Hindawi Publishing Corporation International Journal of Mathematics and Mathematical Sciences Volume 2015, Article ID 982812, 7 pages http://dx.doi.org/10.1155/2015/982812
Research Article The Peak of Noncentral Stirling Numbers of the First Kind Roberto B. Corcino,1 Cristina B. Corcino,1 and Peter John B. Aranas2 1
Mathematics and ICT Department, Cebu Normal University, 6000 Cebu City, Philippines Department of Mathematics, Mindanao State University, Main Campus, 9700 Marawi City, Philippines
2
Correspondence should be addressed to Roberto B. Corcino;
[email protected] Received 18 September 2014; Accepted 20 November 2014 Academic Editor: Serkan Araci Copyright Β© 2015 Roberto B. Corcino et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. We locate the peak of the distribution of noncentral Stirling numbers of the first kind by determining the value of the index corresponding to the maximum value of the distribution.
1. Introduction
and have initial conditions
In 1982, Koutras [1] introduced the noncentral Stirling numbers of the first and second kind as a natural extension of the definition of the classical Stirling numbers, namely, the expression of the factorial (π₯)π in terms of powers of π₯ and vice versa. These numbers are, respectively, denoted by π π (π, π) and ππ (π, π) which are defined by means of the following inverse relations: π
1 ππ [ π (π‘)V ] (π‘ β π)π , π! ππ‘ π‘=π π=0
(π‘)π = β
(1)
π
1 π [Ξ (π‘ β π)π ]π‘=0 (π‘)π , π! π=0
(π‘ β π)π = β
(2)
where π, π‘ are any real numbers, π is a nonnegative integer, and π π (π, π) = ππ (π, π) =
π
1 π [ (π‘) ] , π! ππ‘π V π‘=π 1 π [Ξ (π‘ β π)π ]π‘=0 . π!
(3)
π π (0, 0) = 1, ππ (0, 0) = 1,
π π (π, 0) = (π)π , ππ (π, 0) = (βπ)π ,
π π (0, π) = 0, ππ (0, π) = 0,
π, π =ΜΈ 0, π, π =ΜΈ 0. (6)
It is worth mentioning that for a given negative binomial distribution π and the sum π = π1 + π2 + β
β
β
+ ππ of π independent random variables following the logarithmic distribution, the numbers π π (π, π) appeared in the distribution of the sum π = π + π, while the numbers ππ (π, π) Μ+π Μ where appeared in the distribution of the sum π = π Μ π is the sum of π independent random variables following Μ is the truncated Poisson distribution away from zero and π a Poisson random variable. More precisely, the probability distributions of π and π are given, respectively, by π [π = π] =
ππ (β1)πβπ π βπ (π, π) , π π! βπ (1 β π) (β log (1 β π))
π [π = π] =
ππ (β1)πβπ πβπ/π (π, π) . π π! π π π (π β 1)
π!
π!
(7)
The numbers satisfy the following recurrence relations: π π (π + 1, π) = π π (π, π β 1) + (π β π) π π (π, π) ,
(4)
ππ (π + 1, π) = ππ (π, π β 1) + (π β π) ππ (π, π)
(5)
For a more detailed discussion of noncentral Stirling numbers, one may see [1]. Determining the location of the maximum of Stirling numbers is an interesting problem to consider. In [2], Mez¨o
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International Journal of Mathematics and Mathematical Sciences
obtained results for the so-called π-Stirling numbers which are natural generalizations of Stirling numbers. He showed that the sequences of π-Stirling numbers of the first and second kinds are strictly log-concave. Using the theorem of ErdΒ¨os and Stone [3] he was able to establish that the largest index for which the sequence of π-Stirling numbers of the first kind assumes its maximum is given by the approximation (1) = π + [log ( πΎπ,π
πβ1 1 ) β + π (1)] . πβ1 π
(8)
Following the methods of Mez¨o, we establish strict logconcavity and hence unimodality of the sequence of noncentral Stirling numbers of the first kind and, eventually, obtain an estimating index at which the maximum element of the sequence of noncentral Stirling numbers of the first kind occurs.
2. Explicit Formula In this section, we establish an explicit formula in symmetric function form which is necessary in locating the maximum of noncentral Stirling numbers of the first kind. Let ππ (π₯), π = 1, 2, . . . , π be differentiable functions and let πΉπ (π₯) = βππ=1 ππ (π₯). It can easily be verified that, for all π β₯ 3, πΉπσΈ (π₯) =
β 1β€π1