PUBLICATIONS DE L’INSTITUT MATHÉMATIQUE Nouvelle série, tome 102(116) (2017), 149–153 DOI: https://doi.org/10.2298/PIM1716149B

ON A PROPERTY OF STIRLING POLYNOMIALS

Farid Bencherif and Tarek Garici

Abstract. We positively answer a question posed in 1960 by D. S. Mitrinović and R. S. Mitrinović (Publ. Fac. Electrotech. Univ. Belgrade, Ser. Math. Phys. 34 (1960), 1–23) about the Stirling numbers of the first kind.

1. Introduction

We bring a positive answer to a question posed by D. S. Mitrinović and R. S. Mitrinović in 1960 [4] about the Stirling numbers of the first kind s(n, k) defined by x(x − 1) ⋯ (x − n + 1) = ∑ _{k⩾0} s(n, k)x^{k}. In what follows, we denote by ⌊x⌋ the integer part of a real number x and by mod(n, 2) the remainder of the division of n by 2. We recall that a primitive polynomial is a polynomial over an integral domain R such that no non-invertible element of R divides all its coefficients at once.

Theorem 1.1. Let (m_{n})_{n⩾0} be the sequence defined by

(1.1)  m_{n} := \frac{1}{(n + 1)!} ∏_{p prime and p⩽n+1} p^{⌊^{\frac{n}{p−1}} ⌋+⌊\frac{n}{p(p−1)} ⌋+⌊\frac{n}{p^{2} (p−1)} ⌋+···} .

Then for any n ⩾ 0, m_{n} is a non-negative integer and for any positive integer k, we have

s(n, n − 2k) = \frac{1}{m_{2k}} \binom{n}{2k + 1} P_{2k}(n), (n ⩾ 2k) (1.2)

s(n, n − 2k − 1) = \frac{1}{m_{2k+1}} \binom{n}{2k + 2} n(n − 1)P_{2k+1} (n) (n ⩾ 2k + 1), (1.3)

where P_{2k}(x) and P_{2k+1} (x) are two primitive polynomials over ℤ satisfying

(1.4)  P_{2k}(0) = P_{2k+1} (0).

The sequence (m_{n})_{n⩾0} = (1, 1, 4, 2, 48, 16, 576, …) is the sequence A163176 in the OEIS [6]. The first few expressions of the polynomials P_{n}(x) for 2 ⩽ n ⩽ 9 are

2010 Mathematics Subject Classification: 11B68; 11B39; 05A40.

Key words and phrases: Stirling numbers, Nörlund polynomials, Bernoulli numbers.

Communicated by Gradimir Milovanović.

P_{2}(x) = 3x − 1,
P_{3}(x) = −1,
P_{4}(x) = 15x^{3} − 30x^{2} + 5x + 2,
P_{5}(x) = −3x^{2} + 7x + 2,
P_{6}(x) = 63x^{5} − 315x^{4} + 315x^{3} + 91x^{2} − 42x − 16,
P_{7}(x) = −9x^{4} + 54x^{3} − 51x^{2} − 58x − 16,
P_{8}(x) = 135x^{7} − 1260x^{6} + 3150x^{5} − 840x^{4} − 2345x^{3} − 540x^{2} + 404x + 144,
P_{9}(x) = −15x^{6} + 165x^{5} − 465x^{4} − 17x^{3} + 648x^{2} + 548x + 144.

In [4], Mitrinović and Mitrinović state the relations (1.2), (1.3) and (1.4) for k ∈ {1, 2, 3, 4, 5, 6}, and then raise the question of whether these equalities are true in the general case. Theorem 1.1, main result of our paper, brings a positive answer to this question.

2. Proof of Theorem 1.1

To prove the theorem, it is useful to note the following three lemmas. The proof is mainly based on the properties of the Nörlund polynomials and the sequence (m_{n})_{n⩾0} . The Nörlund polynomials B_{n}^{(x)} are defined by [5]

( \frac{z}{e^{z} − 1} )^{x} = ∑_{n=0}^{∞} B_{n}^{(x)} \frac{z^{n}}{n!} .

For any positive integer n, B_{n}^{(x)} is a polynomial over the rational number field of degree n and it is divisible by x.

The Bernoulli numbers B_{n} are defined for n ⩾ 0 by B_{n} = B_{n}^{(1)}. It is well known that

(2.1)  B_{2n+1} = 0 (n ⩾ 1).

Lemma 2.1. For any positive integer n, we have

(2.2)  \begin{aligned}(B_{n}^{(x)} ) & = (−1)^{n−1} \frac{B_{n}}{n} , \\ [x^{2}](B_{2n+1}^{(x)} ) & = \frac{2n + 1}{4n} B_{2n} \end{aligned}

Proof. Let n ⩾ 1. In [3, Theorems 1 and 2], Liu and Srivastava have determined explicitly the coefficients of the polynomial B_{n}^{(x)} . They proved that for 1 ⩽ k ⩽ n, the coefficient of the x^{k} term in B_{n}^{(x)} is given by

[x^{k}]B_{n}^{(x)} = (−1)^{n−k} \frac{n!}{k!} ∑ \frac{B_{ν_{1}} ⋯ B_{ν_{k}}}{(ν_{1} ⋯ ν_{k})ν_{1}! ⋯ ν_{k}!},

where the sum is taken over all positive integers ν_{1}, …, ν_{k} that satisfy ν_{1}+⋯ +ν_{k} = n. Applying this for k = 1, we obtain easily (2.2), and for k = 2 we get

[x^{2}](B_{2n+1}^{(x)} ) = −\frac{1}{2} ∑_{j=1}^{2n} \binom{2n + 1}{j} \frac{B_{j} B_{2n+1−j}}{j(2n + 1 − j)}

= −\frac{1}{2}\binom{2n + 1}{1} \frac{B_{1}B_{2n}}{2n} − \frac{1}{2}\binom{2n + 1}{2n} \frac{B_{2n} B_{1}}{2n}

− \frac{1}{2} ∑_{j=2}^{2n−1} \binom{2n + 1}{j} \frac{B_{j} B_{2n+1−j}}{j(2n + 1 − j)}

= \frac{2n + 1}{4n} B_{2n} − \frac{1}{2} ∑_{j=2}^{2n−1} \binom{2n + 1}{j} \frac{B_{j} B_{2n+1−j}}{j(2n + 1 − j)} .

The proof will be complete if we can show that

∑_{j=2}^{2n−1} \binom{2n + 1}{j} \frac{B_{j} B_{2n+1−j}}{j(2n + 1 − j)} = 0.

As the case n = 1 is obvious, let us consider the case n ⩾ 2. Firstly we remind that after B_{1}, all the Bernoulli numbers with odd index vanish. Note that j and (2n + 1 − j) have a different parity, so we deduce that at least one of B_{j} and B_{2n+1−j} has an odd index greater than 1 and therefore B_{j} B_{2n+1−j} = 0 for any 2 ⩽ j ⩽ 2n − 1. □

In the proof of Theorem 1.1, the following lemma is essential.

Lemma 2.2. For any integer n ⩾ 2, we have

(2.3)  \binom{x − 1}{n} B_{n}^{(x)} = \frac{1}{m_{n}} \binom{x}{n + 1} (x(x − 1))^{mod (n,2)} P_{n}(x)

where P_{n}(x) is a primitive polynomial over ℤ.

Proof. Let n be a positive integer. For any prime number p, let r_{p}(n) be the highest power of p that divides n!. Adelberg [1, Corollary 3] shows that if we set n_{p} = p⌊\frac{n}{p−1} ⌋ and

(2.4)  d_{n} = \frac{1}{n!} ∏_{p prime and p⩽n+1} p^{r_{p} (n_{p})} ,

then d_{n}B_{n}^{(x)} is a primitive polynomial over ℤ.

By the Legendre’s formula [7, p. 31], we obtain that for any prime number p less than or equal to n + 1

(2.5)  r_{p}(n_{p}) = ∑_{k⩾0} ⌊ \frac{1}{p^{k}}⌊ \frac{n}{p − 1} ⌋⌋ = ∑_{k⩾0} ⌊ \frac{n}{p^{k}(p − 1)} ⌋.

From (2.4), (2.5) and (1.1), we obtain d_{n} = (n + 1)m_{n}. On the other hand, from the definition of the Nörlund polynomials, it is easy to see that B_{n}^{(x)} is divisible by x. Moreover, from (2.2) and (2.1) we have that for any odd integer n ⩾ 3, [x](B_{n}^{(x)} ) = (−1)^{n−1} \frac{B_{n}}{n} = 0 and B_{n}^{(1)} = B_{n} = 0. It follows that in ℤ[x], the primitive polynomial (n + 1)m_{n}B_{n}^{(x)} is divisible by x(x(x − 1))^{mod (n,2)} for n ⩾ 2. The quotient P_{n}(x) of these two polynomials is also a primitive polynomial over ℤ and then we have

(2.6)  (n + 1)m_{n}B_{n}^{(x)} = x(x(x − 1))^{mod (n,2)} P_{n}(x), (n ⩾ 2).

Multiplying both sides of (2.6) by \frac{1}{(n+1)m_{n}} \binom{x−1}{n}, we deduce (2.3). □

The following lemma states some properties of the sequence (m_{n})_{n⩾0} defined by (1.1).

Lemma 2.3. For any non-negative integer n, we have

(1) m_{n} is an integer

(2) m_{2n} = (n + 1)m_{2n+1}

Proof. Let n be a non-negative integer. For any prime number p ⩽ n + 1 and for any integer k ⩾ 0, we have \frac{n}{p^{k} (p−1)} − \frac{n+1}{p^{k+1}} = \frac{n+1−p}{p^{k+1} (p−1)} ⩾ 0 and then ⌊\frac{n}{p^{k} (p−1)} ⌋− ⌊\frac{n+1}{p^{k+1}} ⌋ ⩾ 0. By the Legendre’s formula, we have

v_{p}(m_{n}) = ∑_{k⩾0} ( ⌊ \frac{n}{p^{k}(p − 1)} ⌋ − ⌊ \frac{n + 1}{p^{k+1}} ⌋) ⩾ 0.

It follows that m_{n} is an integer.

Let p ⩽ n + 1 be a prime number. It is clear that for all positive integers x and y we have

⌊ \frac{x + 1}{y} ⌋ − ⌊ \frac{x}{y}⌋ = \{ \begin{matrix}1 if y divides x + 1, \\ 0 otherwise. \end{matrix}

Thus

v_{p}( \frac{(n + 1)m_{2n+1}}{m_{2n}} ) = v_{p}(2^{−1} ) + ∑_{k⩾0} ⌊ \frac{2n + 1}{p^{k}(p − 1)} ⌋ − ⌊ \frac{2n}{p^{k}(p − 1)} ⌋.

Since, p^{k}(p − 1) divide 2n + 1 if and only if p = 2 and k = 0, then for any prime p, v_{p}( \frac{(n+1)m_{2n+1}}{m_{2n}} ) = 0, which is equivalent to state that \frac{(n+1)m_{2n+1}}{m_{2n}} = 1. □

Now we can prove the theorem by using these lemmas.

Let k be a positive integer. It is well known that (cf. [2, p. 329])

(2.7)  s(n, n − j) = \binom{n − 1}{j} B_{j}^{(n)} , for n ⩾ j ⩾ 0.

By Lemma 2.2, (2.7) can be written as

s(n, n − j) = \frac{1}{m_{j}} \binom{n}{j + 1} (n(n − 1))^{mod (j,2)} P_{j} (n) (n ⩾ j ⩾ 2),

Since P_{j} (x) is a primitive polynomial over ℤ, applying this result for j = 2k (resp. j = 2k + 1) gives (1.2), (resp. (1.3)). Furthermore, by taking at first n = 2k in (2.6) and then n = 2k + 1 we get

(2k + 1)m_{2k}B_{2k}^{(x)} = xP_{2k}(x),
(2k + 2)m_{2k+1} B_{2k+1}^{(x)} = x^{3}P_{2k+1} (x) − x^{2}P_{2k+1} (x).

It follows that

P_{2k}(0) = [x]((2k + 1)m_{2k}B_{2k}^{(x)} ),
P_{2k+1} (0) = [x^{2}](−(2k + 2)m_{2k+1} B_{2k+1}^{(x)} ).

By Lemma 2.1, these last two equalities can be written as

P_{2k}(0) = −m_{2k}(2k + 1)\frac{B_{2k}}{2k} ,
P_{2k+1} (0) = −(k + 1)m_{2k+1} (2k + 1)\frac{B_{2k}}{2k} .

Given this and the second relation of Lemma 2.3, we can deduce that P_{2k+1} (0) = P_{2k}(0). This establishes (1.4). The proof of the theorem is now complete.

Remark 2.1. As we know that for any non-negative integer n, B_{n}^{(x)} is a polynomial of degree n, by using relation (2.6) we deduce that for any k ⩾ 1, we have deg(P_{2k}) = 2k − 1 and deg(P_{2k+1} ) = 2k − 2.

Acknowledgment. Supported by LA3C, Laboratoire d’Arithmétique Codage Combinatoire et Calcul formel.

References

1. A. Adelberg, Arithmetic properties of the Nörlund polynomial B_{n}^{(x)} , Discrete Math. 204 N1 (1999), 5–13.

2. Ch. A. Charalambides, Enumerative Combinatorics, Chapman and Hall, 2002.

3. G. -D. Liu, H. M. Srivastava, Explicit Formulas for the Nörlund Polynomials B_{n}^{(x)} and b_{n}^{(x)} , Comput. Math. Appl. 51 N9-10 (2006), 1377–1384.

4. D. S. Mitrinović, R. S. Mitrinović, Tableaux qui fournissent des polynômes de Stirling, Publ. Fac. Electrotech. Univ. Belgrade, Ser. Math. Phys. 34 (1960), 1–23.

5. N. E. Nörlund, Vorlesungen über Differenzenrechnung, Springer, Berlin, 1924; Reprinted by Chelsea, New York, 1954.

6. Sloane, Online Encylopedia of Integer Sequences, http://oeis.org/

7. G. Tenenbaum, Introduction à la théorie analytique et probabiliste des nombres, Belin, 2008.

Faculty of Mathematics (Received 18 09 2014) USTHB Algiers Algeria

    fbencherif@usthb.dz
    tgarici@usthb.dz