-The paper is devoted to the elementary discussion on the triangles of Stirling numbers of the first kind and the rStirling numbers as well. Aim of our investigations was to extract the numerical sequences connected with these triangles, to verify their presence in OEIS and to try to generate some properties of these numbers. Moreover, we have supplemented with new results the paper written by S. Falcon [4], the research object of which was the triangle of numbers given by the binomial transformations of k-Fibonacci numbers. Index Terms-Pascal triangle, Stirling numbers of the first kind, r-Stirling numbers of the first kind, Catalan numbers, Fibonacci numbers
I. INTRODUCTION
Creation of this paper was directly influenced by article [17] which intrigued us and, in fact, imposed the subject of our investigations. We became interested in the state of knowledge concerning the Pascal triangles for the Stirling numbers of the first kind and the r-Stirling numbers of the first kind. Thus, as the goal of our research we took the extraction of numerical sequences connected with these triangles (that is mainly the sequences of sums of elements in the rows and along the antidiagonals of the given triangle). Moreover, we verified the presence of such sequences in OEIS and we undertook the attempt to generate some properties of the investigated numbers. It is worth to emphasize that we have also supplemented with completely new results the paper written by Sergio Falcon [4] concerning the triangle of numbers given by the binomial transformations of k-Fibonacci numbers.
II. THE STIRLING TRIANGLE
We begin by presenting the definition of Stirling numbers of the first kind.
Definition 2.1: Stirling numbers of the first kind describe the number of permutations on the n element set possessing k cycles (that is the permutations which may be decomposed into exactly k separated cycles). There is no one standard notation for these numbers and for denoting them one can use one of the following symbols: s(n; k), Sn(k), S1(n; k), n k , where n 2 N, k 2 N0, k n.
In this elaboration we will use the last one from the above listed notations.
We will use here one more, alternative, definition of the Stirling numbers of the first kind nk , it means: h n i = sum of all possible products of (n k k) different integers taken from among the n - initial nonnegative integers, that is from among numbers 0; 1; : : : ; n 1.
So we have n 2 1 2 h n i 0 Y 0; i = h n i
1 n 1 X (n k=1 (n 1)! 1)!; = (n 1)!Hn 1; k=1 1 i n 1
i6=k n(n 1) = A000217(n 1); n = 1; 2; : : : where Hn 1 denotes the (n 1)th harmonic number, that is Hn := Pkn=1 k1 . Moreover we take nn := 1 for every n 2 N.
Let us notice that from this definition, almost immediately after multiplication of the monomials located on the left hand side, the following equality results x (1 + x)(2 + x) : : : (n 1 + x) = xn := k=0 n X h n i k
x ; (
For comparison we have the following widely known Newton binomial formula (1 + x)n = Xn n
k k=0 xk; describing the generating function for the binomial coefficients.
Let us notice that by using definition 2.1 one can easily derive the formula (recurrence relation for the Stirling numbers of the first kind): h n i k = 1)
1
(
Let us construct now the Pascal triangle for the Stirling
numbers of the first kind which will be called henceforward
as the Stirling triangle of the first kind
Hence, as well as on the basis of formula (
n Fn+1 = Xb2 c n k=0 k k ; where Fn denotes the nth Fibonacci number.
1 : : : 1
Next, by executing in the Stirling triangle of the first kind the summation of elements along the antidiagonals, as it is presented in the following scheme (the first column containing zeros and one digit one at the zero level is omitted here): we obtain the sequence of natural numbers labeled by symbol A237653 in the Sloane’s OEIS encyclopaedia. In other words we have h n i + 0 n
1 1 + n
2 2 + : : : + <8 1;(r +w1h)e!n; n = 2r;
: when n = 2r + 1; =A237653(n) for every n = 1; 2; : : :.
Remark 2.1: In the classic Pascal triangle the numbers at
the given level n (starting with the zero level) represent the
coefficients in the expansion of number (b + 1)n in the given
numerical base b (we assume that all the coefficients at level
n are b) which results from the binomial formula (
Whereas in the Stirling triangle of the first kind the numbers
at the given level n (starting with the zero level) represent the
n 1
coefficients in the expansion of number Q (bk + 1) in the
k=0
given numerical base b (under the assumptions that that all the
coefficients at level n are b) which results from formula (
What is interesting, Slone’s OEIS reports the sequences: which form the binomial transforms of rows of the following triangle of k-Fibonacci numbers where Fk;n; k 2 R; k > 0, and n 2 N0 : for every n 2 N. The following results give a supplement for the Falcon’s paper [4]. We find that the polynomials pn(k) satisfy the double recurrence relation pn+1(k) qn+1(k) = = (k + 1)pn(k) + qn(k); pn(k) + qn(k); for every n 2 N, where p1(k) = q1(k) 1; p2(k) = k + 2; q2(k) 2.
Hence, after simple algebra we get the recursive relation for polynomials pn(k) : pn+2(k) = (k + 2)pn+1(k)
kpn(k); pn+1(k) = 1 + kpn(k) + and the recurence relation for pn(k) The triangle T has the form 1 1 2 1 3 4 1 4 8 8 1 5 13 20 1 6 19 38 1 7 26 63 1 8 34 96 1 : : : : : : : : :
We observe also that the sum of elements on antidiagonals of T form three known sequences defined in the OEIS, that is the sequence of all sums
f1; 1; 3; 4; 9; 14; 28; 47; 89; 155; 286; : : :g = A006053 which satisfies the linear recurrence relation of the third order where A006053(n) := an, n = 1; 2; : : : And next the following sequences obtained by bisection of fang : fa2n 1g fa2ng = = f1; 3; 9; 28; 89; 286; : : :g = A094790 f1; 4; 14; 47; 155; : : :g = A094789 which satisfy both the same recurrence relation of the third order an = 5an 1 6an 2 + an 3: Let us set T 2 = [tkn]N N. If n > k then tkn = 0. We also verified that n X tnk k=1
A030240(n 1); where equality holds only for n = 1; 2; 3; 4 and
5
X t5k = A030240(
Remark 2.3: In the context of problems discussed in this section it is also worth to mention the Narayana numbers defined in the way given below (see [6]):
N (n; k) =
N (0; 0) = 1; 1 n n n k k 1 ; N (n; 0) = 0; k; n 2 N; k n:
Creating the Narayana triangle we find one more beautiful result. The sums of elements in the rows of the Narayana triangle are equal to the Catalan numbers n X N (n; k) = Cn: k=1 In result of summation over the antidiagonals of Narayana triangle we get the generalized Catalan numbers n 1 (Cn+1 = Cn + X Ck Cn 1 k; n 2 N0, the sequence k=1 A004148(n) in OEIS), i.e.
b n2 1 c X N (n k=0 k; k + 1) = Cn: III. R-STIRLING TRIANGLE OF THE FIRST KIND Let us notice that = number of all permutations of set f1; 2; : : : ; ng so that the numbers 1; : : : ; r are in different, mutually disjoint cycles.
Moreover, by using the definition of r-Stirling numbers of the
first kind we easily derive the recurrence relation
(
k r
k=0
= (n
for every r 1; which is the generalization of identity (
1)(m + 2); > 2 > > > :>> when n = 2m + 1; n for every n = 2; 3 : : :. Notation A;;;;;k; k 2 N; k 9; with the empty sets, is the notation invented by us to denote the sequences fA;;;;;k(n); n 2 N; g, k 2 N; k 9 not included in OEIS.
We have verified numerically (although basing on premises resulting from the algebraic estimation) that for 5 n 500 the above sequence satisfies the following inequalities we obtain the sequence of natural numbers not existing in OEIS. We have
Remark 3.1: In case of the r-Stirling numbers the numbers
at one level n represent the coefficients in the expansion of
n 1
number Q (bk + 1) in the given numerical base b (under the
k=r
assumption that all the coefficients at level n are b) which
results from formula (
IV. TRIANGLES FOR THE POWERS OF STIRLING NUMBERS
AND THE BINOMIAL COEFFICIENTS
Properties of the classic Pascal triangle are studied in [8] together with the presentation of Fibonacci sequence with the aid of binomial coefficients. Let us turn our attention into the Pascal triangle for the powers of binomial coefficients. For example, for the squares of binomial coefficients we receive : : : : : :
: : : This is the sequence labeled by symbol A051286 in the Sloane’s OEIS encyclopaedia. In other words we have n 2 n 1 2 n 2 2
+ + + : : : 0 > > > :
1 8 > 1; when > > + < (m + 1)2; Summing the elements along the antidiagonals in the analogical triangle for the cubes of binomial coefficients we obtain 1; 2; 9; 29; 92; 343; 1281; 4720; : : :, that is the sequence denoted by A181545 in Sloane’s OEIS encyclopaedia. In other words for every n = 1; 2; : : :.
Moreover, we have checked numerically that for 4 500 the above sequence satisfies the inequalities n n! 5 < A;;;;;3(n) < n! 5 2 :
Next, by summing the elements along the antidiagonals in the Pascal triangle for n 3 we receive the sequence k 0 + 1 = 1; 0 + 1 = 1; 8 + 1 = 9; 216 + 27 = 243; 13824 + 1331 + 1 = 15156; and so on, that is
In the paper the number of elementary properties of the numerical triangles connected with the Stirling numbers and the k-Fibonacci numbers is presented. We linked the discussed sequences with the sequences included in OEIS. The obtained results summarize some state of research. In the future we intend to refer to these numerical triangles from the algebraic side, similarly like, among others, the authors of papers [2], [3], [9], [10] and [14] did it in case of the classic Pascal triangle. In the future we also plan to analyze paper [13] and the drawn conclusions will be included in our next publication. Acknowledgments
The Authors want to express their sincere thanks to the Referees for their constructive reports with many essential remarks and advices concerning the references. [12] T. Koshy, Fibonacci and Lucas Numbers with Applications, Wiley, New
York 2001. [13] W. Lang, On generalizations of the Stirling number triangles, J. Integer
Seq. 3 (2000), Article 00.2.4. [14] B. Lewis, Revisiting the Pascal Matrix, Amer. Math. Monthly 117 (2010), 50–66. [15] G. Capizzi, G. Lo Sciuto, C. Napoli, E. Tramontana, M. Wozniak, Automatic classification of fruit defects based on co-occurrence matrix and neural networks, IEEE Federated Conference on Computer Science and Information Systems (FedCSIS), 861–867. [16] A.Sofo, Finite sums in Pascal’s triangle, The FIbonacci Quarterly 50
No. 4 (2012), 337–345. [17] T. Wright, Pascal’s triangle gets its genes from Stirling numbers of the first kind, College Math. J. 26 No. 5 (1995), 368–371.