{0} + s_{1}t + s_{2}t^{2} + ⋯ + s_{k}t^{k} + ⋯,
G(t) = r_{0} + r_{1}t + r_{2}t^{2} + ⋯ + r_{k}t^{k} + · · · .

Note that,

tU(t) = s_{0}t + s_{1}t^{2} + s_{2}t^{3} + ⋯ + s_{k}t^{k+1} + ⋯,
t^{2}U(t) = s_{0}t^{2} + s_{1}t^{3} + s_{2}t^{4} + ⋯ + s_{k}t^{k+2} + ⋯,
t^{3}U(t) = s_{0}t^{3} + s_{1}t^{4} + s_{2}t^{5} + ⋯ + s_{k}t^{k+3} + · · · .

Therefore

(1 − t − t^{2} − t^{3})U(t) − G(t) = (s_{0} − r_{0}) + (s_{1} − s_{0} − r_{1})t + (s_{2} − s_{1} − s_{0} − r_{2})t^{2}.

Then the Equation (19) follows.

Theorem 8. Let Q_{s}(z) be the generating function of the incomplete tribonacci numbers t_{n}(s). Then

Q_{s}(z) = \frac{t_{2s+1} + (t_{2s+2} − t_{2s+1})z + (t_{2s+3} − t_{2s+2} − t_{2s+1} − 2)z^{2} − (z^{2} + z^{3})\frac{(1 + z)^{s}}{(1 − z)^{s+1}}}{1 − z − z^{2} − z^{3}} .

Proof. Let s be a fixed positive integer and t_{n}(s) the n-th incomplete tribonacci number. Since Q_{s}(z) is the generating function of the t_{n}(s), we have Q_{s}(z) = ∑_{i=0}^{∞} t_{i}(s)z^{i}.

From (7) with x = 1 and (15), we get t_{n}(s) = 0 for 0 ≤ n < 2s + 1, t_{2s+1}(s) = t_{2s+1}, t_{2s+2}(s) = t_{2s+2} and t_{2s+3}(s) = t_{2s+3} − 1, and that

t_{n}(s) = t_{n−1}(s) + t_{n−2}(s) + t_{n−3}(s) − (B(n − 3 − s, s) + B(n − 4 − s, s)) .  (20)

Now let

s_{0} = t_{2s+1}(s), s_{1} = t_{2s+2}(s), s_{2} = t_{2s+3}(s) and

s_{n} = t_{n+2s+1}(s).

Also let r_{0} = r_{1} = 0, r_{1} = 1 and

r_{n} = B(n + s − 2, s) + B(n + s − 3, s)

= ∑_{j=0}^{s} \binom{s}{j}\binom{n + s − 2 − j}{s} + ∑_{j=0}^{s} \binom{s}{j}\binom{n + s − 3 − j}{s}.

The generating function of the sequence (r_{n})_{n≥0} is computed using the methods expounded in [23, page 127]. Hence the generating function is equal to

(z^{2} + z^{3})\frac{(1 + z)^{s}}{(1 − z)^{s+1}} .

Thus, from Lemma 7, we get the generating function Q_{s}(z) of sequence (t_{n}(s))_{n=0}^{∞}.

Example 9. The generating functions of the incomplete tribonacci numbers for s = 1, 2, 3, 4 are

Q_{1}(z) = \frac{2}{(z − 1)^{2}} = 2 + 4z + 6z^{2} + 8z^{3} + 10z^{4} + 12z^{5} + 14z^{6} + 16z^{7} + 18z^{8} + ⋯

Q_{2}(z) = \frac{−5z^{2} + 8z − 7}{(z − 1)^{3}} = 7 + 13z + 23z^{2} + 37z^{3} + 55z^{4} + 77z^{5} + 103z^{6} + 133z^{7} + ⋯

Q_{3}(z) = \frac{−12z^{3} + 48z^{2} − 52z + 24}{(z − 1)^{4}} = 24 + 44z + 80z^{2} + 140z^{3} + 232z^{4} + 364z^{5} + ⋯

Q_{4}(z) = \frac{−45z^{4} + 192z^{3} − 338z^{2} + 256z − 81}{(z − 1)^{5}} = 81 + 149z + 273z^{2} + 493z^{3} + 865z^{4} + ⋯

An open problem is to find the generating function of the incomplete tribonacci polynomials.

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2010 Mathematics Subject Classification: Primary 11B39; Secondary 11B83, 05A15. Keywords: incomplete tribonacci number, incomplete tribonacci polynomial, tribonacci number.

(Concerned with sequences \underline{A000045}, \underline{A000032}, \underline{A000073}, and \underline{A073778}.)

Received November 7 2013; revised versions received December 10 2013; January 22 2014. Published in Journal of Integer Sequences, February 16 2014.

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