FIRST REMARK ON Aζ-ANALOGUE OF THE STIRLING NUMBERS
M. Dziemia´nczuk
Institute of Informatics, University of Gda´nsk, Poland [email protected]
Received: 1/12/10, Revised: 10/9/10, Accepted: 12/9/10, Published: 1/31/11
Abstract
The so-called ζ-analogues of the Stirling numbers of the first and second kind are considered. These numbers cover ordinary binomial and Gaussian coefficients,p, q -Stirling numbers and other combinatorial numbers studied with the help of object selection, Ferrers diagrams and rook theory.
Our generalization includes these and now also thep, q-binomial coefficients. This special subfamily of F-nomial coefficients encompasses among others, Fibonomial ones. The recurrence relations with generating functions of the ζ-analogues are delivered here. A few examples ofζ-analogues are presented.
1. Introduction
Let w = {wi}i≥1 be a vector of complex numbers wi. The generalized Stirling numbers of the first kindCn
k(w) and the second kindS n
k(w) are defined as follows:
Cn k(w) =
�
1≤i1<i2<···<ik≤n
wi1wi2· · ·wik,
Sn k(w) =
�
1≤i1≤i2≤···≤ik≤n
wi1wi2· · ·wik.
(1)
If the elements of the weight vector w are positive integers then the coefficients are interpreted as a selection ofk objects fromk ofnboxes without and with box repetition allowed, respectively. In this case the number of distinct objects in the
s-th box is designated by thes-th element of the weight vector w.
One shows that the numbers Cn
k(w) and Skn(w) cover among others, binomial
binomial coefficients:
Settingwi=qi−1 gives us Gaussian coe
fficients:
In this note, the ordinary Stirling numbers of the first kind are defined in the following way
and the second kind
1 tainp, q-Stirling numbers considered by Wachs and White [12]
p(n2)Sn
which satisfy the following recursive relation
�
We refer the reader also to Wagner [13], M´edicis and Leroux [11].
Notice, that the weight vectorw in the definition of the coefficientsCn
k(w) and Sn
k(w) is constant and independent of the numbern. Theζ-analogueof the Stirling
numbers introduced in the next section do not require this assumption. We define the weight vectorwn(ζ) dependent on the numbernand a complex numberζ.
2. A ζ-analogue of the Stirling Numbers
Take a vectorwn(ζ) ofncomplex numberswiζn−i, wherei= 1,2, ..., n, i.e.,
wn(ζ) =
�
w1ζn−1, w2ζn−2, . . . , wn−1ζ, wn�
. (2)
We write it as wˆn for short and denote thei-th element ofwˆn by ˆwn,i or just ˆwi
for fixedn∈N. We assumew0(ζ) =∅and ˆw0= 0.
It is important to notice, that thej-th element ofwn(ζ) is not equal to thej-th
element ofwm(ζ) whilen�=min general. Indeed,wjζn−j �=wjζm−j.
Definition 1. For anyn, k ∈ N∪{0}the ζ-analogues of the Stirling numbers of
the first kind ˆCn
k(wˆn) and the second kind ˆSkn(wˆn) are defined as follows:
ˆ
Cn k(wˆn) =
�
1≤i1<i2<···<ik≤n
ˆ
wi1wiˆ2· · ·wiˆk, (3a)
ˆ
Sn k(wˆn) =
�
1≤i1≤i2≤···≤ik≤n
ˆ
wi1wiˆ2· · ·wiˆk, (3b)
with ˆCn
0(wˆn) = ˆS0n(wˆn) = 1 due to the empty product.
2.1. Combinatorial Interpretation
If the ˆwi are positive integers, the coefficients ˆCn
k(wˆn) and ˆSkn(wˆn) denote the
number of ways to selectkobjects fromkofnboxes without box repetition allowed and with box repetition allowed, respectively. In this case, the size of thei-th box is designated by the i-th element of wn(ζ) for i = 1,2, ..., n. However, all the
results in this note holds for any vector wˆ of complex numbers and can be proved algebraically.
Theorem 2. For any n, k∈Nwe have
ˆ
Cn
k(wˆn) =wnζk−1Cˆkn−1−1(wˆn−1) + ζkCˆkn−1(wˆn−1), (4a)
ˆ
Sn
k(wˆn) =wnSˆkn−1(wˆn) + ζkSˆkn−1(wˆn−1), (4b)
whereCˆn
0(wˆn) = ˆSn0(wˆn) = 1andCˆsn(wˆn) = 0fors > n,Sˆk0(wˆ0) = 0fork >0.
Proof. The proof uses terms of the combinatorial interpretation of these coefficients, but still holds for any vectorwˆn of complex numbers. In point of fact, we consider
the sums (3a), (3b) and only play with its summations.
Fix a natural numbernand take the weight vectorwˆn =�w1ζn−1, . . . , wn−1ζ, wn�.
selected (ik< n), respectively (repetition of boxes is not allowed)
Observe that we can rewrite the right-hand side of the above as follows:
wn �
(b) In the same way we prove the case with box repetition allowed.
Notation 3. Letn∈Nandm∈N∪{0}. Denote bywˆ(nm) the vector
i1, i2, . . . , ij take on values from the set {1,2, . . . , n} and the remaining (k−j) variables from{n+ 1, . . . , n+m}, i.e.,
ˆ
Cn+m
k (wˆn+m) = k
�
j=0
�
1≤i1<···<ij≤n
wi1ζ
n+m−i1· · ·wi
jζ
n+m−ij
· �
n+1≤ij+1<···<ik≤n+m
wij+1ζ
n+m−ij+1· · ·wi
kζ
n+m−ik.
Finally, we need to correct the powers ofζ’s as follows: �
1≤i1<i2<···<ij≤n
wi1ζ
n+m−i1· · ·wi
jζ
n+m−ij =ζj·mCˆn
j(wˆn).
(b) The same proof remains valid for the coefficients ˆSnk+m(wˆn+m).
This result provides a more general form of the recurrence relation (4a) for the coefficients ˆCn
k(wˆn). Indeed, letting n = n�−1 andm = 1 in the equation (5a)
gives (4a).
Proposition 5. For anyn, k∈Nwe have
ˆ
Cn k(wˆn) =
n
�
j=k
wjζk(n−j+1)−1Cˆkj−1−1(wˆj−1), (6a)
ˆ
Sn k(wˆn) =
n
�
j=1
wjζk(n−j)Sˆkj−1(wˆj). (6b)
Proof. (a) Consider the sum (3a) from the definition of the coefficient ˆCn
k(wˆn) and
separate it into (n−k+ 1) sums where in thej-th one the last variableik is equal to (k+j) forj = 0,1, . . . , n−k, i.e.,
ˆ
Ckn(wˆn) = n−k
�
j=0
�
1≤i1<···<ik−1<ik=k+j wi1ζ
n−i1· · ·wi
kζ
n−ik (7)
=
n
�
j=k
wjζn−j � 1≤i1<···<ik−1≤j−1
wi1ζ
n−i1· · ·wi
k−1ζ
n−ik−1. (8)
Taking out the common factorζ(n−j+1) from (k
−1) factors (wiζn−i) gives
ˆ
Cn k(wˆn) =
n
�
j=k
wjζk(n−j+1)−1 � 1≤i1<···<ik−1≤j−1
wi1ζj−1−i1· · ·wi
k−1ζ
j−1−ik−1
=
n
�
j=k
(b) The second equation (6a) might be handled in much the same way. Observe only that the variablej takes on values from the set{1,2. . . , n}.
Proposition 6. For anyn, k∈Nwe have
ˆ
Cn+1
k (wˆn+1) = k
�
j=0 ˆ
Ckn−−jj(wˆn−j)ζ(j+1)(k−j)+(
j
2)
j−1 �
i=0
wn+1−i, (9a)
ˆ
Sn+1
k (wˆn+1) = k
�
j=0 ˆ
Sn
k−j(wˆn)ζ(k−j)wjn+1. (9b)
Proof. (a) Consider the sum (3a) of ˆCkn+1(wˆn+1) and observe that it may be
sepa-rated into (k+ 1) sums where in thej-th one (j = 0,1,2, . . . , k) we have
1≤i1<· · ·< ik−j≤n−j; ik−j+1=n+ 1−j+ 1, . . . , ik =n+ 1.
(b) In the case of the coefficient ˆSn+1
k (wˆn+1) we may separate the sum (3b) into
(k+ 1) sums where in thej-th one (j= 0,1,2, . . . , k) we have
1≤i1≤· · ·≤ik−j≤n; ik−j+1=ik−j+2=· · ·=ik =n+ 1.
The rest of the proof is straightforward and goes in much the same way as the proofs of Proposition 4 and Proposition 5.
3. Generating Functions
Letn≥0 and define two generating functions:
An(x, y) =
�
k≥0
(−1)kCˆn
k(wˆn)xkyn−k, (10a)
Bn(x) =
�
k≥0 ˆ
Sn
k(wˆn)xk. (10b)
Theorem 7. Forn≥1we have An(x, y) =
n
�
i=1 �
y−wiζn−ix�
, (11a)
Bn(x) = n
�
i=1
1
(1−wiζn−ix), (11b)
Proof. Applying recurrences (4a) and (4b), respectively, shows thatAn(x, y) and
Solving these recurrence relations proves (11a) and (11b).
Corollary 8. For any n, j ∈ N the coefficients Cˆn
the Cauchy product of power series with (11a) and (11b) finishes the proof.
Letf(x) be a series in powers of x. Then by the symbol [xn]f(x) we will mean
Observe that if we evaluate the above withx= (wiζn−i)−1, all summands except
Proposition 9 we obtain the well-known identity for the Stirling numbers of the second kind:
4. Remarks and Examples
It is clear that ˆCn
called a (p, q)-sequence. In the literature, the elements of (p, q)-sequences are called (p, q)-analogues and are denoted bynp,q ≡[n]p,q (see Briggs and Remmel [1]).
Example 11. (p, q-binomial coefficients)
Thep, q-binomial coefficients generalize binomial, Gaussian and Fibonomial coeffi -cients [2, 3, 4] and are defined as
Therefore, if the weight vectorwn(p) takes the form�pn−1, qpn−2, . . . , qn−2p, qn−1�,
one covers the family of p, q-binomial coefficients [2, 3, 4], i.e., ˆ
binatorial interpretations of itsp, q-binomial coefficients: expressed in the language of cobweb posets partitions [9], tilings of hyper-boxes [4] and now as an object selection from weighted boxes.
Example 12. (Fibonomial coefficients)
It is easy to show that the Fibonacci numbers define a (ϕ,ρ)-sequence whereϕ= (1 +√5)/2 and ρ = (1−√5)/2. Therefore, from the previous example, the ζ-analogue also generalize the Fibonomial coefficients, i.e.,
ˆ
binatorial interpretation in terms of object selection cannot be applied in this case vector ϕn does not consist of only nonnegative integers. Fixing s, n ∈ N, from
Corollary 8 we have also
s
Example 13. (p, q-Stirling numbers)
The ζ-analogue generalizes the p, q-Stirling numbers [12]. Indeed, let us consider the vector in(ζ) =�[1]p,qζn−1,[2]p,qζn−2, . . . ,[n]p,q�, where [i]p,q =�
Finally, by Theorem 2 we have that theζ-analogues ofp, q-Stirling numbers satisfy �
The form of the weight vector wn(ζ) given by (2) is one possible choice and we
ˆ
wi,n = win−i, etc. We leave it for further investigation. Our choice is caused by
unifyingp, q-binomial coefficients and generalized Stirling numbers.
Acknowledgements This note was inspired by the paper of Professor Kwa´sniewski [8] where he asked about relations between F-nomial and Konvalina coefficients. The author thanks the referee for useful suggestions and corrections of this note.
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