On the Dn(S) Polynomials

On the Dn(s) polynomials

Aleksandar Petojevi´c University of Novi Sad, Teacher Training Faculty Podgoriˇcka 4, 25000 Sombor SERBIA and MONTENEGRO E-mail: [email protected]

Abstract In this paper we study a sequence of polynomials

n m m s Dn(s) = n! (−1) , (n ∈ N0). m! mX=0

We give connection between {vMm(s; a, r)}r∈N functions and Dn(s) polynomials, where vMm(s; a, z) is the function deﬁned in [8].

1 Introduction and Deﬁnition

In 1971 Kurepa (see [4, 5]) deﬁned so-called the left factorial !n by:

n−1 !0 = 0, !n = k! (n ∈ N) Xk=0

and extended it to the complex half-plane Re (z) > 0 as

+∞ z t − 1 t !z = e− dt. Z0 t − 1

2000 Mathematics Subject Classiﬁcation : Primary 11B34. Key words and phrases : Dn(s) polynomials, vMm(s; a, z) function, factorial function.

1 Such function can be also extended analytically to the whole complex plane by !z =!(z + 1) − Γ(z + 1), where Γ(z) is the gamma function defined by +∞ z t Γ(z) = t −1e− dt (Re (z) > 0). Z0 Milovanović in [6] defined and studied a sequence of the factorial functions +∞ {Mm(z)}m=−1 where M−1(z) = Γ(z) and M0(z) = !z. Namely,

+∞ z+m t − Qm(t, z) −t Mm(z) = m+1 e dt (Re (z) > −(m + 1)), (1.1) Z0 (t − 1) where the polynomials Qm(t; z), m = −1, 0, 1, 2, . . ., are given by m m + z k Q−1(t, z) = 0, Qm(t, z) = (t − 1) . k Xk=0 For m = −1, 0, 1, 2, ... and Re (z) > v−m−2 in [8] is given the generalization of Milovanovi´c’s factorial function:

v k−1 z + m + 1 − k vMm(s; a, z) = (−1) L[s; 2F1(a, k − z, m + 2; 1 − t)], m + 1 Xk=1 (1.2) where v is a positive integer, s, a, z are complex variables. The hypergeo- metric function 2F1(a, b, c; x) is deﬁned by the series

∞ n (a)n(b)n x 2F1(a, b, c; x) = (|x| < 1), (c)n n! nX=0 and has integral representation 1 Γ(c) b−1 c−b−1 −a 2F1(a, b, c; x) = t (1 − t) (1 − tx) dt, Γ(b)Γ(c − b) Z0 in the x plane cut along the real axis from 1 to ∞, if Re (c) > Re (b) > 0. The symbols (z)n and L[s; F (t)] represent the Pochhammer symbol Γ(z + n) (z)0 = 1, (z)n = z(z + 1)...(z + n − 1) = , Γ(z) and Laplace transform ∞ st L[s; F (t)] = e− F (t)dt. Z0

2 This function is of interest because its special cases include:

Mm(z) = 1Mm(1, 1, z), Γ(z) = 1M−1(1; 1, z), (1.3)

!z = 1M0(1; 1, z), An = nM−1(1; 1, n + 1), (1.4)

z = 1Mm−1(1; 0, z), m

m k s(m, k) · z = 1Mm−1(1; 0, z − m + 1) · 1M1(1; 1, m + 1) (m ∈ N), Xk=0 z where An, m and s(n, m) are the alternating factorial numbers, the figured number (see [2]) and the Stirling number of the first kind respectively, are defined as n n−k An = (−1) k!, Xk=1

z z + m − 1 = , m m

n k x(x − 1)...(x − n + 1) = s(n, k)x . Xk=0

However, apart from n!, !n and An twenty-ﬁve more well-known integer sequences in [9] are special cases of the function vMm(s; a, z).

2 Basic deﬁnition and properties

We now introduce following polynomials:

Deﬁnition 2.1 For n ∈ N0 the polynomials Dn(s) deﬁned by

n m m s Dn(s) = n! (−1) . m! mX=0

The ﬁrst ﬁve Dn(s) polynomials are listed below.

D0(s) = 1

D1(s) = −s + 1

3 2 D2(s) = s − 2s + 2

3 2 D3(s) = −s + 3s − 6s + 6

4 3 2 D4(s) = s − 4s + 12s − 24s + 24

Special cases include

Dn(0) = n! , Dn(1) = Dn , and sequences in [9]: A010843, A000023, A000522, A010842 ... , where Dn is the derangement number (sequence A000166 in [9]). Since n m n−1 m m s n m s n! (−1) = (−s) + n! (−1) m! m! mX=0 mX=0 this polynomial satisﬁes the recurrence relation

n D0(s) = 1, Dn(s) = nDn−1(s) + (−s) , n ∈ N . (2.5)

The well-known relation

n m −x x Γ(n + 1, x) = n! · e , (n ∈ N0) m! mX=0 yields Γ(n + 1, −s) Dn(s) = , (2.6) es where Γ(z, x), the incomplete gamma function, is deﬁned by

+∞ z t Γ(z, x) = t −1e− dt. Zx

We now establish a connection between vMm(s; a, z) function and the polynomials Dn(s).

Theorem 2.2 For r ∈ N we have

r−1 r − 1 Dn(s) 1M−1(s; 1, r) = , (Re (s) > 0) . n sn+1 nX=0

4 Proof . Since

1M−1(s; 1, r) = L[s; 2F1(1, 1 − r, 1; 1 − t)] we have ∞ ∞ n −st (1)n(1 − r)n (1 − t) 1M−1(s; 1, r) = e · dt (1)n n! Z0 nX=0

∞ r−1 st (1 − r)n n = e− · (1 − t) dt n! Z0 nX=0

r−1 ∞ (1 − r)n st n = e− (1 − t) dt . n! Xn=0 Z0

Now use αs z e Γ(z, αs) L[s; (t + α) −1] = (Re (s) > 0) sz and n Γ(z) (1 − z)n = (−1) Γ(z − n) to obtain

r−1 n n (r − 1)! (−1) Γ(n + 1, −s) 1M−1(s; 1, r) = (−1) · . (r − n − 1)! · n! essn+1 nX=0 The relation (2.6) produces

r−1 r − 1 Dn(s) 1M−1(s; 1, r) = . n sn+1 nX=0

Let ∞ 1 ζ(z) = (Re (z) > 1) kz Xk=1 Riemann Zeta-function. Then

Lemma 2.3 For r ∈ N we have ∞ r−1 1 r − 1 Dn(s) ζ(r) = . (r − 1)! n sn+1 Xs=1 Xn=0

5 Proof. Since

z−1 2F1(m + 2, 1 − z, m + 2, 1 − t) = t we have Γ(z) L[s; 2F1(m + 2, 1 − z, m + 2, 1 − t)] = . sz Hence, applying Theorem 2.2 we have

∞ r−1 1 r − 1 Dn(s) ζ(r) = . (r − 1)! n sn+1 Xs=1 Xn=0

Question 2.4 For z ∈ C is it correct that

∞ ∞ 1 z Dn(k) ζ(z) = ? Γ(z + 1) n kn Xk=1 nX=0 3 Acknowledgements

This work was supported in part by the Serbian Ministry of Science, Tech- nology and Development under Grant # 2002: Applied Orthogonal Systems, Constructive Approximation and Numerical Methods.

References

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[8] A. Petojevi´c, The function vMm(s; a, z) and some well-known sequences, Journal of Integer Sequences, Vol.5, (2002), Article 02.1.7, 1–16. [9] N.J.A.Sloane, The On-Linea Encyclopedia of Integer Sequence, pub- lished elec. at http://www.research.att.com/~njas/sequences/