On Relations Between the Stirling Numbers of First and Second Kind
aa r X i v : . [ m a t h . G M ] M a r On Relations Between the Stirling Numbers ofFirst and Second Kind
Henrik Stenlund ∗ March 25th, 2019
Abstract
Four new relations have been found between the Stirling numbers offirst and second kind. They are derived directly from recently publishedrelations. Stirling numbers of first and second kind
Mathematics Subject Classification 2010: 11B73
Only a few relations are known to exist between the Stirling numbers of first andsecond kind. Recently two new equations were introduced for this purpose. TheStirling numbers are extensively studied for instance in the field of combinatorics[1], [2], [6]. [7] has displayed a nice historic review of the Stirling numbers.The equations below seem to be the only nontrivial relations between theStirling numbers of first and second kind [4]. S ( m )1 n = n − m X k =0 ( − k (cid:0) n − kn − m + k (cid:1)(cid:0) n − mn − m − k (cid:1) S ( k )2 n − m + k (1) S ( m )2 n = n − m X k =0 ( − k (cid:0) n − kn − m + k (cid:1)(cid:0) n − mn − m − k (cid:1) S ( k )1 n − m + k (2) ∗ The author is with Visilab Signal Technologies in Finland and is grateful for receivingsupport for this work. Visilab Report j,k = max ( k,j )+1 X l =0 S ( j )1 l S ( l )2 k (3) δ k,j = max ( k,j )+1 X l =0 S ( l )1 k S ( j )2 l (4)The following two equations were recently found by the author [8]1 = m X j =1 S ( j )1 m j X k =1 S ( k )2 j (5)1 = m X j =1 S ( j )2 m j X k =1 S ( k )1 j (6)In the following, one is using mostly arithmetic operations in order to derivenew expressions. Therefore it is not necessary to make too strict assumptionson number domains and differentiability etc. In general, the applied indexes ∈ N + and the variable x ∈ C for instance. S ( m )1 i is the Stirling number of first kind. The recurrence relation is the following S ( m )1 n +1 = S ( m − n − nS ( m )1 n (7)and the special values are S (0)1 n = δ n , S ( n )1 n = 1 , S (1)1 n = ( − n − ( n − S ( m )2 i is the Stirling number of second kind. The recursion relation is S ( m )2 i +1 = mS ( m )2 i + S ( m − i (9)and the special values are S (0)2 n = δ n , S (1)2 n = 1 , S ( n )2 n = 1 (10)For both kinds of numbers, they are determined from the recursion relationsabove. Numbers outside the number triangles are zero. The basic features ofthe Stirling numbers of first and second kind can be found in most handbooks,see [3], [4], [5]. 2 Identities For the Stirling Numbers
The equations below were found by the author [8] while establishing the equa-tions (5) and (6). x m = m X j =1 S ( j )1 m j X k =1 S ( k )2 j x k (11) x j = j X m =1 S ( m )2 j m X k =1 S ( k )1 m x k (12)The parameter x can have any value whatsoever. At the point x = 1 one willget the results mentioned above. However, to derive new results one must startfrom the equations (11) and (12). The summation (11) can be opened up to thefollowing polynomial x m = S (1)1 m S (1)2 1 x + S (2)1 m [ S (1)2 2 x + S (2)2 2 x ]+ (13) ... + S ( m )1 m [ S (1)2 m x + S (2)2 m x + ... + S ( m )2 m x m ] (14)By using the known properties of the Stirling numbers one recognizes that thehighest powers x m are eliminated from this equation leaving only powers lowerthan m . 0 = m − X j =1 S ( j )1 m j X k =1 S ( k )2 j x k + S ( m )1 m m − X k =1 S ( k )2 m x k (15)This leads at x = 1 to − m − X j =1 S ( j )1 m j X k =1 S ( k )2 j = m − X k =1 S ( k )2 m (16)Identical steps starting from (12) lead to0 = j − X m =1 S ( m )2 j m X k =1 S ( k )1 m x k + S ( j )2 j j − X k =1 S ( k )1 j x k (17)Correspondingly, at x = 1 one obtains − j − X m =1 S ( m )2 j m X k =1 S ( k )1 m = j − X k =1 S ( k )1 j (18)By differentiating equations (15) and (17), one will get at x = 0 immediately S (1)2 m = − m − X j =1 S ( j )1 m S (1)2 j (19) S (1)1 j = − j − X m =1 S ( m )2 j S (1)1 m (20)3 Discussion
The Stirling numbers of first and second kind form Pascal-type triangles withrather complicated closed-form equations for the numbers themselves. Not somany relations exist between these two kinds of Stirling numbers.The process started from the original equations (11) and (12) containingthe variable x by studying them in detail. The results were evaluated at x =1. Further two equations were produced by differentiating the equations (15)and (17) and evaluating them at x = 0 which clears out all its powers. It isobvious that more complicated relations between the Stirling numbers of firstand second kind can be generated by differentiating the equations (15) and (17)and evaluating at some value. These two polynomial equations seem to be in akey position while generating new equations.The results indicate that the the two kinds of Stirling numbers are connectedin a complicated way and there exist several complex relations. Equations (15),(16), (17), (18), (19) and (20) are believed to be new. References [1]
Jordan, C. : ”On Stirling’s Numbers” , Tohoku Mathematical Journal,First Series 37, p. 254 - 278[2] Bailey, W. N.
Generalised Hypergeometric Series . Cambridge, England:University Press, [3]
Abramowitz, M., Stegun, I.A. : Handbook of Mathematical Functions ,Dover , 9th Edition[4]
Gradshteyn, I.S., Ryzhik, I.M. : Table of Integrals, Series and Products ,Academic Press , 7th Edition[5]
Jeffrey, A., Hui-Hui Dai : Handbook of Mathematical Formulas andIntegrals , Elsevier , 4th Edition[6]
Khristo N. Boyadzhiev : ”Exponential Polynomials, Stirling Numbers,and Evaluation of Some Gamma Integrals” , Hindawi Publishing Corpora-tion Abstract and Applied Analysis Volume , Article ID 168672, 18pages doi:10.1155/2009/168672[7] Mohammad-Noori, M. : ”Some remarks about the derivative operator andgeneralized Stirling numbers”’ arXiv:1012.3948v3 [math.CO] [8] Stenlund, Henrik : ”On Some Relations between the Stirling Numbers ofFirst and Second kind” , International Journal of Mathematics and Com-puter Research, Vol. 07, Issue 03 March-2019, p. 1948-19502019