On the Numbers Related to the Stirling Numbers of ... - facta universitatis

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In this paper we give the new explicit expression for the numbers cnk treated .... the classical Stirling numbers s(n, k) and S(n, k) are in fact S(n, k; 1, 0, 0) and S(n ...
ˇ FACTA UNIVERSITATIS (NIS) Ser. Math. Inform. Vol. 22, No. 2 (2007), pp. 105–108

ON THE NUMBERS RELATED TO THE STIRLING NUMBERS OF THE SECOND KIND Nenad P. Caki´ c

Abstract. In this paper we give the new explicit expression for the numbers cnk treated by Mijajlovi´c and Markovi´c [Facta Univ. Ser. Math. Inform. 13 (1998), 7–17]. The new combinatorial identities are also given.

1.

Introduction

In this paper, we consider the numbers cnk , which may be defined by the operational relation n X (xα+1 D)n = cnk xnα+k Dk , k=1

and by the recurrence relation (triangular or Pascal-type relation) cn+1k = cnk−1 + (nα + k)cnk , 1 ≤ k ≤ n, c11 = 1, cnk = 0 for k > n, where α ∈ R (see [5]). It is clear that for α = 0, cnk are the usual Stirling numbers of the second kind. The aim of this paper is to prove the new explicit expression for the numbers cnk . Also, we give some additional remarks and comments. 2.

Main Result

In [5] Mijajlovi´c and Markovi´c studied the numbers cnk and proved that (x| − α)n =

n X

cnk (x|1)k ,

k=0

where n ∈ N, α ∈ R and (x|α)n denotes the generalized factorial of the form (x|α)n =

n−1 Y

(x − jα), n ≥ 1, (x|α)0 = 1.

j=0

Received June 20, 2007. 2000 Mathematics Subject Classification. Primary 11B73, 11B75; Secondary 05A19.

105

106

Nenad P. Caki´c

and in particular (x|1)n = (x)(n) = x(x − 1) · · · (x − n + 1), (x)(0) = 1. Theorem. The explicit expression for the numbers cnk defined in [5] is given by:

! ! 1+α (n − 1)(1 + α) − i1 − · · · − in−2 ··· . i1 in−1

X

(2.1) cnk =

i1 + · · · + in−1 = n − k ij ∈ {0, 1}

Proof. Starting from (2.1), for in ∈ {0, 1} we get cn+1k = ! ! 1+α n(1 + α) − i1 − · · · − in−1 ··· i1 in

X i1 + · · · + in = n + 1 − k ij ∈ {0, 1}

! ! 1+α (n − 1)(1 + α) − i1 − · · · − in−2 ··· + i1 in−1

X

=

i1 + · · · + in = n + 1 − k ij ∈ {0, 1}

X

(nα + k)

i1 + · · · + in−1 = n − k ij ∈ {0, 1}

! ! 1+α (n − 1)(1 + α) − i1 − · · · − in−2 ··· i1 in−1

i.e., cn+1k = cnk−1 + (nα + k)cnk , because n(1 + α) − i1 − · · · − i(n − 1) = n(1 + α) − n + k = nα + k. This completes the proof.

2

Corollary. Unsigned Lah numbers, L(n, k) (see, for example [2]and [3] ) defined by triangular recurrence relation L(n, k) = L(n − 1, k − 1) + (n + k − 1)L(n − 1, k), L(n, n) = L(n, 1) = 1, L(n, k) = 0 for n < k, have the explicit representation

L(n, k) = X i1 + i2 + · · · + in−1 = n − k ij ∈ {0, 1}

2 i1

!

4 − i1 i2

!

6 − i1 − i2 i3

This expression follows from (2.1) for α = 1.

!

! 2(n − 1) − i1 − · · · − in−2 ··· . in−1

107

On the Numbers Related to the Stirling Numbers of the Second Kind Naturally, it is well-known that (see [3]) n! L(n, k) = k!



n−1 k−1

« .

So, the combinatorial identity

n! n − 1 k! k − 1 X

! = 2 i1

i1 + · · · + in−1 = n − k ij ∈ {0, 1}

!

4 − i1 i2

!

6 − i1 − i2 i3

!

2(n − 1) − i1 − · · · − in−2 ··· in−1

!

is valid. Remark 2.1. Explicit expression for the numbers cnk from [5] is

cnk =

(2.2)

! n 1 X k (−1)k−i i(i + α) · · · (i + (n − 1)α), k! i k=1

and therefore, if we identify the right sides of (2.1) and (2.2), we get the following combinatorial identity

X i1 + · · · + in−1 = n − k ij ∈ {0, 1}

1+α i1

!

2(1 + α) − i1 i2

!

(n − 1)(1 + α) − i1 − · · · − in−2 ··· in−1

!

! n 1 X k−i k = (−1) i(i + α) · · · (i + (n − 1)α). k! i k=1

Remark 2.2. For α = 0 we get the explicit formula for the Stirling numbers of the second kind (see [1] and comment by L. Carlitz therein)

S(n, k) =

X i1 + · · · + in−1 = n − k ij ∈ {0, 1}

1 i1

!

2 − i1 i2

!

! n − 1 − i1 − · · · − in−2 ··· . in−1

Remark 2.3. The numbers cnk are the special case of a generalized Stirling numbers S(n, k; α, β, γ), introduced by Hsu and Shiue ([4]). Namely, cn,k = S(n, k; −α, 1, γ), where the following relations holds (x|α)n =

n X

S(n, k; α, β, γ)(x − γ|β)k ,

k=0

S(n + 1, k; α, β, γ) = S(n, k − 1) + (kβ − nα + γ)S(n, k; α, β, γ).

108

Nenad P. Caki´c

Evidently, the classical Stirling numbers s(n, k) and S(n, k) are in fact S(n, k; 1, 0, 0) and S(n, k; 0, 1, 0) respectively. Also, the binomial coefficients are given by S(n, k; 0, 0, 1). In a forthcoming paper we give an explicit formula for generalized Stirling numbers S(n, k; α, β, γ).

REFERENCES ´: On some combinatorial identities. Univ. Beograd. Publ. Elek1. N. P. Cakic trotehn. Fak. Ser Mat. Fiz. No. 678–715 (1980), 91-94. 2. L. Comtet: Advanced Combinatorics: The Art of Finite and Infinite Expansions. D.Reidel Publ. Co., Holland 1974. 3. R.L. Graham, D.E. Knuth and O. Patashnik: Concrete Mathematics. AddisonWesley, Reading, Mass. 1989. 4. L.C. Hsu and P.J.-S. Shiue: A unified approach to generalized Stirling numbers. Advanced in Applied Math. 20 (1998), 366–384. ˇ Mijajlovic ´ and Z. Markovic ´: Some recurrence formulas related to the dif5. Z. ferential operator θD. Facta Univ. Ser. Math. Inform. 13 (1998), 7–17.

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