2 3 47 6 23 Journal of Intgr Squncs, Vol. 6 (2003), Articl 03..2 Drangmnts and Applications Mhdi Hassani Dpartmnt of Mathmatics Institut for Advancd Studis in Basic Scincs Zanjan, Iran mhassani@iasbs.ac.ir Abstract In this papr w introduc som formulas for th numbr of drangmnts. Thn w dfin th drangmnt function and us th softwar packag MAPLE to obtain som intgrals rlatd to th incomplt gamma function and also to som hyprgomtric summations. Introduction and motivation A prmutation of S n = {, 2, 3,, n} that has no fixd points is a drangmnt of S n. Lt D n dnot th numbr of drangmnts of S n. It is wll-known that D n = n! n ( ) i, () D n = n! ( dnots th narst intgr). (2) W can rwrit (2) as follows: D n = n! + 2. W can gnraliz th abov formula rplacing by vry m [, ]. In fact w hav: 2 3 2
2 Thorm. Suppos n is an intgr, w hav { n! D n = + m, n is odd, m [0, ]; 2 n! + m 2, n is vn, m 2 [, ]. (3) 3 For a proof of this thorm, s Hassani [3]. anothr proof of it. At th nd of th nxt sction w giv On th othr hand, th ida of proving (2) lads to a family of formulas for th numbr of drangmnts, as follows: w hav n! D n (n + ) + (n + )(n + 2) + (n + )(n + 2)(n + 3) +. Lt M(n) dnot th right sid of abov inquality. W hav and thrfor M(n) < (n + ) + (n + ) + = 2 n, Also w can gt a bttr bound for M(n) as follows and similarly D n = n! + (n 2). (4) n M(n) < n + ( + (n + 2) + (n + 2) 2 + ) = n + 2 (n + ) 2, D n = n! + n + 2 (n 2). (5) (n + ) 2 Th abov ida is xtnsibl, but bfor xtnding w rcall a usful formula (s [2, 3]). For vry positiv intgr n, w hav n n! = n!. (6) 2 Nw familis and som othr formulas Thorm 2. Suppos m is an intgr and m 3. distinct objcts (n 2) is Th numbr of drangmnts of n (n + m 2)! n + m D n = ( + (n + m 2)! (n + m )(n + m )! + )n! n!. (7)
3 Proof: For m 3 w hav n! D n < (n + ) ( + (n + 2) ( + (n + m ) ( n + m n + m ) )). Lt M m (n) dnot th right sid of th abov inquality; w hav (n + )(n + 2)(n + 3) (n + m )M m (n) = (n + 2)(n + 3) (n + m ) + (n + 3) (n + m ) + + (n + m ) + n + m n + m, and dividing by (n + )(n + 2)(n + 3) (n + m ) w obtain Thrfor n+m 2 n + m M m (n) = n!( (n + m )(n + m )! + ). i=n+ D n = n! n+m 2 + n!( n + m (n + m )(n + m )! + ). (8) Now considr (6) and rwrit (8) by using n+m 2 i=n+ complt. Corollary 2.2 For n 2, w hav Proof: W giv two proofs. Mthod. Bcaus (7) holds for all m 3, w hav i=n+ = n+m 2 n. Th proof is D n = ( + )n! n!. (9) + m 2)! n + m D n = lim ( (n + m (n + m 2)! (n + m )(n + m )! + )n! n! = ( + )n! n!. Mthod 2. By using (6), w hav n M(n) = n!( ) = n! n! = {n!} (n, { } dnots th fractional part), and th proof follows. Now and if w put M (n) = n and M 2(n) = n+2 (n+) 2 lim M m(n) = M(n), m M m+ (n) < M m (n) (n ). (s formulas (4) and (5)), thn Now w find bounds sharpr than {n!} for n! D n and consquntly anothr family of formulas for D n. This family is an xtnsion of (9).
4 Thorm 2.3 Suppos m is an intgr and m. distinct objcts (n 2) is Th numbr of drangmnts of n Proof: Sinc m w hav n! D n ( ) n+ = n! {(n + 2m)!} D n = ( + (n + 2m)! m i= n + 2i (n + 2i)! m ( (n + 2i )! (n + 2i)! ) < n!( i= i= + )n!. (0) n + 2i (n + 2i)! + i=2m+ Lt N m (n) dnot th right mmbr of abov inquality. Considring (6), w hav N m (n) = n!( m i= n + 2i (n + 2i)! + {(n + 2m)!} ), (n + 2m)! and for (n 2), D n = n! + N m (n). This complts th proof. Corollary 2.4 For all intgrs m, n, w hav (n + i)! ). N m+ (n) < N m (n), N (n) < {n!}. Thrfor w hav th following chain of bounds for n! D n n! D n < < N 2 (n) < N (n) < {n!} < < M 2 (n) < M (n) < (n 2). Qustion. Can w find th following limit? lim N m(n). m Bfor going to th nxt sction w giv our proof of Thorm. Th ida of prsnt proof is hiddn in Apostol s analysis [], whr h provd th irrationality of by using (). And now, Proof: (Proof of Thorm ) Suppos k b an intgr, w hav 0 < 2k ( ) i < (2k)! () so, for vry m, w hav m < (2k )! + m 2k ( ) i (2k )! < m + 2
5 if 0 m 2, thn 2k ( ) i (2k )! Similarly sinc (), for vry m 2 w hav (2k )! = + m. Now, if m 2 3, thn m 2 < (2k)! 0 < (2k)! thrfor, if 3 m 2, w obtain + m 2 + m 2 2k 2k ( ) i (2k)! ( ) i (2k)! < m 2. This complts th proof. 2k ( ) i (2k)! = (2k)! + m 2. In th nxt sction thr ar som applications of th provn rsults. 3 Th drangmnt function, incomplt gamma and hyprgomtric functions Lt s find othr formulas for D n. Th computr algbra program MAPLE yilds that and D n = ( ) n hyprgom([, n], [ ], ), D n = Γ(n +, ), whr hyprgom([, n], [ ], ) is MAPLE s notation for a hyprgomtric function. Mor gnrally, hyprgom([a a 2 a p ], [b b 2 b q ], x) is dfind as follows (s [4]), [ ] a a 2 a p pf q ; x = t b b 2 b k x k q k 0 whr t k+ t k = (k + a )(k + a 2 ) (k + a p ) (k + b )(k + b 2 ) (k + b q )(k + ) x. Also Γ(n +, ) is an incomplt gamma function and gnrally dfind as follows: Γ(a, z) = z t t a dt (R(a) > 0),
6 Now, bcaus w know th valu of D n, w can stimat som summations and intgrals. To do this, w dfin th drangmnt function, a natural gnralization of drangmnts, dnotd by D n (x), for vry intgr n 0 and vry ral x as follows: { n! n x i D n (x) =, x 0; n!, x = 0. It is asy to obtain th following gnralizd rcursiv rlations: D n (x) = (x + n)d n (x) x(n )D n 2 (x) = x n + nd n (x), (D 0 (x) =, D (x) = x + ). Not that D n (x) is a nic polynomial. Its valu for x = is D n, for x = 0 is th numbr of prmutations of n distinct objcts and for x = is w n+2 = th numbr of distinct paths btwn vry pair of vrtics in a complt graph on n + 2 vrtics, and A natural qustion is D n () = n! (n ), (s [3]). Qustion 2. Is thr any combinatorial maning for th valu of D n (x) for othr valus of x? Th abov dfinitions yild [ D n (x) = x n 2F 0 n ; x ] (x 0), and D n (x) = x Γ(n +, x). (2) W obtain [ 2F 0 and [ n 2F 0 Also w hav som corollaris. n ] ; = n!, ] n! + ; = ( ) n. Corollary 3. For vry ral x 0 w hav [ ] n + F n + 2 ; x = (n + )(n! x D n (x)). x n+ Proof: Obvious.
7 Corollary 3.2 For vry intgr n w hav n! + t t n dt =, 0 t t n dt = n!, t t n dt = n!, and t t n dt = {n!}, 0 0 { { t t n n! dt = } n is odd, { n! n is vn. } t t n dt = ( + )n! ( + ) n!, Proof: Us rlations (3), (6), (9), (2) and th dfinition of drangmnt function in th cas x = 0. [ ] n Qustion 3. Ar thr any similar formulas for 2 F 0 ;? In othr words, givn x any ral numbr x, is thr an intrval I (dpndnt on x) such that n! n 4 Acknowldgmnts x i = x n! + m (m I x )? I would lik to xprss my gratitud to Dr. J. Rooin for his valuabl guidanc. Also I thank th rfr for his/hr priclss commnts on th third sction. Rfrncs [] T.M. Apostol, Mathmatical Analysis, Addison-Wsly, 974. [2] P.J. Camron, Combinatorics: Topics, Tchniqus, Algorithms, Cambridg Univrsity Prss, 994. [3] M. Hassani, Cycls in graphs and drangmnts, Math. Gaztt, to appar. [4] M. Ptkovšk, H.S. Wilf and D. Zilbrgr, A = B, A. K. Ptrs, 996. Also availabl at http://www.cis.upnn.du/~wilf/aqb.html.
8 2000 Mathmatics Subjct Classification: 05A0, 33B20, 33C20. Kywords:, drangmnts, drangmnt function, incomplt gamma function, hyprgomtric function (Concrnd with squnc A00066.) Rcivd Fbruary 7, 2003; rvisd vrsion rcivd Fbruary 24, 2003. Publishd in Journal of Intgr Squncs Fbruary 25, 2003. Rturn to Journal of Intgr Squncs hom pag.