Some Formulas for Numbers of Restricted Words

Milan Janjić Department of Mathematics and Informatics University of Banja Luka Banja Luka, 78000 Republic of Srpska, Bosnia and Herzegovina

                    agnus@blic.net

Abstract

For an arithmetic function f_{0}, we consider the number c_{m}(n, k) of weighted compositions of n into k parts, where the weights are the values of the (m − 1)^{th} invert transform of f_{0}. We connect c_{m}(n, k) with c_{1}(n, k) via Pascal matrices. We then relate c_{m}(n, k) to the number of certain restricted words over a finite alphabet. In addition, we develop a method which transfers some properties of restricted words over a finite alphabet to words over a larger alphabet.

Several examples illustrate our findings. Some examples concern binomial coefficients and Fibonacci numbers. Some examples also extend the classical results about weighted compositions of Hoggatt and Lind. In each example, we derive an explicit formula for c_{m}(n, k).

1 Introduction

For a given initial arithmetic function f_{0}, Janjić [5] examined some properties of the function f_{m}, which is the m^{th} invert transform of f_{0}. In that paper, as well as in Birmajer et al. [2], some cases in which f_{m} counts the number of restricted words over a finite alphabet were considered. In the present paper, we consider the function c_{m}(n, k), which is the number of weighted compositions of n into k parts, where the weights are {f_{m−1}(1), f_{m−1}(2), …}. Note that, in Janjić [6], properties of c_{1}(n, k) were investigated.

As mentioned by Birmajer at al. in a recent preprint [3], a formula for the number c_{1}(n, k) was firstly given by Hoggatt and Lind [4], based on results by Moser and Whitney [7]. For a sequence of weights {f_{m−1}(1), f_{m−1}(2), …}, the number of f_{m−1}-weighted compositions of n into k parts is

c_{m}(n, k) = ∑_{π_{k}(n)} \frac{k!}{k_{1}! ⋯ k_{n}!}f_{m−1}(1)^{k_{1}} ⋯ f_{m−1}(1)^{k_{n}},

where the sum runs over all solutions of

k_{1} + 2k_{2} + ⋯ + nk_{n} = n such that k_{1} + ⋯ + k_{n} = k and k_{j} ∈ ℕ_{0} for all j.

In other words,

c_{m}(n, k) = \frac{k!}{n!}B_{n,k}(1!f_{m−1}(1), 2!f_{m−1}(2), …),  (1)

where B_{n,k}(x_{1}, x_{2}, …) is the Bell partial polynomial. Moreover, as discussed in Hoggatt and Lind [4], this formula is equivalent to

c_{m}(n, k) = ∑_{γ_{k}(n)} f_{m−1}(a_{1}) ⋯ f_{m−1}(a_{k}),  (2)

where γ_{k}(n) indicates summations over k-part compositions a_{1} + a_{2} + ⋯ + a_{k} of n. As an immediate consequence, we obtain the following two results:

Corollary 1. The function c_{m}(n, k) satisfies the recurrence

c_{m}(0, 0) = 1, c_{m}(n, 0) = 0, (n > 1),

and

c_{m}(n, k) = ∑_{i=1}^{n−k+1} f_{m−1}(i)c_{m}(n − i, k − 1), (1 ≤ k ≤ n).  (3)

Corollary 2. The following formula holds:

f_{m}(n) = ∑_{k=1}^{n} c_{m}(n, k).  (4)

Note that, throughout the paper, letters m, n, k will have the meaning as in the definition of c_{m}(n, k). Using Birmajer et al. [1, Corollary 10], we derive a formula connecting c_{m}(n, k) with c_{m−1}(n, k). The formula may be written in terms of the lower triangular Pascal matrices. We then extend this result to obtain a relation between c_{m}(n, k) and c_{1}(n, k).

For the particular case f_{0}(1) = 1, we develop a method which allows us to derive an interpretation of c_{m}(n, k) in terms of restricted words, when we know the number of restricted words counted by f_{m−1}. We finish the paper with a number of examples illustrating our results. Some examples extend the classical results on weighted compositions given by Hoggatt and Lind [4].

It is important to note that quantities f_{m}(n) and c_{m}(n, k) depend only on the initial arithmetic function f_{0}.

2 A connection of c_{m}(n, k) and c_{m−1}(n, k)

Let C_{m}(n) be the lower triangular matrix of order n, whose (i, j) entry is c_{m}(i, j), (i = 1, 2, …, n; 1 ≤ j ≤ i). We let L_{n} denote the lower triangular Pascal matrix of order n. Hence, the (i, j) entry of L_{n} is \binom{i−1}{j−1}, (1 ≤ j ≤ i). First, we prove the following:

Proposition 3. For each m > 1, we have

C_{m}(n) = C_{m−1}(n) · L_{n}.

Proof. It is easy to see that the statement is equivalent to the equation

c_{m}(n, k) = ∑_{i=k}^{n} \binom{i − 1}{k − 1}c_{m−1}(n, i).  (5)

In our terminology, Birmajer et al. [1, Corollary 10] may be written in the form

∑_{j_{1}+j_{2}+···+j_{k}=n} f_{m−1}(j_{1}) ⋯ f_{m−1}(j_{k}) = ∑_{i=1}^{n} \binom{i + k − 1}{i}c_{m−1}(n, i),

where the sum is taken over nonnegative j_{1}, …, j_{k}. Since at most k − 1 of j_{t} may be equal 0, (2) yields

∑_{j=0}^{k−1} \binom{k}{j}c_{m}(n, k − j) = ∑_{i=1}^{n} \binom{i + k − 1}{i}c_{m−1}(n, i).

Replacing k − j by t, and denoting ∑_{i=1}^{n} \binom{i+k−1}{i}c_{m−1}(n, i) = a_{k}, implies

∑_{t=1}^{k} \binom{k}{t}c_{m}(n, t) = a_{k}, (k = 1, 2, …, n).

Denoting X = (c_{m}(n, 1), c_{m}(n, 2), …, c_{m}(n, n))^{T}, and A = (a_{1}, a_{2}, …, a_{n})^{T}, this system may be written in the matrix form

Q · X = A,

where Q is obtained from the Pascal matrix L_{n+1} by omitting the first row and the first column. It follows that X = Q^{−1} · A, where Q^{−1} = ((−1)^{i+j}\binom{i}{j})_{n×n} . For k = 1, 2, …, n, we obtain

c_{m}(n, k) = ∑_{i=1}^{n} (∑_{j=1}^{n}(−1)^{j+k}\binom{k}{j}\binom{i + j − 1}{i}) c_{m−1}(n, i).  (6)

Equation (6) holds for each m > 1, as well as for any arbitrary arithmetic function f_{0}. In particular, taking f_{0}(1) = 1, f_{0}(i) = 0, (i > 1), we obviously have c_{1}(n, n) = 1, and c_{1}(n, k) = 0 for k < n. Also, f_{1}(n) = 1 for all n. In this case, c_{2}(n, k) is the number of compositions of n into k parts, that is, c_{2}(n, k) = \binom{n−1}{k−1}. Therefore, (6) becomes

\binom{n − 1}{k − 1} = ∑_{j=1}^{n}(−1)^{j+k}\binom{k}{j}\binom{n + j − 1}{n}.  (7)

Hence the expression in the square brackets in (6) equals \binom{i−1}{k−1}, which proves (5). Remark 4. As a byproduct, we proved the binomial identity (7). Remark 5. Replacing i − k by t in (5), we obtain

c_{m}(n, k) = ∑_{t=0}^{n−k} \binom{k + t − 1}{t}c_{m−1}(n, k + t).  (8)

From the equation C_{m}(n) = C_{m−1}(n) · L_{n} follows

C_{m}(n) = C_{m−1}(n) · L_{n} = C_{m−2}(n) · L_{n}^{2} = ⋯ = C_{1}(n) · L_{n}^{m−1}.

We thus obtain, Proposition 6. The following matrix equation holds:

C_{m}(n) = C_{1}(n)L_{n}^{m−1}.

Or, explicitly,

c_{m}(n, k) = ∑_{i=k}^{n}(m − 1)^{i−k}\binom{i − 1}{k − 1}c_{1}(n, i), (1 ≤ k ≤ n).  (9)

Proof. The assertion is true since (i, k) entry of L_{n}^{m−1} is (m − 1)^{i−k}\binom{i−1}{k−1}.

Now, we derive a formula in which f_{m}(n) is expressed in terms of c_{1}(n, k). Proposition 7. The following formula holds:

f_{m}(n) = ∑_{i=1}^{n} m^{i−1}c_{1}(n, i).  (10)

Proof. Equation (4) yields

f_{m}(n) = ∑_{k=1}^{n} ∑_{i=k}^{n}(m − 1)^{i−k}\binom{i − 1}{k − 1}c_{1}(n, i).

Changing the order of summation gives

f_{m}(n) = ∑_{i=1}^{n} (∑_{k=1}^{i}(m − 1)^{i−k}\binom{i − 1}{k − 1}) c_{1}(n, i).

Using the binomial theorem, we obtain (10).

Note 8. Equation (10) appears in Birmajer et al. [2] with a combinatorial proof based on the enumeration of certain restricted words.

As an immediate consequence of (1) and (9), we obtain the following identity for the partial Bell polynomials:

Identity 9. If the sequence y_{1}, y_{2}, … is the invert transform of the sequence x_{1}, x_{2}, …, then

k!B_{n,k}(y_{1}, 2! · y_{2}, 3! · y_{3}, …) = ∑_{i=k}^{n} \binom{i − 1}{k − 1}i!B_{n,i}(x_{1}, 2! · x_{2}, 3! · x_{3}, …).

We next prove the following result.

Proposition 10. Assume that f_{0}(1) = 1 and m > 1. Assume next that, for n ≥ 1, f_{m−1}(n) is the number of some words of length n − 1 over a finite alphabet α. Let x be a letter which is not in α. Then c_{m}(n, k) is the number of words of length n − 1 over the alphabet α ∪ {x}, in which exactly k − 1 letters equal x.

Proof. Since f_{0}(1) = 1, it follows from Janjić [5, Corollary 2] that f_{m−1}(1) = 1. We use induction on k. For k = 1, (2) yields c_{m}(n, 1) = f_{m−1}(n). Since f_{m−1}(n) is the number of words of length n − 1 over α not containing x, we conclude that the statement is true for k = 1. Assume that the claim is true for k−1. Consider the first term f_{m−1}(1)c_{m}(n−1, k−1) in (3). By the induction hypothesis, c_{m}(n − 1, k − 1) is the number of words of length n − 2 having k − 2 letters equal to x. Adding x at the beginning of each such word, we obtain all the words of length n − 1 over α ∪ {x}, having k − 1 letters equal to x, and all beginning with x.

Consider now the term f_{m−1}(i)·c_{m}(n−i, k−1), (i > 1) in (3). By the induction hypothesis, c_{m}(n − i, k − 1) is the number of words of length n − i − 1 with k − 2 letters equal to x. We first insert x at the beginning of each such word. In front of x, we insert an arbitrary word of length i−1 over α, which are f_{m−1}(i) in number. We thus obtain all words of length n−1 over α ∪ {x}, such that the first appearance of x is at the ith position. It follows that the right-hand side of (3) counts all the desired words.

Remark 11. We stress the fact that the preceding method may be applied only when we know the number of (n − 1)-length words counted by f_{m−1}(n). This is always true when f_{0}(1), f_{0}(2), … is a binary sequence. Namely, there is a bijection between the compositions counted by c_{1}(n, k) and the binary words of length n − 1 with k − 1 ones. This bijection is given by the correspondence

1 → 1, 2 → 10, 3 → 100, ….  (11)

In this way, the compositions of n into k parts designate binary words of length n and having k ones, all of which begin with 1. The converse is also true. Omitting the leading 1, we obtain the desired correspondence. Equation (4) implies that f_{1} and c_{1}(n, k) both count some binary words of length n − 1. We thus may apply Proposition 10 to obtain a combinatorial interpretation of c_{m}(n, k) as well as f_{m}(n). Note that this method may be applied in non-binary cases also.

Remark 12.

1. Janjić [5, 6] derive several formulas concerning correspondence (11).

2. Also, Birmajer at al. [2, Theorem 3, Corollary 5] developed a method to obtain the words counted by f_{m−1} starting with the initial function f_{0}.

We now illustrate our method by a simple example.

Example 13. Assume that f_{0}(i) = 1 for i = 1, 2, . . .. Then, we have f_{m−1}(n) = m^{n−1}. It yields that f_{m−1}(n) is the number of all words of length n − 1 over α = {0, 1, …, m − 1}. Then, c_{1}(n, k) = \binom{n−1}{k−1}. This means that C_{1}(n) = L_{n}. It follows that C_{m}(n) = L_{n}^{m}. From the well-known formula for the terms of L_{n}^{m}, we obtain

c_{m}(n, k) = m^{n−k}\binom{n − 1}{k − 1}.  (12)

Equation (12) is in accordance with Proposition 10. Namely, according to Proposition 10, c_{m}(n, k) is the number of words of length n − 1 over {0, 1, …, m} with k − 1 letters equal to m. These k − 1 letters may be chosen in \binom{n−1}{k−1} ways. The remaining letters may be arbitrary letters from {0, 1, …, m − 1}, which are m^{n−k} in number.

As a byproduct, using (9), we obtain the following binomial identity:

Identity 14. For m > 1, we have

m^{n−k}\binom{n − 1}{k − 1} = ∑_{j=0}^{n−k}(m − 1)^{j}\binom{n − 1}{k + j − 1}\binom{k + j − 1}{j}.  (13)

This simple case is related to the coefficients of the Tchebychev polynomials U_{n}(x) of the second kind.

Corollary 15. The number |[x^{n−k}](U_{n+k−2}(x))| is the number of words of length n − 1 over the alphabet {0, 1, 2} having k − 1 twos.

Proof. It is a well known that (−1)^{k}2^{n−k}\binom{n−1}{k−1} is the coefficient of U_{n+k−2}(x) by x^{n−k}. We thus obtain

[x^{n−k}](U_{n+k−2}(x)) = (−1)^{n−k} ∑_{j=0}^{n−k} \binom{n − 1}{k + j − 1}\binom{k + j − 1}{j}.

This is the case when m = 2 in (13).

3 More examples

Firstly, we revise the result from Janjić [6, Corollary 9].

Example 16. We define f_{0}(1) = f_{0}(2) = 1, and f_{0}(n) = 0 otherwise. According to Janjić [5, Corollary 33], f_{m−1}(n) is the number of words of length n−1 over the alphabet {0, 1, …, m− 1} having all zeros isolated.

Corollary 17. The number c_{m}(n, k) is the number of words of length n−1 over the alphabet {0, 1, …, m}, which have k − 1 letters equal to m and all zeros isolated. Also,

c_{1}(n, k) = \binom{k}{n − k},

and

c_{m}(n, k) = ∑_{j=⌈ \frac{n}{2} ⌉−k}^{n−k}(m − 1)^{j}\binom{j + k − 1}{k − 1}\binom{j + k}{n − j − k}, (m > 1, ⌈\frac{n}{2}⌉ ≥ k) .

Proof. The first formula is Hoggatt and Lind [4, Case (ii)]. Since c_{1}(n, k +j) = \binom{k+j}{n−k−j}, (j = 0, …, n − k), we have k + j ≥ n − k − j, which yields 2j ≥ n − 2k and n ≥ 2k. The second formula is true according to (9).

The arrays \underline{A030528} and \underline{A154929} in Sloane [8] are related to Example 16.

Next, we reexamine the result in Janjić [5, Corollary 28].

Example 18. We define f_{0}(n) = 1 when n is odd, and f_{0}(n) = 0 otherwise. According to Janjić [5, Corollary 28], f_{m−1}(n) is the number of words of length n − 1 over the alphabet {0, 1, …, m − 1}, avoiding runs of zeros of odd lengths. From Janjić [6, Proposition 24], it follows that

c_{1}(n, k) = \begin{cases}\binom{\frac{n−k}{2} +k−1}{k−1}, & if n − k is even; \\ 0, & if n is odd. \end{cases}

The number c_{1}(n, k) is the number of binary words of length n − 1 with k − 1 ones, and avoiding runs of zeros of odd lengths. This follows from bijection (11). We add a short combinatorial proof.

Proposition 19. The number c_{1}(n, k) is the number of binary words of length n − 1 with k − 1 ones, avoiding runs of zeros of odd lengths.

Proof. Assume that n and k are of different parities. Since a word of length n − 1 with k − 1 ones must have n − k zeros, and since n − k is odd, we conclude that such a word must have an odd run of zeros. It follows that c_{1}(n, k) = 0. If n and k are of the same parity, then n − k is even. This means that there are \frac{n−k}{2} pairs of zeros. Of these \frac{n−k}{2} pairs and k − 1 ones, we may form \binom{\frac{n−k}{2} +k−1}{k−1} words of length n − 1 having k − 1 ones and avoiding runs of zeros of odd lengths.

Remark 20. The formula for c_{1}(n, k) appears in Hoggatt and Lind [4, Case (iv)]. Using induction and Proposition 10, we obtain

Corollary 21. The number c_{m}(n, k) is the number of words of length n−1 over {0, 1, …, m} with k − 1 letters equal to m, avoiding runs of zeros of odd lengths.

From (9), we obtain an explicit formula for c_{m}(n, k).

The arrays \underline{A037027} and \underline{A054456} in Sloane [8] are related to Example 18.

Example 22. We define f_{0}(i) = i, (i = 1, 2, …). According to Janjić [5, Corollary 37], f_{m−1}(n) is the number of 01-avoiding words of length n − 1 over the alphabet {0, 1, …, m}.

Applying Proposition 10 several times, we obtain

Corollary 23. The number c_{m}(n, k) is the number of words of length n−1 over {0, 1, …, m+ 1} having k − 1 letters equal to m + 1 and avoiding 01.

Corollary 24. The following formula holds:

c_{1}(n, k) = \binom{n + k − 1}{2k − 1}.  (14)

Also,

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

Proof. Formula (14) may be found in Hoggatt and Lind [4, Case (iii)].

Since \binom{n+k−1}{2k−1} is obviously the number of binary words of length n + k − 1 with 2k − 1 zeros, we obtain the following Euler-type identity:

Identity 25. The number of binary words of length n + k − 1 with 2k − 1 zeros equals the number of ternary words of length n − 1, having k − 1 letters equal to 2 and avoiding 01.

The arrays \underline{A125662}, \underline{A207823}, and \underline{A207824} in Sloane [8] are related to Example 22.

The last two examples concern the case f_{0}(1) = 0. Note that in these examples, Proposition 10 can not be used. The first example is an extension of the result from Janjić [6, Proposition 13].

Example 26. We define f_{0}(1) = 0, and f_{0}(n) = 1 otherwise. It follows from Janjić [5, Corollary 24] that, for n > 3, f_{m}(n) is the number of words of length n−3 over {0, 1, …, m}, where no two consecutive letters are nonzero. From Janjić [6, Proposition 13], we obtain c_{1}(n, k) = \binom{n−k−1}{k−1} for (1 ≤ k ≤ ⌊\frac{n}{2}⌋), and c_{1}(n, k) = 0 otherwise. Equation (5) implies that c_{m}(n, k) = 0 when k > ⌊\frac{n}{2}⌋. Remark 27. Note that the formula for c_{1}(n, k) also appears in Hoggatt and Lind [4, Case (iii)].

Corollary 28. For n > 3 and 1 ≤ k ≤ ⌊\frac{n}{2}⌋, the number c_{m}(n, k) is the number of words of length n − 3 over {0, 1, …, m} with k − 1 ones, and all nonzero letters are isolated. An explicit formula for c_{m}(n, k) is

c_{m}(n, k) = ∑_{j=0}^{⌊\frac{n}{2}⌋−k}(m − 1)^{j}\binom{j + k − 1}{k − 1}\binom{n − k − j − 1}{k + j − 1}, (1 ≤ k ≤ ⌊\frac{n}{2}⌋) ,

and c_{m}(n, k) = 0 when k > ⌊\frac{n}{2}⌋.

Proof. We know that c_{1}(n, k) is the number of compositions of n into k parts, each of which is greater than 1. Using the bijection (11), we conclude that, for n > 3, c_{1}(n, k) is the number of binary words of length n beginning with 10, ending with 0 and all ones are isolated. Omitting 10 at the beginning, and 0 at the end of each word, we conclude that c_{1}(n, k), (n > 3) is the number of binary words of length n − 3 with k − 1 ones, all of which are isolated. Hence, the statement is true for m = 1. Assume that the statement is true for m − 1. In (8), by the induction hypothesis, c_{m−1}(n, k + t) is the number of words of length n − 3 with k + t − 1 ones, in which all nonzero letters are isolated. Replacing t ones with m, we obtain the desired words. The number t may be chosen in \binom{k+t−1}{t} ways. Hence, the right-hand side of (8) counts all the desired words. The formula follows from (9) and the fact that n − k − j − 1 ≥ k + j − 1.

The arrays \underline{A037027}, \underline{A249139}, and \underline{A006130} in Sloane [8] are related to Example 26.

Example 29. Define f_{0} in the following way: f_{0}(2) = f_{0}(3) = 1, and f_{0}(n) = 0 otherwise. Janjić [6, Proposition 5] proved that c_{1}(n, k) = \binom{k}{n−2k}, (⌈\frac{n}{3}⌉ ≤ k ≤ ⌊\frac{n}{2}⌋) and c_{1}(n, k) = 0 otherwise.

We know that c_{1}(n, k) is the number of compositions of n into k parts equal to either 2 or 3. In other words, for n ≥ 3 and 1 ≤ k ≤ ⌊\frac{n}{2}⌋, c_{1}(n, k) is the number of binary words of length n − 1 with k − 1 ones. These words begin with 0 and end with 0. Also, zero avoids a run of length greater than 2, and all ones are isolated.

Corollary 30. For n > 3, the number c_{m}(n, k) is the number of words of length n − 1 over the alphabet {0, 1, …, m} with k − 1 ones, which begin and end with 0. Also, 0 avoids a run of length greater than 2 and all nonzero letters are isolated.

Proof. The statement holds for m = 1. Assume that the statement is true for m−1. Consider the term \binom{k+t−1}{t}c_{m−1}(n, k +t) in (8). The number c_{m−1}(n, k +t) is the number of the desired words of length n − 1 over {0, 1, …, m − 1} with k + t − 1 ones. We replace t of k + t − 1 ones with m and then sum over t to obtain c_{m}(n, k).

From (5) follows that c_{m}(n, k) = 0 if k > ⌊\frac{n}{2}⌋. Otherwise, from (9), we obtain

c_{m}(n, k) = ∑_{j=0}^{⌊ \frac{n}{2} ⌋}(m − 1)^{j}\binom{j + k − 1}{k − 1}\binom{k + j}{n − 2k − 2j}, (1 ≤ k ≤ ⌊\frac{n}{2}⌋) .

4 Acknowledgment

I would like to thank the referee for a very careful reading of the manuscript and many useful suggestions. A special thanks for providing me with historical facts about weighted compositions.

References

[1] D. Birmajer, J. B. Gill, and M. D. Weiner, Linear recurrence sequences and their convolutions via Bell polynomials, J. Integer Seq. 18 (2015), Article 15.1.2.

[2] D. Birmajer, J. B. Gill, and M. D. Weiner, On the enumeration of restricted words over a finite alphabet, J. Integer Seq. 19 (2016), Article 16.1.3.

[3] D. Birmajer, J. B. Gill, and M. D. Weiner, Composition colored by symplicial polytopic

   numbers, preprint 2006, http//arxiv.org/abs/1601.01595.

[4] V. E. Hoggatt and D. A. Lind, Fibonacci and binomial properties of weighted compositions, J. Combin. Theory 4 (1968), 121–124.

[5] M. Janjić, On linear recurrence equation arising from compositions of positive integers J. Integer Seq. 18 (2015), Article 15.4.7.

[6] M. Janjić, Binomial coefficients and enumeration of restricted words, J. Integer Seq. 19 (2016), Article 16.7.3.

[7] L. Moser and E. L. Whitney, Weighted compositions, Canad. Math. Bull. 4 (1961), 39–43.

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

2000 Mathematics Subject Classification: Primary 05A10; Secondary 11B39. Keywords: binary word, integer composition, restricted word, Pascal matrix.

(Concerned with sequences \underline{A000027}, \underline{A006130}, \underline{A030528}, \underline{A037027}, \underline{A054456}, \underline{A125662}, \underline{A154929}, \underline{A207823}, \underline{A207824}, and \underline{A249139}.)

Received October 3 2016; revised versions received March 26 2017; June 1 2017. Published in Journal of Integer Sequences, June 25 2017.

Return to Journal of Integer Sequences home page.