A itroductio to the Smaradache Double factorial fuctio Felice Russo Via A. Iifate 7 67051 Avezzao (Aq) Italy I [1], [2] ad [3] the Smaradache Double factorial fuctio is defied as: Sdf() is the smallest umber such that Sdf()!! is divisible by, where the double factorial is give by [4): m.. " - 1 x 3 xjx -... m, I if m IS a dd, m.. "-246- x x x... m, if I m IS eve. I this paper we will study this fuctio ad several examples, theorems, cojectures ad problems will be preseted. The behaviour of this fuctio is similar to the other Sraradache fuctios itroduced i the chapter I. I the table below the first 100 values of fuctio Sdf() are give: Sdf() Sdf() Sdf() Sdf() Sdf() 1 1 21 7 41 41 61 61 81 15 2 2 22 22 42 14 62 62 82 82 3 3 23 23 43 43 63 9 83 83 4 4 24 6 44 22 64 8 84 14 5 5 25 15 45 9 65 13 85 17 6 6 25 26 46 46 66 22 86 86 7 7 27 9 47 47 67 67 87 29 8 4 28 14 48 6 68 34 88 22 9 9 29 29 49 21 69 23 89 89 10 10 30 10 50 20 70 14 90 12 11 11 31 31 51 17 71 71 91 13 12 6 32 8 52 26 72 12 92 46 13 13 33 11 53 53 73 73 93 31 14 14 34 34 54 18 74 74 94 94 15 5 35 7 55 11 75 15 95 19 16 6 36 12 56 14 76 38 96 8 17 17 37 37 57 19 77 11 97 97 18 12 38 38 58 58 78 26 98 28 19 19 39 13 59 59 79 79 99 11 20 10 40 10 60 10 80 10 100 20 158
Accordig to the experimetal data the followig two cojectures ca be foulated: co Cojecture 4.1 The series L Sdf() is asymptotically equal to a b where a =1 ad b are close to 0.8834., ad 1.759.. respectively. co Cojecture 4.2 The series ~ 1 L.J Sdf() is asymptotically equal to a b where a =l ad b are close to 0.9411.. ad 0.49.. respectively. Let's start ow with the proof of some theorems. Theorem 4.1. Sdf(P)=p where p is ay prime umber. Proof For p=2, of course Sdf(2)=2. For p odd istead observes that oly for m=p the factorial of first m odd itegers is a multiple of p, that is 1 3 5 7K K P = (p-2)!! p. Theorem 4.2. For ay squarefree eve umber, Sdf() = 2 max{pi>p2,p3,k K Pk} where pj,p2,pj,kk Pk are the prime factors of. Proof Without loss of geerality let's suppose that = PJ.P2,P3 where PJ > P2 > PI ad PI = 2 < Give that the factorial of eve itegers must be a multiple of of course the smallest iteger m such that 2 4 6K < m is divisible by is 2 P3. Ifact for m = 2. PJ we have : 2 4 6K 2 P2K 2 P3 = (2 P2. PJ)' (4 6K 2K 2) = k (2 P2. PJ) where ken Theorem 4.3. For ay squarefree composite odd umber, Sdf() = max{pl>p2,k Pk} where PJ,P2,K Pk are the prime factors of. 159
Proof Without loss of geerality let suppose that = PI. P2 where PI ad P2 are two distict primes ad P2' > PI. Of course the factorial of odd itegers up to P2 is a multiple of because beig PI < P2 the factorial will cotai the product PI. P2 ad therefore : 1 3 5K. PI K. P2 Theorem 4.4. ~ 1 L.JSdfC) =1 a:j diverges. Proof. This theorem is a direct cosequece of the divergece of sum L.J ~pl where p is ay prime umber. a:j a:j I fact ~ 1 > ~...!..- accordig to the theorem 4.1 ad this proves the L.J Sdf(k) L.J p k=i p=2 theorem. p Theorem 4.5 The Sdf() fuctio is ot additive that is SdfC+m):;;Sdf()+Sdf(m) for (,m)=!. Proof I fact for example Sdf(2 + 15):;; Sdf(2) + Sdf(l5). Theorem 4.6 The Sdf() fuctio IS ot multiplicative, that IS Sdf( m) :;; Sdf() Sdf(m) for (,m) = 1. Proof I fact for example Sdf(3 4) :;; Sdf(3) Sdf( 4). Theorem 4.7 Sdf():::; Proof If is a square free umber the based o theorems 4.1, 4.2 ad 4.3 Sdf():::;. Let's ow cosider the case whe is ot a squarefree umber. Of course the maximum value of the Sdf() fuctio caot be larger tha because whe we arrive i the factorial to for shure it is a multiple of. 160
CXl Theorem 4.8 L Sd~() diverges. =1 >' 00 00 ~ SdfkCk) Sd f ) pcp Proof I fact L..J L..J where p is ay prime umber ad of k=1 p=2,sdfp(p) course L..J diverges because the umber of primes IS ifiite [5] ad Sdf(p)=p. p Theorem 4.9 Sdf() ~ 1 for ~ 1 Proof This theorem is a direct cosequece of the Sdf() fuctio defiitio. I fact for =l, the smallest m such that 1 divide Sdf(l) is trivially 1. For ::j; 1, m must be greater tha 1 becuase the factorial of 1 caot be a multiple of. Theorem 4.10 o < Sdf() ~ 1 for ~ 1 Pool The theorem is a direct cosequece of theorem 4.7 ad 4.9. Theorem 4. 11 (primorial) [4]. Sdf(Pk#) = 2 Pk where Pk# is the product of first k primes Proof The theorem is a direct cosequece of theorem 4.2. Theorem 4<12 The equatio Sdf() = 1 has a ifiite umber of solutios. 161
Proof The theorem is a direct cosequece of theorem 4.1 ad the well-kow fact that there is a ifiite umber of prime umbers [5]. Theorem 4.13 The eve (odd respectively) umbers are ivariat uder the applicatio of Sdf fuctio, amely Sdf( eve)=eve ad Sdf( odd)=odd Proof Of course this theorem is a direct cosequece of the Sdf() fuctio defiitio. Theorem 4.14 solutios. The diophatie equatio Sdf() = SdJ( + 1) does't admit Proof I fact accordig to the previous theorem if is eve (odd respectively) the Sdf() also is eve (odd respectively). Therefore the equatio Sdf()=Sdf(+l) ca ot be satisfied because Sdf() that is eve should be equal to Sdf(+ 1) that istead is odd. Cojecture 4.3 ]0,1]. The fuctio Sdf() is ot distributed uiformly i the iterval Cojecture 4.4 For ay arbitrary real umber > 0, there is some umber ~ 1 such that Sdf() < Let's ow start with some problems related to the Sdf() fuctio. Problem 1. Use the otatio FSdf()=m to deote, as already doe for the Zt() ad Zw() fuctios, that m is the umber of dijjeret itegers k such that Zw(k) =. Example FSdf(1)=l sice Sdf(1)=l ad there are o other umbers such that Sdf() =1 Study thefuctio Fsdf(). 162
Evaluate m :L FS r(k2 lim -ck :..::=;!...I m~ m Problem 2. Is the differece ISdf(+ 1)-Sdf() I bouded or ubouded? Problem 3. Fid the solutios of the equatios: Sdf(+l) = k Sdf() where k is ay positive iteger ad > 1 for the first equatio. Sdf() = k Sdf(+ 1) Cojecture 4.5 The previous equatios do't admits solutios. Problem 4. Aalyze the iteratio of Sdf() for all values of. For iteratio we ited the repeated applicatio ofsdf(). For example the k-th iteratio ofsdf() is: Sdfk() = Sdf(Sdf(KK (Sdf())K) where Sdfis repeated k times. For all values of, will each iteratio of Sdf() produces always a fixed poit or a cycle? Problem 5. Fid the smallest k such that betwee Sdf() ad Sdf(k+), for > 1, there is at least a prime. Problem 6. Is the umber 0.1232567491011.... where the sequece of digits is Sdf() for 21 a irratioal or trascedetal umber? (We call this umber the Pseudo-Smaradache-Double Factorial costat). Problem 7. Is the Smaradache Euler-Mascheroi sum (see chapter 11 for defiitio) coverget for Sdf() umbers? lfyes evaluate the covergece value. 163
00 Problem 8. Evaluate L(-ll Sdf(k)-l k=l Problem 9. Evaluate 1 Sdf() =l 00 Problem 10. Evaluate lim Sdf(k) k~oo B(k) where B(k) = Ll(Sdf(» 5,k Problem 11. Are there m,, k o-ull positive itegers for which Sdf( m) = m k. Sdf()? Problem 12. Are there itegers k> 1 ad > 1 such that (Sdf()/ = k Sdf( k)? Problem 13. Solve the problems from 1 up to 6 already formulated for the Zw() fuctio also for the Sdf() fuctio. Problem 14. Fid all the solutio of the equatio Sdf()!=Sdf(!) Problem 15. Fid all the solutios of the equatio Sdf( k ) = k Sdf() for k>1ad >1. Problem 16. Fid all the solutios of the equatio Sdf( k ) = Sdf(k) for k> 1. 164
Problem 17. Fid all the solutios of the equatio Sdf( k ) = m. Sdf(m) where k>j ad, m >0. Problem 18. For the first values of the Sdf() fuctio the followig iequality is true: 1 --~ _. + 2 for 1 ~ ~ 1000 Sdf() 8 Is this still true for > 1 000? Problem 19. true: For the first values of the Sdf() fuctio the followig iequality is _Sd...:..if-.:.(...:..) ~ o17~ for 1 ~ ~ 1000. J Is this still true for all values of > J 000? Problem 20. For the first values of the Sdf() fuctio the followig iequality hold: Is this still true for > 1 000? 1 1 1 - + < 4 for 2 < ~ 1000 Sdf() Problem 21. holds: For the first values of the Sdf() fuctio the followig iequality 1 ---< 4 Sdf() 5 for l~~looo 165
Is this iequality still true for > 1000? Problem 22. Study the covergece of the Smaradache Double factorial harmoic series: 00 "" 1 where a > 0 ad a E R ~Sdfa() =l Problem 23. Study the covergece of the series: where x is ay icreasig sequece such that lim X = 00 ~oo Problem 24. Evaluate "" ~(Sd[(})2 L...J lim k=7 -- X: I(k) Is this limit coverget to some kow mathematical costat? Problem 25. Solve the fuctioal equatio: where r is a iteger 2:: 2. Wath about the juctioal equatio: Sdf(/ +Sdf(/- l +A A Sdf() = Sdf(/ + Sdf(/- 1 +A /\. Sdf() = k 166
where r ad k are two itegers ~ 2. Problem 26. Is there ay relatioship betwee SdJ TI mkj '1 l k=1 m ad L Sdj(mk)? k=i Refereces. [1] F. Smaradache, "Oly Problems, ot Solutios!", Xiqua, Publishig House, Phoeix-Chicago, 1990, 1991, 1993; [2] F. Smaradache, "The Floreti Smaradache papers" special collectio, Arizoa State Uiversity, Hayde Library, Tempe, Box 871006, AZ 85287-1006, USA; [3] C. Dumitrescu ad V. Seleacu, Some otios ad questios i Number Theory, Erhus Uiv. Press, Gledale 1994 [4] E. Weisstei, erc Cocise Ecyclopedia of Mathematics, CRC Press, 1999 [5] P. Ribeboim, The book a/prime umbers records, Secod editio, New York, Spriger-Verlag, 1989 167