Generalized Binomial Expansions and Bernoulli Polynomials, INTEGERS: The Electronic

In this paper we investigate generalized binomial expansions that arise from two-dimensional sequences satisfying a broad generalization of the triangular recurrence for binomial coefficients. In particular, we present a new combinatorial formula for such sequences in terms of a ’shift by rank’ quasi-expansion based on ordered set partitions. As an application, we give a new proof of Dilcher’s formula for expressing generalized Bernoulli polynomials in terms of classical Bernoulli polynomials.


Introduction
The binomial expansion formula where n k denotes the binomial coefficients given by the formula is well known to all mathematicians.In this paper, we consider instead a generalized binomial expansion of two sequences a(k) and b(k) where the first few cases are given by the quasi-expansion that we shall refer to as 'shift by rank': Can the reader guess the correct expansion for the next case, (a(k) + b(k)) 4 , and more generally the formula for (a(k) + b(k)) n where n is any non-negative integer?Observe that if a(k) and b(k) are constant sequences, then the pattern reduces to that for binomial coefficients.
As we shall explain in this paper, the 'shift-by-rank' expansion above appears in a formula for twodimensional sequences x(n, k) satisfying the generalized triangular recurrence where a(n, k) and b(n, k) are arbitrary two-dimensional sequences.If a(n, k) and b(n, k) are constant and equal to one, then (4) reduces to the classic triangular recurrence satisfied by the binomial coefficients: This recurrence is commonly illustrated by arranging the binomial coefficients into the much-celebrated figure known as Pascal's triangle: Moreover, the recurrence (5) uniquely defines the binomial coefficients if we initialize the values on the boundary of Pascal's triangle to be 1, i.e.
since all other entries can then be generated by this recurrence.
Of course, the mathematical literature contains many extensions of (5) which are special cases of (4) (see [1], [2], and [5]).For example, the unsigned Stirling numbers of the first kind are defined by the recurrence and Stirling numbers of the second kind are defined by An even more interesting recurrence is one satisfied by the generalized Bernoulli polynomials We shall consider this recurrence further in Section 3. Our approach to studying x(n, k) relies on a different initialization of Pascal's triangle, which will appear quite natural if we realize it as an infinite rectangular array (also referred to as Pascal's matrix or square; see [3] and [6]), where infinitely many 0's are appended to both ends of each row as shown in (11): In this case, it suffices to initialize only those values along the first row (n = 0) as follows: where is Kronecker's delta.All other binomial coefficients in this array can then be generated from the triangular recurrence as before.
As an example, let x(n, k) be any two-dimensional sequence which satisfies the classical triangular recurrence, i.e.
x(n, k) but whose values along row n = 0 are prescribed by an arbitrary bi-directional sequence f (k), i.e.
It is straightforward to show that x(n, k) satisfies the following convolution formula: This demonstrates that binomial coefficients can be considered as fundamental building blocks for generating all two-dimensional sequences which satisfy the triangular recurrence.
In this paper we shall describe the fundamental building blocks for two-dimensional sequences x(n, k) satisfying the generalized triangular recurrence (4).This will require us to introduce the notion of rank for comparing the relative size of an integer with respect to a given subset.As a result, we obtain what seems to be a new combinatorial formula for x(n, k) (Theorem 1) involving ordered set partitions; an extensive search of the literature, including the standard reference [1], which mentions (4) but only discusses special cases, did not reveal any formula similar to ours.As an application, we give a different proof of Dilcher's formula ([1], Theorem 2), which expresses generalized Bernoulli polynomials in terms of classical Bernoulli polynomials.We conclude by presenting a generalization of our results to three-term recurrences.

Generalized Triangular Recurrences
We begin our study of generalized triangular recurrences by first considering the simple generalization where a and b are constants.Cadogan has proven in [2] that which shows that the coefficients in (18) can be generated by expanding the binomial (ax + by) n .As a next step, let us assume that the more general triangular recurrence holds: where a(n) and b(n) are one-dimensional sequences.In this case, the situation becomes much more interesting.To gain some intuition, let's write out x(n, k) explicitly for the first several values of n: Guided by the combinatorial definition of binomial coefficients as counting the number of subsets of a given set, we look to describe the coefficients of f in (20) by ordered partitions of {1, 2, 3}.
Definition 2.1.Let A n = {1, 2, ..., n}.We define P = (σ 1 , σ 2 ) to be a 2-block ordered partition of A n if σ 1 and σ 2 are disjoint sets whose union is A n .Morever, let P 2 (A n ) denote the set of all 2-block ordered partitions of A n .
Observe that in the definition above we allow either σ 1 or σ 2 to be empty (but not both), which differs from the standard definition of set partitions (see [8]).Also, the correspondence (σ 1 , σ 2 ) ↔ σ 2 gives a bijection between P 2 (A n ) and the power set of A n , i.e., the set of all subsets of A n , since σ 1 and σ 2 are complements of each other in A n .Thus, we can express any 2-block ordered partition as (σ, σ), where σ is a subset of A n .Moreover, P 2 (A n ) has cardinality 2 n .Definition 2.2.Given (σ, σ) ∈ P 2 (A n ) where the elements of σ and σ are denoted explicitly by {i 1 , i 2 , ..., i m } and {j 1 , j 2 , ..., j n−m }, respectively, we define the product It follows from this definition that x(n, k) can be described by the formula or equivalently, where A n (m) denotes the set of m-element subsets of A n .We shall prove a more general formula later on.
Let us now assume that x(n, k) satisfies the recurrence where a(k) and b(k) are again one-dimensional sequences as in ( 19), but this time dependent on the second index k instead of the first index n.How does this affect our formula for x(n, k)?Again, we write out x(n, k) explicitly for the first several values of n: The pattern in this case appears more complicated than the previous case due to shifts in certain indices; in fact, (25) is essentially the 'shift and rank' expansion (modulo f ) discussed at the beginning of this paper.To shed some light on this pattern, we first fix an ordering for how the factors a and b should appear in each coefficient: indices should be written left to right in ascending order and if two factors have the same index, then they are written left to right in reverse-alphabetical order.For example, the expansion in (25) is written using this ordering.
Next, we compare the shifts of each index with the corresponding positions of the factor a and b.In particular, if we denote by s j the amount of shift in the index of the j -th factor of a given coefficient of f , then what pattern does s j satisfy?The answer is that the shift in the index of a given factor depends on the number of factors involving b that are higher in position, i.e. further to the right.
We formalize this pattern by introducing a rank function to indicate the relative size of an integer with respect to a given subset.
We define the rank of a positive integer j relative to σ, and denote it by R σ (j), to be the number of elements in σ that are greater than j, i.e.R σ (j) = |{i ∈ σ : i > j}| For example, if σ = {1, 2, 4} ∈ A 5 (3), then R σ (3) = 1 since there is only one element in σ, namely 4, that is greater than 3.We also have R σ (5) = 0.In general, R σ (n) = 0 for every subset σ ∈ A n .Then given (σ, σ) ∈ P 2 (A n ), we define the product (analogous to (21)) As a result, we are now able to express x(n, k) by the shift-by-rank expansion formula This brings us to the last case where we assume that x(n, k) satisfies the most general recurrence given by (17) in terms of arbitrary two-dimensional sequences a(n, k) and b(n, k).Again, given (σ, σ) ∈ P 2 (A n ), we define the product By combining the results of the two previous cases, we obtain the following main theorem, which we prove rigorously.
Theorem 1: Suppose a two-dimensional sequence x(n, k) satisfies the generalized triangular recurrence where n is a non-negative integer and k ∈ Z. Then Proof.For n = 0, formula (29) reduces to x(0, k) = f (k).By induction on n, we have Next, we simplify the inner summation in each of double summations above as follows.First, assume σ = {i 1 , i 2 , ..., i m } ∈ A n (m) so that σ = {j 1 , j 2 , ..., j n−m } ∈ A n (n − m).Then observe that we can view σ as a subset of A n+1 (m); moreover, the complement of σ in A n+1 (m) equals σ ∪ {n + 1}.Set j n−m+1 = n + 1 and recall that R σ (n + 1) = 0.It follows that the inner summation of the first term can be rewritten as As for the second term, since R σ (j) = R σ (j) + 1 for σ ∈ A n (m), where σ = σ ∪ {n + 1} and 1 ≤ j < n + 1, we can simplify it as It follows that But the two summations above can be viewed as a partition of A n+1 into those subsets σ that contain the integer n + 1 and those subsets σ that do not.Then by re-indexing the second summation on the right hand side above and making the limits of summation the same for both, which involve inserting empty cases, we can combine both summations into one as follows: This completes the proof.

Application to Bernoulli Polynomials
As an application of Theorem 1, we prove Dilcher's formula ( [1]) for expressing generalized Bernoulli polynomials in terms of classical Bernoulli polynomials B n (z).The latter arises in many important applications involving special functions such as the Riemann zeta function.
Let p be a positive integer.
where z = z 1 + ... + z p , it follows from equating (30) and (31) that In [3], Nörlund proved that the generalized Bernoulli polynomials satisfy the generalized triangular recurrence for n > p ≥ 1.By applying Theorem 1 to this recurrence, we obtain the following corollary.
Corollary 3.1.The generalized Bernoulli polynomials are given in terms of the classical Bernoulli polynomials by Proof.Formula (34) follows immediately from (29) by setting The following lemma describes certain properties of the rank function will be useful in demonstrating that formula (34) reduces to Dilcher's formula, which we show afterwards.