x(x 1)(x 2)... (x k + 1) = [x] k n+m 1


 Marvin Dawson
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1 1 Coutig mappigs For every real x ad positive iteger k, let [x] k deote the fallig factorial ad x(x 1)(x 2)... (x k + 1) ( ) x = [x] k k k!, ( ) k = 1. 0 I the sequel, X = {x 1,..., x m }, Y = {y 1,..., y }. If we write X (resp., Ỹ ), that meas that we caot distiguish the elemets of X (resp., Y ). Propositio 1 The total umber of all mappigs f : X Y is m ad the umber of ijective mappigs f : X Y is [] m. Propositio 2 The umber of bijective mappigs f : X Y is! whe = m ad 0 otherwise. Propositio 3 The umber of ijective mappigs f : X Y is ( ) m. PROOF. Associate with every ijective mappig f : X Y the image, f(x), of X. By Propositio 1, the umber of ijective mappigs f : X Y is [] m. By Propositio 2, every subset Z of Y with Z = m is the image of X for m! ijective mappigs, ad these mappigs are idistiguishable i the model f : X Y. Thus, the umber i questio is [] m /m! = ( ) m. Propositio 4 The umber of surjective mappigs f : X Y is ( ) m 1 1. The total umber of all mappigs f : X Y is ( ) +m 1 1. PROOF. Assig a set of 1 vertical strips betwee m poits poits o a lie to every surjective mappig f : X Y as follows. If f 1 (y i ) = a i, the put the ith strip betwee the poits with the umbers a a i ad a a i +1. Observe that this is a bijectio betwee the set of surjective mappigs f : X Y ad the set of arragemets of 1 strips betwee m poits o a lie such that each iterval betwee two cosecutive poits ca cotai at most oe strip. But the umber of such arragemets is ( ) m 1 1. Similarly, we have a bijectio betwee the set of all mappigs f : X Y ad the set of all possible arragemets of 1 strips betwee m poits o a lie. Ad the umber of such arragemets is ( ) +m 1 1. The umber S(m, k) of the partitios of a melemet set ito k oempty subsets is a Stirlig umber of the secod kid. Propositio 5 (1) The umber of surjective mappigs f : X Ỹ is S(m, ). (2) The umber of surjective mappigs f : X Y is!s(m, ). (3) The umber of all mappigs f : X Ỹ is S(m, k). 1
2 PROOF. (1) is easy. Every surjective mappig f : X Y ca be costructed i two steps. O the first oe we partitio X ito parts that will be mapped ito the same elemet of Y (by the defiito, there is S(m, ) ways to do this), ad o the secod step we choose the image for every of these parts (! ways). This proves (2). Now, (3) follows from (1). Propositio 6 (1) S(m, m) = 1 for m 0. (2) S(m, 0) = 0 for m > 0. (3)S(m, k) = S(m 1, k 1) + ks(m 1, k) for 0 < k < m. PROOF. (1) ad (2) are evidet. To see (3), observe that the set R of all partitios of X = {x 1,..., x m } ito oempty parts is the uio of R 1 = {π R {x m } is a part of the partitio } ad R 2 = R\R 1. Clearly, R 1 = S(m 1, k 1). Assig every π R 2 the partitio π of X\{x m }, obtaied from π by deletig x m. Notice that every partitio π of X\{x m } ito k parts is assiged to k differet partitios of X. This shows that R 2 = ks(m 1, k). This propositio lets us cout S(m, k) recursively: m \ k Propositio 7 For every k 2 S(m, k) = m 1 i=k 1 ( ) m 1 S(i, k 1). i PROOF. Let (X 1,..., X k ) be a arbitrary partitio of X ito k oempty parts ad x m X k. There are ( ) m 1 r 1 ways to choose Xk with X k = r ad x m X k. Now there are S(m r, k 1) ways to partitio the rest ito k 1 oempty parts. Hece, S(m, k) = m k+1 r=1 ( ) m 1 S(m r, k 1) = r 1 Theorem 8 For every iteger m 0, m 1 i=k 1 x m = S(m, k)[x] k. 2 ( ) m 1 S(i, k 1). i
3 PROOF. Let be a positive iteger. By Propositio For every A Y, A, let X A be the set of mappigs f : X Y such that f(x) = A. If A = k, the X A = k!s(m, k). Hece by Propositio 5 (3), m = ( k)s(m, k)k! = S(m, k)[] k. Thus the values of the polyomials x m ad m S(m, k)[x] k coicide for x = 1, 2, 3,.... It follows that this is the same polyomial. Now, cosider mappigs X ito Ỹ. I order to distiguish two such mappigs f 1 ad f 2, we eed the sets { f1 1 (y 1 ), f1 1 (y 2 ),..., f1 1 (y ) } ad { f2 1 (y 1 ), f2 1 (y 2 ),..., f2 1 (y ) } to be differet. Thus, the umber R(m, ) of all mappigs of X ito Ỹ is the umber of partitios of the umber m ito the sum of oegative itegers. Let P (m, k) deote the umber of partitios of the umber m ito the sum of k positive itegers, ad P (m) deote the umber of partitios of the umber m ito the sum of positive itegers. Clearly, P (m) = P (m, k) m > 0, R(m, ) = P (m, k). Helpful tools i studyig P (m, k) are Ferrer s diagrams. A Ferrer s diagram for the partitio m = a 1 + a a k (assumig a 1 a 2... a k ) cosists of k rows, ad the ith row cotais a i poits i colums 1,..., a i. The example below shows the Ferrer s diagram for the partitio 16 = Ferrer s diagrams imply the followig propositio. Propositio 9 (1) The umber of partitios of m ito the sum of exactly k positive itegers equals the umber of partitios of m ito the sum of positive itegers the maximum of which is k. (2) P (m, ) = P (m, 1) + P (m, 2) P (m, ). Now we prove a deeper fact Propositio 10 The umber of partitios of m ito the sum of distict positive itegers equals the umber of partitios of m ito the sum of odd positive itegers. 3
4 PROOF. Let Q be the set of partitios of m ito the sum of distict positive itegers ad P be the set of partitios of m ito the sum of odd positive itegers. Costruct the mappig F : Q P as follows. Let m = a 1 + a a k ad a 1 > a 2 >... > a k. Write every a i i the form a i = 2 c i b i, where b i is odd. There is oly oe way to do so. Map the partitio m = a 1 + a a k to the partitio m = b 1 + b b }{{ 1 + b } 2 + b b b }{{} k + b k b k. }{{} 2 c 1 times 2 c 2 times 2 c k times This is our mappig F. Now we costruct the mappig G : P Q as follows. Let a partitio m = b 1 + b b }{{ 1 + b } 2 + b b b }{{} s + b s b }{{} s d 1 times d 2 times d s times of m ito the sum of odd positive itegers be give, ad let there be exactly s distict summads. Every of d i s ca be writte i the form Now the partitio d i = 2 e i,1 + 2 e i, e i,t(i), e i,1 > e i,2 >... > e i,t(i). (2) s m = (2 e i,1 b i + 2 e i,2 b i e i,t(i) b i ) (3) i=1 cosists of distict summads. Ideed, for the same i, the summads are distict sice e i,j are distict, ad for distict i, the maximum odd divisors are distict. Let G map every partitio of the kid (1) to the partitio of the kid (3). Observe that if we apply G to a partitio of the kid (1) ad the apply F to the obtaied partitio, the we get the iitial partitio. Sice the form (2) is uiquely determied, F maps differet partitios of m ito the sum of distict positive itegers ito differet partitios of m ito the sum of odd positive itegers. It follows that F ad G are bijectios ad P = Q. Part of the obtaied results is summarized i the followig table. All Ijective Surjective Bijective mappigs mappigs mappigs mappigs X Y m [] m!s(m, )! (1) X Y ( ) ( ) ( ) +m 1 m 1 m m 1 1 X Ỹ X Ỹ S(m, k) 0, if m > S(m, ) 1 1, if m P (m, k) 0, if m > P (m, ) 1 1, if m 4
5 2 Catala s umbers The Catala s umbers {c } =0 are defied as follows: c 0 = 1 ad for every > 0, c := c 0 c 1 + c 1 c c 1 c 0. (4) The sequece begis: 1, 1, 2, 5, 14, 42, 132, 429,... Example 1. A plaar rooted tree is a tree with a special vertex (called the root ) where the sos of every vertex are liearly ordered. To every plaar rooted tree T with the root r, we ca correspod the pair (T 1, T 2 ) obtaied from T by deletig the edge (r, s) coectig r with the oldest so s. We assume that T 1 cotais r, ad T 2 cotais s. It is ot hard to check that this is a 11 correspodece. Hece, for the umber t of plaar rooted trees with edges, we have t 0 = 1 ad for every > 0, t := t 0 t 1 + t 1 t t 1 t 0. I other words, t = c for every. Example 2. Let a plaar rooted tree T be placed with root up o the plae ad so that the sos of every vertex are placed from left to right i their order. The the walk aroud this tree startig from the edge rs ad leavig the tree o the left had all the time correspods to a sequece of 1s ad 1s as follows: whe we come from a father to a so, we put 1, otherwise, we put 1. We will get a sequece of oes ad miusoes, where for every 1 i 2, the sum of the first i etries is oegative. Clearly, every sequece with these properties correspods to plaar rooted tree with edges, ad this correspodece is 11. It follows that the umber p of such sequeces is also c. To calculate c, write p i the form p = w v, where w is the umber of all sequeces with oes ad miusoes, ad v is the umber of sequeces (a 1,..., a 2 ) of oes ad miusoes such that i : a a i 1. (5) Clearly, w = ( ) 2. I order to evaluate v, we correspod to every sequece A = {a j } 2 j=1 i the family V of sequeces of oes ad miusoes satisfyig (5), the sequece F (A) of + 1 oes ad 1 miusoes as follows. Let i(a) be the smallest positive iteger i, such that a a i = 1. The we get F (A) by chagig i A the sigs of the first i(a) members. Oe ca observe (please, try it) that so defied F is a 11 correspodece betwee V ad the set Z of all sequeces of + 1 oes ad 1 miusoes. Cosequetly, V = Z = ( 2 1 ) c = p = ( ) ( ) 2 2 = ( 2. ) ad thus Now we will derive this formula agai usig the geeratig fuctio C(t) of these umbers which we will fid by the defiitio (4): C(t) = c t = 1 + (c 0 c c 1 c 0 )t = =0 =1 5
6 = 1 + t (c 0 c c c 0 )t = 1 + t(c(t)) 2. =0 Solvig the quadratic with respect to C(t), we obtai C(t) = 1 ± 1 4t. 2t Sice we kow that C(0) = 1, we must take the mius i the formula. Let us calculate the th derivative of 1 4t: d dt 1 4t = ( ) 1/2!(1 4t) 1/2 ( 4) = 2 ( 1)(2 3)!!(1 4t) 1/2. Hece ad Observe that 1 1 4t = C(t) = 2 (2 1)!! ( + 1)! =0 =1 2 (2 3)!! t! 2 (2 1)!! t. ( + 1)! = (2)!!( + 1)! = 1 ( )
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