INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 6 (2006), #A05

THE AKIYAMA-TANIGAWA ALGORITHM FOR CARLITZ’S q-BERNOULLI NUMBERS

Jiang Zeng

Institut Camille Jordan, Université Claude Bernard (Lyon I), 69622 Villeurbanne Cedex, France

                          zeng@math.univ-lyon1.fr

Received: 9/9/05, Accepted: 2/4/06, Published: 2/22/06

Abstract

We show that the Akiyama-Tanigawa algorithm and Chen’s variant for computing Bernoulli numbers can be generalized to Carlitz’s q-Bernoulli numbers. We also put these algorithms in the larger context of generalized Euler-Seidel matrices.

1. Introduction

Carlitz [?] introduced the q-Bernoulli numbers β_{n} (n ≥ 1) by the recurrence:

q(qβ + 1)^{n} − β_{n} = \{ \begin{matrix}1, & if n = 1; \\ 0, & if n > 1; \end{matrix}  (1)

where β_{0} = 1 and β_{k} = β^{k} after expansion. The first few values of β_{n} are

β_{0} = 1, β_{1} = −\frac{1}{[2]}, β_{2} = \frac{q}{[2][3]}, β_{3} = −\frac{(q − 1)q}{[3][4]} ,

where [n] = (1 − q^{n})/(1 − q) and [n]! = [1][2] … [n] for n ≥ 0. More generally we have the following explicit formula (see [?]):

(q − 1)^{n}β_{n} = ∑_{i=0}^{n}(−1)^{n−i}\binom{n}{i}\frac{i + 1}{[i + 1]}.

Recently, Akiyama and Tanigawa’s amazing algorithm for computing Bernoulli numbers [?] has attracted the attention of several authors [?, ?, ?]. One of our aims is to show that there is an analogue algorithm for Carlitz’s q-Bernoulli numbers as follows: start with the 0-th row 1, \frac{1}{[2]}, \frac{1}{[3]}, \frac{1}{[4]}, \frac{1}{[5]}, … and define the first row by

1 · (1 − \frac{1}{[2]}) , [2] · (\frac{1}{[2]} − \frac{1}{[3]}) , [3] · (\frac{1}{[3]} − \frac{1}{[4]}) , …

which produces the sequence \frac{q}{[2]}, \frac{q^{2}}{[3]}, \frac{q^{3}}{[4]}, . . .. Then define the next row by

1 · (\frac{q}{[2]} − \frac{q^{2}}{[3]}) , [2] · (\frac{q^{2}}{[3]} − \frac{q^{3}}{[4]}) , [3] · (\frac{q^{3}}{[4]} − \frac{q^{4}}{[5]}) , …

thus giving \frac{q}{[2][3]}, \frac{[2]q^{2}}{[3][4]}, \frac{[3]q^{3}}{[4][5]}, … as the second row. In general, denoting the m-th (m = 0, 1, 2, …) coefficient in the n-th (n = 0, 1, 2, …) row by a_{n,m}, then the following recurrence relation holds:

a_{n,m} = [m + 1] · (a_{n−1,m} − a_{n−1,m+1}) (m ≥ 0, n ≥ 1).  (2)

We claim that the 0-th component a_{n,0} of each row is just the n-th q-Bernoulli number β_{n} for n ≥ 2.

Chen [?] gave a variant of the Akiyama and Tanigawa algorithm, which generates the Bernoulli numbers starting from n = 1. We have also a q-version of Chen’s algorithm for q-Bernoulli numbers as follows: if we replace (??) by the following equation

a_{n,m} = [m]a_{n−1,m} − [m + 1]a_{n−1,m+1} (m ≥ 0, n ≥ 1),  (3)

then the 0-th component a_{n,0} of each row is just the n-th q-Bernoulli number β_{n} for n ≥ 1.

The validity of these algorithms is based on two facts: the first one (Theorem 1) relates the 0-th component a_{n,0} of each row to the initial sequence a_{0,m} by means of q-Stirling numbers of second kind, and the second one gives two explicit formulae (Theorem 2) of the q-Bernoulli numbers in terms of q-Stirling numbers of second kind.

Recall [?] that the q-Stirling numbers of second kind \left\{{n \atop k}\right\}_{q} are defined by the recurrence:

\left\{{n \atop k}\right\}_{q} = \left\{{n − 1 \atop k − 1}\right\}_{q} + [k] \left\{{n − 1 \atop k}\right\}_{q}  for n ≥ k ≥ 1,  (4)

where \left\{{n \atop 0}\right\}_{q} = \left\{{0 \atop k}\right\}_{q} = 0 except \left\{{0 \atop 0}\right\}_{q} = 1.

Theorem 1. Let (a_{n})_{n≥0} be a sequence in a commutative ring. If we define the array (a_{n,m})_{m,n≥0} by a_{0,m} = a_{m} for m ≥ 0 and the recurrence (??), then

a_{n,0} = ∑_{k=0}^{n}(−1)^{k}[k]! \left\{{n + 1 \atop k + 1}\right\}_{q} a_{0,k}  (n ≥ 0),  (5)

and if we use the recurrence (??) instead of (??), then

a_{n,0} = ∑_{k=0}^{n}(−1)^{k}[k]! \left\{{n \atop k}\right\}_{q} a_{0,k}  (n ≥ 0).  (6)

We shall give the first proof of Theorem 1 in Section 2 by using q-differential operator and generating functions, and the second one in Section 4 by applying Theorem 3 in Section 4.

Theorem 2. We have

β_{n} = ∑_{k=0}^{n}(−1)^{k} \left\{{n \atop k}\right\}_{q} \frac{[k]!}{[k + 1]}  (n ≥ 1),  (7)

and

β_{n} = ∑_{k=0}^{n}(−1)^{k} \left\{{n + 1 \atop k + 1}\right\}_{q} \frac{[k]!}{[k + 1]}  (n ≥ 2).  (8)

Note that Eq. (??) was already given by Carlitz [?]. For the sake of completeness we shall include a proof of Theorem 2 in Section 3, which is essentially due to Carlitz [?].

The definition of the Akiyama-Tanigawa algorithm is reminiscent of the so-called Euler-Seidel matrix, a term coined by Dumont [?]. Recall that the Euler-Seidel matrix associated to a sequence (a_{n}) is an infinite matrix (a_{n,m}) (n ≥ 0, m ≥ 0) given by the recurrence a_{n,0} = a_{n} (n ≥ 0) and

a_{n,m} = a_{n−1,m} + a_{n−1,m+1} (m ≥ 0, n ≥ 1).

The sequence (a_{0,m}), first row of the matrix, is called initial sequence, while the sequence (a_{n,0}), first column of the matrix, is called the final sequence. Note that the Euler-Seidel matrix may be used as a simple device for computing its initial and final sequences quickly, see Arnold [?, ?] and Dumont [?].

In the following theorem we shall unify the Akiyama-Tanigawa type algorithms and the classical Euler-Seidel matrices and prove a general formula about the corresponding coefficients. Theorem 3. Let (x_{m}), (y_{m}) and (z_{m}) (m ≥ 0) be three sequences in a commutative ring. The generalized Euler-Seidel matrix (a_{n,m}) (n, m ≥ 0) associated to (x_{m}) is defined by a_{0,m} = x_{m} (m ≥ 0) and

a_{n,m} = y_{m}a_{n−1,m} + z_{m}a_{n−1,m+1}  (m ≥ 0, n ≥ 1).  (9)

Then

a_{n,m} = ∑_{k=0}^{n} x_{m+k} (∏_{j=0}^{k−1} z_{m+j}) h_{n−k}(y_{m}, y_{m+1}, …, y_{m+k}),  (10)

where h_{n}(z_{1}, …, z_{r}) is the n-th complete symmetric function of z_{1}, …, z_{r} defined by

∑_{n≥0} h_{n}(z_{1}, …, z_{r})t^{n} = \frac{1}{(1 − z_{1}t)(1 − z_{2}t) ⋯ (1 − z_{r}t)}.

In particular, we have

a_{n,0} = ∑_{k=0}^{n} x_{k} (∏_{j=0}^{k−1} z_{j}) h_{n−k}(y_{0}, y_{1}, …, y_{k}).  (11)

We will prove Theorem 3 and give some applications in Section 4. In particular, we shall derive an alternative proof of Theorem 1.

INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 6 (2006), #A05 4

2. Proof of Theorem 1 For any formal power series f(x) denote by D_{q} the q-derivative operator:

D_{q}f(x) = \frac{f(qx) − f(x)}{(q − 1)x}.

We will need the two associated operators:

Δ_{q} = 1 − (1 − xq)D_{q},
δ_{q} = (x − 1)D_{q}.

For n ≥ 0 define (x; q)_{0} = 1 and (x; q)_{n} = (1 − x)(1 − xq) … (1 − xq^{n−1}). The following formulas are easy to prove once discovered. Lemma 1. For n ≥ 1 we have

Δ_{q}^{n} = ∑_{k=0}^{n}(−1)^{k} \left\{{n + 1 \atop k + 1}\right\}_{q} (xq; q)_{k}D_{q}^{k},  (12)
δ_{q}^{n} = ∑_{k=0}^{n}(−1)^{k} \left\{{n \atop k}\right\}_{q} (x; q)_{k}D_{q}^{k}.  (13)

Proof. We proceed by induction on n ≥ 1. It is easy to check (??) and (??) for n = 1. For example we have

Δ_{q} = 1 − (1 − xq)D_{q} = \left\{{2 \atop 1}\right\}_{q} − \left\{{2 \atop 2}\right\}_{q} (1 − xq)D_{q} = 1 − (1 − xq)D_{q}.

Note that D_{q}(x^{n}) = [n]x^{n−1} and

D_{q} ((xq; q)_{n}) = −q[n]_{q}(xq^{2}; q)_{n−1},
D_{q} ((x; q)_{n}) = −[n]_{q}(xq; q)_{n−1}.

Suppose the formulas are true for n. Then, using the rule D_{q}(f(x)g(x)) = f(x)D_{q}(g(x)) + g(qx)D_{q}(f(x)),

INTEGERS: ELECTRONIC JOURNAL OF COMBINATORIAL NUMBER THEORY 6 (2006), #A05 5

and the induction hypothesis we have

Δ_{q}^{n+1} = (1 − (1 − xq)D_{q})) ∑_{k=0}^{n}(−1)^{k} \left\{{n + 1 \atop k + 1}\right\}_{q} (xq; q)_{k}D_{q}^{k}
= ∑_{k=0}^{n}(−1)^{k} \left\{{n + 1 \atop k + 1}\right\}_{q} [k + 1]_{q}(xq; q)_{k}D_{q}^{k}

− ∑_{k=0}^{n}(−1)^{k} \left\{{n + 1 \atop k + 1}\right\}_{q} (xq; q)_{k+1}D_{q}^{k+1}

= \left\{{n + 1 \atop 1}\right\}_{q} + ∑_{k=1}^{n}(−1)^{k} ([k + 1] \left\{{n + 1 \atop k + 1}\right\}_{q} + \left\{{n + 1 \atop k}\right\}_{q}) (xq; q)_{k}D_{q}^{k}

+(−1)^{n+1} \left\{{n + 1 \atop n + 1}\right\}_{q} (xq; q)_{n+1}D_{q}^{n+1}

= ∑_{k=1}^{n+1}(−1)^{k} \left\{{n + 2 \atop k + 1}\right\}_{q} (xq; q)_{k}D_{q}^{k},

and

δ_{q}^{n+1} = (x − 1)D_{q} ∑_{k=0}^{n} \left\{{n \atop k}\right\}_{q} (x − 1)(xq − 1) … (xq^{k−1} − 1)D_{q}^{k}
= ∑_{k=0}^{n} \left\{{n \atop k}\right\}_{q} [k](x − 1)(xq − 1) … (xq^{k−1} − 1)D_{q}^{k}

+(x − 1) ∑_{k=1}^{n} \left\{{n \atop k}\right\}_{q} (xq − 1)(xq^{2} − 1) … (xq^{k} − 1)D_{q}^{k+1}

= ∑_{k=0}^{n+1} \left\{{n + 1 \atop k}\right\}_{q} (x − 1)(xq − 1) … (xq^{k−1} − 1)D_{q}^{k}.

This completes the proof. Remark: A q-analogue of (x\frac{d}{dx})^{n} = ∑_{k=0}^{n} \left\{{n \atop k}\right\} x^{k} (\frac{d}{dx})^{k} is the following formula:

(xD_{q})^{n} = ∑_{k=0}^{n} q^{\binom{k}{2}} \left\{{n \atop k}\right\}_{q} x^{k}D_{q}^{k},

which can be verified easily by induction on n.

We are now ready to prove Theorem 1. For fixed n ≥ 0, consider the generating function of a_{n,k} (k ≥ 0) defined by (??):

g_{n}(x) = ∑_{k=0}^{∞} a_{n,k}x^{k} = ∑_{k=0}^{∞}[k + 1](a_{n−1,k} − a_{n−1,k+1})x^{k}

= D_{q} (∑_{k=0}^{∞} a_{n−1,k}x^{k+1}) − D_{q} (∑_{k=0}^{∞} a_{n−1,k+1}x^{k+1})
= D_{q}(xg_{n−1}(x)) − D_{q}(g_{n−1}(x) − a_{n−1,0})
= g_{n−1}(x) + (xq − 1)D_{q}(g_{n−1}(x))
= Δ_{q}(g_{n−1}(x)).

By iteration g_{n}(x) = Δ_{q}^{n}(g_{0}(x)) and Lemma 1 implies that

g_{n}(x) = ∑_{k=0}^{n}(−1)^{k} \left\{{n + 1 \atop k + 1}\right\}_{q} (1 − xq)(1 − xq^{2}) … (1 − xq^{k})D_{q}^{k}(g_{0}(x)).

Putting x = 0 in the above equation yields (??).

Similarly, we consider the generating function of a_{n,k} (k ≥ 0) defined by (??):

h_{n}(x) = ∑_{k=0}^{∞} a_{n,k}x^{k} = ∑_{k=0}^{∞}([k]a_{n−1,k} − [k + 1]a_{n−1,k+1})x^{k}

= xD_{q} (∑_{k=0}^{∞} a_{n−1,k}x^{k}) − D_{q} (∑_{k=0}^{∞} a_{n−1,k+1}x^{k+1})
= xD_{q}(h_{n−1}(x)) − D_{q}(h_{n−1}(x) − a_{n−1,0})
= δ_{q}(h_{n−1}(x)).

Thus h_{n}(x) = δ_{q}^{n}(h_{0}(x)) and Lemma 1 implies that

h_{n}(x) = ∑_{k=0}^{n} \left\{{n \atop k}\right\}_{q} (x − 1)(xq − 1) … (xq^{k−1} − 1)D_{q}^{k}(h_{0}(x)).

Putting x = 0 in the last equation yields (??).

Now, if we take a_{0,m} = 1/[m + 1] in Theorem 1 and apply algorithm (??), then it follows from (??) and (??) that a_{n,0} = β_{n} for n ≥ 2; while applying algorithm (??) will yield that a_{n,0} = β_{n} for n ≥ 1 by (??) and (??).

3. Proof of Theorem 2

Let [x] = (q^{x} − 1)/(q − 1). For integer s ≥ 0 define [x]_{s} = [x][x − 1] ⋯ [x − s + 1] and the q-binomial coefficient \left[{x \atop s}\right] = [x]_{s}/[s]!.

Let η_{0}, η_{1}, η_{2}, … be a sequence such that by η_{0} = 1, η_{1} = 0 and

∑_{i=0}^{m} \binom{m}{i}q^{i}η_{i} = η_{m} (m > 1).  (14)

We define the polynomials η_{m}(x) (m ≥ 0) in q^{x} by

η_{m}(x) = ∑_{i=0}^{m} \binom{m}{i}η_{i}[x]^{m−i}q^{ix}.  (15)

Then η_{m}(0) = η_{m} and

η_{m}(x + y) = ∑_{i=0}^{m} \binom{m}{i}η_{i}(y)[x]^{m−i}q^{ix}.  (16)

Indeed, substituting η_{i}(y) by (??) and exchanging the order of summations in the right-hand side of (??) yields

∑_{j=0}^{m} ∑_{i=j}^{m} \binom{m}{i}\binom{i}{j}[x]^{m−i}[y]^{i−j}q^{ix+jy} = ∑_{j=0}^{m} \binom{m}{j}η_{j}q^{j(x+y)} ∑_{i=0}^{m−j} \binom{m − j}{i}[x]^{m−j−i}(q^{x}[y])^{i}

= ∑_{j=0}^{m} \binom{m}{j}η_{j}q^{j(x+y)} ([x] + q^{x}[y])^{m−j}

= ∑_{j=0}^{m} \binom{m}{j}η_{j}[x + y]^{m−j}q^{j(x+y)},

which is equal to η_{m}(x + y) by (??).

Setting x = 1 in (??) we see that η_{m}(1) = η_{m} for m > 1. It follows from (??) with y = 1 that for m ≥ 0,

η_{m}(x + 1) − η_{m}(x) = ∑_{i=0}^{m} \binom{m}{i}(η_{i}(1) − η_{i})[x]^{m−i}q^{ix} = mq^{x}[x]^{m−1}.  (17)

It follows that for k ≥ 1,

∑_{i=0}^{k−1} q^{i}[i]^{m} = \frac{1}{m + 1} ∑_{i=0}^{k−1}(η_{m}(i + 1) − η_{m}(i))

= \frac{1}{m + 1}(η_{m+1}(k) − η_{m+1})

= \frac{1}{m + 1} ∑_{i=1}^{m+1} \binom{m + 1}{i}[k]^{i}q^{(m+1−i)k}η_{m+1−i} + (q^{(m+1)k} − 1)\frac{η_{m+1}}{m + 1}.  (18)

On the other hand, it is readily seen by induction on n ≥ 1 that

x^{n} = ∑_{k=1}^{n} \left\{{n \atop k}\right\}_{q} x(x − [1]) … (x − [k − 1]),  (19)

which, by substitution x → [x], yields

[x]^{n} = ∑_{k=1}^{n} q^{k(k−1)/2} \left\{{n \atop k}\right\}_{q} [x]_{k}.

Therefore

∑_{i=0}^{k−1} q^{i}[i]^{m} = ∑_{i=0}^{k−1} q^{i} ∑_{s=1}^{m} q^{s(s−1)/2} \left\{{m \atop s}\right\}_{q} [i]_{s} = ∑_{s=0}^{m} q^{s(s+1)/2} \left\{{m \atop s}\right\}_{q} \frac{[k]_{s+1}}{[s + 1]},  (20)

where we used the identity for q-binomial coefficients:

∑_{i=s}^{k−1} q^{i−s}\left[{i \atop s}\right] = \left[{k \atop s + 1}\right].

Combining (??) and (??) we obtain a polynomial identity on q^{k}. Dividing both sides by [k] and setting k = 0 leads to

η_{m} + (q − 1)η_{m+1} = ∑_{s=0}^{m} q^{s(s+1)/2} \left\{{n \atop s}\right\}_{q} \frac{[−1]_{s}}{[s + 1]}.

Now, it remains to prove β_{m} = η_{m} + (q − 1)η_{m+1}. Indeed, the sequence η_{n} + (q − 1)η_{n+1} satisfies the recurrence (??) for n > 1:

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

= qη_{n} + (q − 1) ∑_{i=1}^{n+1} (\binom{n + 1}{i} − \binom{n}{i}) q^{i}η_{i}

= qη_{n} + (q − 1) (η_{n+1} − η_{n})

= η_{n} + (q − 1)η_{n+1}.

This completes the proof of (??).

For n ≥ 2, simplifying x in (??) and setting x = 0 we get

0 = ∑_{m=1}^{n}(−1)^{m} \left\{{n \atop m}\right\}_{q} [m − 1]!.  (21)

Now, using (??) we have

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

= ∑_{k=0}^{n}(−1)^{k} \left\{{n \atop k}\right\}_{q} \frac{[k]!}{[k + 1]} + ∑_{k=0}^{n}(−1)^{k} \left\{{n \atop k + 1}\right\}_{q} [k]!.

Formula (??) follows then from (??) and (??).

4. Proof of Theorem 3 and applications

By induction on n ≥ 0. The formula is clear for n = 0 and n = 1. Suppose that the formula is true until n ≥ 1. Then

a_{n+1,m} = y_{m}a_{n,m} + z_{m}a_{n,m+1}

= y_{m} ∑_{k=0}^{n} x_{m+k}z_{m} … z_{m+k−1}h_{n−k}(y_{m}, …, y_{m+k})

+ z_{m} ∑_{k=0}^{n} x_{m+k+1}z_{m+1} … z_{m+k}h_{n−k}(y_{m+1}, …, y_{m+k+1})

= x_{m}y_{m}^{n+1} + ∑_{k=1}^{n} x_{m+k}z_{m} … z_{m+k−1}(y_{m}h_{n−k}(y_{m}, …, y_{m+k}) + h_{n−k+1}(y_{m+1}, …, y_{m+k}))

+ x_{m+n+1}z_{m} … z_{m+n}.

Since y_{m}h_{n−k}(y_{m}, …, y_{m+k})+h_{n−k+1}(y_{m+1}, …, y_{m+k}) = h_{n+k−1}(y_{m}, …, y_{m+k}), we are done.

The following examples are special cases of Theorem 3:

• if y_{m} = z_{m} = 1, we recover the so-called Euler-Seidel matrix (see [?]) associated to the initial sequence x_{m} (m ≥ 0).

• if z_{m} = xq^{m} and y_{m} = 1, then we recover the q-Seidel matrix introduced by Clarke, Han and Zeng [?].

• if z_{m} = −y_{m}, then Theorem 3 reduces to a result of Lascoux [?].

• if y_{m} = −z_{m} = [m + 1] and x_{m} = 1/[m + 1], then this is our q-analogue of the Akiyama-Tanigawa algorithm. Indeed, it is readily seen that

∑_{n≥k} \left\{{n \atop k}\right\}_{q} t^{n} = \frac{t^{k}}{(1 − [1]t)(1 − [2]t) … (1 − [k]t)},

which yields the explicit formula: \left\{{n \atop k}\right\}_{q} = h_{n−k}([1], …, [k]), so

\left\{{n + 1 \atop k + 1}\right\}_{q} = h_{n−k}([1], …, [k + 1]).

Eq. (??) follows then from (??) with the above specializations.

• if y_{m} = [m] and z_{m} = −[m + 1] and x_{m} = 1/[m + 1], this is our q-analogue of Chen’s algorithm and (??) follows directly from (??).

It may be worth pointing out that it is possible to write explicitly the general coefficients a_{n,m} in Theorem 1, because

h_{n−k}([m], …, [m + k]) = ∑_{i=0}^{k}(−1)^{k−i}q^{−k(m+i)+\binom{i+1}{2}} \frac{[m + i]^{n}}{[i]![k − i]!}.  (22)

Indeed, there holds

\frac{1}{(1 − z_{0}t)(1 − z_{1}t) … (1 − z_{k}t)} = ∑_{i=0}^{k} \frac{∏_{j=0,j≠i}^{k}(1 − z_{j}/z_{i})^{−1}}{1 − z_{i}t} .

Equating the coefficients of t^{n} (n ≥ 0) in the two sides yields

h_{n}(z_{0}, …, z_{k}) = ∑_{i=0}^{k} ∏_{j=0,j≠i}^{k}(1 − z_{j}/z_{i})^{−1} z_{i}^{n},

which gives (??) by taking z_{i} = [m + i] for i = 0, …, k.

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