Hyper-Sums of Powers of Integers and the Akiyama-Tanigawa Matrix

Yoshinari Inaba Toba Senior High School Nishikujo, Minami-ku Kyoto 601-8449 Japan

                     inava@kyoto-be.ne.jp

Abstract In this short essay, we consider hyper-sums of powers of integers, namely sums of power sums. We can obtain easily their formulae as polynomials by using formulae for ordinary sums of powers of integers. The coefficient of the first-degree term in each polynomial coincides with the matrix element of the Akiyama-Tanigawa matrix.

1 The Akiyama-Tanigawa matrix

The Akiyama-Tanigawa algorithm which is reformulated by K. W. Chen [1] and M. Kaneko [3] is described by the sequence {a_{n, m}};

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

with an initial sequence a_{0, m} (m = 0, 1, 2, …). Let B_{n}(x) be the n-th Bernoulli polynomial

\frac{te^{xt}}{e^{t} − 1} = ∑_{n=0}^{∞} B_{n}(x)\frac{t^{n}}{n!}, |t| < 2π

When the initial sequence is a_{0, m} = 1/(m + 1) then this algorithm yields B_{n}(1) as the leading element a_{n, 0}. Here B_{n}(1) are, in fact, the Bernoulli numbers B_{n}, with the sole exception of n = 1, for which B_{1}(1) = −B_{1} holds. We assume this situation here and the Akiyama-Tanigawa matrix {a_{n, m}} is then

a_{n, 0} a_{n, 1} a_{n, 2} a_{n, 3} a_{n, 4} a_{n, 5} a_{n, 6} a_{n, 7} a_{n, 8} a_{n, 9} a_{n, 10}
1 \frac{1}{2}  \frac{1}{3}  \frac{1}{4}  \frac{1}{5}  \frac{1}{6}  \frac{1}{7}  \frac{1}{8}  \frac{1}{9}  \frac{1}{10}  \frac{1}{11} ⋯
\frac{1}{2}  \frac{1}{3}  \frac{1}{4}  \frac{1}{5}  \frac{1}{6}  \frac{1}{7}  \frac{1}{8}  \frac{1}{9}  \frac{1}{10}  \frac{1}{11} ⋯
\frac{1}{6}  \frac{1}{6}  \frac{3}{20}  \frac{2}{15}  \frac{5}{42}  \frac{3}{28}  \frac{7}{72}  \frac{4}{45} \frac{9}{110} ⋯
0 \frac{1}{30}  \frac{1}{20}  \frac{2}{35}  \frac{5}{84}  \frac{5}{84} \frac{7}{120} \frac{28}{495} ⋯
−\frac{1}{30} −\frac{1}{30} −\frac{3}{140} −\frac{1}{105}  0 \frac{1}{140} \frac{49}{3960} ⋯
0 −\frac{1}{42} −\frac{1}{28} −\frac{4}{105} −\frac{1}{28} −\frac{29}{924} ⋯
\frac{1}{42}  \frac{1}{42}  \frac{1}{140} −\frac{1}{105} −\frac{5}{231} ⋯
0 \frac{1}{30}  \frac{1}{20}  \frac{8}{165} ⋯
−\frac{1}{30} −\frac{1}{30}  \frac{1}{220} ⋯
0 −\frac{5}{66} ⋯
\frac{5}{66} ⋯
⋯

Each matrix element a_{n, m} is determined uniquely by the recursive rule (1) and the initial values.

2 Explicit formulation

In this section, we determine an explicit expression of a_{n, m}.

Proposition 1.

a_{n, m} = \frac{1}{m!} ∑_{i=0}^{m}(−1)^{i} [\begin{matrix}m + 1 \\ i + 1 \end{matrix}] B_{n+i}(1),  n ∈ ℕ.  (2)

where \left[{n \atop k}\right] are the Stirling numbers of the first kind (Sloane’s A008275, cf. [2], Ch. 6.1).

Proof. First, we shall see that this a_{n, m} satisfies the recursive rule (1). To do this, we use the well known recurrence relation \left[{n+1 \atop k}\right] = n\left[{n \atop k}\right] + \left[{n \atop k−1}\right].

(m + 1)(a_{n−1, m} − a_{n−1, m+1})

= (m + 1)(\frac{1}{m!} ∑_{i=0}^{m}(−1)^{i} [\begin{matrix}m + 1 \\ i + 1 \end{matrix}] B_{n+i−1}(1) − \frac{1}{(m + 1)!} ∑_{i=0}^{m+1}(−1)^{i} [\begin{matrix}m + 2 \\ i + 1 \end{matrix}] B_{n+i−1}(1))

= \frac{1}{m!} ∑_{i=0}^{m+1}(−1)^{i}((m + 1) [\begin{matrix}m + 1 \\ i + 1 \end{matrix}] − [\begin{matrix}m + 2 \\ i + 1 \end{matrix}])B_{n+i−1}(1)

= \frac{1}{m!} ∑_{i=0}^{m+1}(−1)^{i+1} [\begin{matrix}m + 1 \\ i \end{matrix}] B_{n+i−1}(1)

= a_{n, m}, where we have used the fact that \left[{m+1 \atop 0}\right] = 0.

Next, to show that the initial sequence a_{0, m} = 1/(m + 1), we use an expression that can be found in Kaneko [3]. Let \left\{{n \atop k}\right\} be the Stirling numbers of the second kind (Sloane’s A008277, cf. [2], Ch. 6.1). Then

B_{n}(1) = ∑_{i=0}^{n} \frac{(−1)^{i}i!}{i + 1} \begin{cases}n + 1 \\ i + 1 \end{cases}\} , n ≥ 0,

By the well known inversion formula (cf. [2], Ch. 6.1),

∑_{k=m}^{n} [\begin{matrix}n \\ k \end{matrix}] \begin{cases}k \\ m \end{cases}\} (−1)^{n−k} = \begin{cases}1, & m = n \\ 0, & m ≠ n \end{cases} ,

we can compute what we want easily.

(m + 1)a_{0, m} = \frac{m + 1}{m!} ∑_{i=0}^{m}(−1)^{i} [\begin{matrix}m + 1 \\ i + 1 \end{matrix}] B_{i}(1)

= \frac{m + 1}{m!} ∑_{k=0}^{m} \frac{k!(−1)^{k}}{k + 1} ∑_{i=k}^{m}(−1)^{i} [\begin{matrix}m + 1 \\ i + 1 \end{matrix}] \begin{cases}i + 1 \\ k + 1 \end{cases}\}

= \frac{(m + 1)(−1)^{m}}{m!} ∑_{k=0}^{m} \frac{k!(−1)^{k}}{k + 1} ∑_{i=k}^{m}(−1)^{m−i} [\begin{matrix}m + 1 \\ i + 1 \end{matrix}] \begin{cases}i + 1 \\ k + 1 \end{cases}\}

= 1

From the recursive rule (1) and initial values a_{0, m}, the proof of this proposition is done and we have an explicit expression of a_{n, m}.

3 Hyper-sum polynomials

Now we shall consider “hyper-sums” of powers of integers, namely sums of power sums. Let P_{k}^{(0)}(n) = ∑_{i=1}^{n} i^{k}, P_{k}^{(1)}(n) = ∑_{i=1}^{n} P_{k}^{(0)}(i), P_{k}^{(2)}(n) = ∑_{i=1}^{n} P_{k}^{(1)}(i), …, and P_{k}^{(m)}(n) = ∑_{i=1}^{n} P_{k}^{(m−1)}(i), with k and n positive integers. P_{k}^{(m)}(n) is a (k+m+1)-th degree polynomial in n and P_{k}^{(0)}(n) is the ordinary power sum 1^{k} + 2^{k} + 3^{k} + ⋯ + n^{k}. Let c_{k, m} be the coefficient of the first-degree term in P_{k}^{(m)}(n). Since the formulae for P_{k}^{(0)}(n) have been investigated for a long time and abundant methods of determining them have been developed, we can use some of them. Here we choose the formula including the Stirling numbers of the second kind, which is for instance seen in the paper of Srivastava, Joshi and Bisht [4].

P_{k}^{(0)}(n) = ∑_{i=1}^{k} i!\binom{n + 1}{i + 1} \begin{cases}k \\ i \end{cases}\} , k ∈ ℕ

where the binomial coefficient \binom{n+1}{i+1} is taken to be zero for n < i. Here we use the well known identity (cf. [2], Ch. 5.1),

∑_{k=0}^{n} \binom{k}{m} = \binom{n + 1}{m + 1},

where the binomial coefficient \binom{k}{m} is zero for k < m. This yields

∑_{n=0}^{N} \binom{n + 1}{i + 1} = ∑_{n=i}^{N} \binom{n + 1}{i + 1} = ∑_{n=i+1}^{N+1} \binom{n}{i + 1} = \binom{N + 2}{i + 2}.

So we can get easily

P_{k}^{(1)}(n) = ∑_{i=1}^{k} i!\binom{n + 2}{i + 2} \begin{cases}k \\ i \end{cases}\} , k ∈ ℕ.

Thus, by means of the successive computations, we arrive at

P_{k}^{(m)}(n) = ∑_{i=1}^{k} i!\binom{n + m + 1}{i + m + 1} \begin{cases}k \\ i \end{cases}\} , k ∈ ℕ.

From the fact that \binom{n+m+1}{i+m+1} has the factor n for i ≥ 1, P_{k}^{(m)}(n) is divisible by n. Therefore, to find c_{k, m}, we shall use the following relation.

c_{k, m} = \frac{P_{k}^{(m)}(n)}{n}|_{n=0}

Since

\frac{1}{n}\binom{n + m}{k + m}|_{n=0} = \frac{m!(k − 1)!(−1)^{k−1}}{(k + m)!} .

We obtain

c_{k, m} = ∑_{i=1}^{k} \frac{(m + 1)!(i − 1)!i!(−1)^{i−1}}{(i + m + 1)!} \begin{cases}k \\ i \end{cases}\} ,  k ∈ ℕ.  (3)

Now only c_{0, m} has been left for us. But it is not difficult to see it, because the identity P_{0}^{(1)}(n) = P_{1}^{(0)}(n) = ∑_{i=1}^{n} i yields P_{0}^{(m)}(n) = P_{1}^{(m−1)}(n) immediately. Thus we have c_{0, m} = c_{1, m−1} = 1/(m+1). By combining these consequences, we can state the following proposition.

Proposition 2.

c_{k, m} = \{ \begin{matrix}\frac{1}{m + 1}, & k = 0 \\ ∑_{i=1}^{k} \frac{(m + 1)!(i − 1)!i!(−1)^{i−1}}{(i + m + 1)!} \left\{{k \atop i}\right\}, & k ≥ 1 \end{matrix}

4 Equality between a_{n, m} and c_{n, m}

As one of the important consequences in this essay, let us see the equality between a_{n, m} and c_{n, m}. To do this, we shall see the recurrence relation whom c_{n, m} satisfies.

Lemma 1.

c_{n, m} = (m + 1)(c_{n−1, m} − c_{n−1, m+1}),  n ≥ 1

Proof. For the initial case n = 1, by proposition 2 we have c_{0, m} = 1/(m + 1) and c_{1, m} = 1/(m + 2) without difficulty. So we can confirm that c_{1, m} = (m + 1)(c_{0, m} − c_{0, m+1}) easily.

For the case n ≥ 2, we use the well known recurrence relation \left\{{n \atop k}\right\} = k\left\{{n−1 \atop k}\right\} + \left\{{n−1 \atop k−1}\right\}.

c_{n, m} = ∑_{i=1}^{n} \frac{(m + 1)!(i − 1)!i!(−1)^{i−1}}{(i + m + 1)!} \begin{cases}n \\ i \end{cases}\}

= ∑_{i=1}^{n} \frac{(m + 1)!(i − 1)!i!(−1)^{i−1}}{(i + m + 1)!} (i \begin{cases}n − 1 \\ i \end{cases}\} + \begin{cases}n − 1 \\ i − 1 \end{cases}\})

= ∑_{i=1}^{n−1} \frac{(m + 1)!i!i!(−1)^{i−1}}{(i + m + 1)!} \begin{cases}n − 1 \\ i \end{cases}\} − ∑_{i=1}^{n−1} \frac{(m + 1)!i!(i + 1)!(−1)^{i−1}}{(i + m + 2)!} \begin{cases}n − 1 \\ i \end{cases}\}

= ∑_{i=1}^{n−1} \frac{(m + 1)!i!i!(−1)^{i−1}}{(i + m + 1)!} · \frac{m + 1}{i + m + 2} \begin{cases}n − 1 \\ i \end{cases}\}

On the other hand,

(m + 1)(c_{n−1, m} − c_{n−1, m+1})

= (m + 1)(∑_{i=1}^{n−1} \frac{(m + 1)!(i − 1)!i!(−1)^{i−1}}{(i + m + 1)!} \begin{cases}n − 1 \\ i \end{cases}\} − ∑_{i=1}^{n−1} \frac{(m + 2)!(i − 1)!i!(−1)^{i−1}}{(i + m + 2)!} \begin{cases}n − 1 \\ i \end{cases}\})

= (m + 1)(∑_{i=1}^{n−1} \frac{(m + 1)!(i − 1)!i!(−1)^{i−1}}{(i + m + 1)!} · \frac{i}{i + m + 2} \begin{cases}n − 1 \\ i \end{cases}\})

= ∑_{i=1}^{n−1} \frac{(m + 1)!i!i!(−1)^{i−1}}{(i + m + 1)!} · \frac{m + 1}{i + m + 2} \begin{cases}n − 1 \\ i \end{cases}\} .

Thus this lemma and the fact c_{0, m} = 1/(m + 1) yield the next consequence immediately.

Proposition 3.

c_{n, m} = a_{n, m}.

Corollary 1.

\frac{1}{m!} ∑_{i=0}^{m}(−1)^{i} [\begin{matrix}m + 1 \\ i + 1 \end{matrix}] B_{n+i}(1) = ∑_{i=1}^{n} \frac{(m + 1)!(i − 1)!i!(−1)^{i−1}}{(i + m + 1)!} \begin{cases}n \\ i \end{cases}\} , n ∈ ℕ.

Proof. It is clear from (2) and (3).

5 Acknowledgements

I should like to thank the referees for several comments and suggestions.

References

[1] K. W. Chen, Algorithms for Bernoulli numbers and Euler numbers, Article 01.1.6, J. Integer. Seq 4 (2001), 1–7.

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

[3] M. Kaneko, The Akiyama-Tanigawa algorithm for Bernoulli numbers, Article 00.2.9, J. Integer. Seq 3 (2000), 1–6.

[4] H. M. Srivastava, J. M. C. Joshi and C. S. Bisht, Fractional calculus and the sum of powers of natural numbers, Stud. Appl. Math. 85 (1991), 183–193.

2000 Mathematics Subject Classification: Primary 11Y55; Secondary 11B68 .

Keywords: Bernoulli numbers, Stirling numbers, power sums.

(Concerned with sequences \underline{A008275} and \underline{A008277}.)

Received January 5 2005; revised version received March 9 2005; April 1 2005; April 29 2005. Published in Journal of Integer Sequences, May 15 2005.

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