Scientia Magna Vol. 3 (2007), No. 2, 104-108 On the F.Smarandache function and its mean value Zhongtian Lv Department of Basic Courses, Xi’an Medical College Xi’an, Shaanxi, P.R.China Received December 19, 2006 Abstract For any positive integer n, the famous F.Smarandache function S(n) is defined asthesmallestpositiveintegermsuchthatn|m!. Thatis,S(n)=min{m: n|m!, n∈N}. The main purpose of this paper is using the elementary methods to study a mean value probleminvolvingtheF.Smarandachefunction,andgiveasharperasymptoticformulaforit. Keywords F.Smarandache function, mean value, asymptotic formula. §1. Introduction and result For any positive integer n, the famous F.Smarandache function S(n) is defined as the smallest positive integer m such that n | m!. That is, S(n) = min{m : n | m!, n ∈ N}. For example, the first few values of S(n) are S(1) = 1, S(2) = 2, S(3) = 3, S(4) = 4, S(5) = 5, S(6) = 3, S(7) = 7, S(8) = 4, S(9) = 6, S(10) = 5, ···. About the elementary properties of S(n), some authors had studied it, and obtained some interesting results, see reference [2], [3] and [4]. For example, Farris Mark and Mitchell Patrick [2] studied the elementary properties of S(n), and gave an estimates for the upper and lower bound of S(pα). That is, they showed that (p−1)α+1≤S(pα)≤(p−1)[α+1+log α]+1. p Murthy [3] proved that if n be a prime, then SL(n) = S(n), where SL(n) defined as the smallest positive integer k such that n | [1, 2, ··· , k], and [1, 2, ··· , k] denotes the least common multiple of 1, 2, ···, k. Simultaneously, Murthy [3] also proposed the following problem: SL(n)=S(n), S(n)(cid:54)=n ? (1) Le Maohua [4] completely solved this problem, and proved the following conclusion: Every positive integer n satisfying (1) can be expressed as n=12 or n=pα1pα2···pαrp, 1 2 r where p , p , ···, p , p are distinct primes, and α , α , ···, α are positive integers satisfying 1 2 r 1 2 r p>pαi, i=1, 2, ···, r. i Vol. 3 OntheF.Smarandachefunctionanditsmeanvalue 105 Dr. XuZhefeng[5]studiedthevaluedistributionproblemofS(n),andprovedthefollowing conclusion: Let P(n) denotes the largest prime factor of n, then for any real number x > 1, we have the asymptotic formula (cid:161) (cid:162) (cid:195) (cid:33) (cid:88)(S(n)−P(n))2 = 2ζ 23 x32 +O x32 , 3lnx ln2x n≤x where ζ(s) denotes the Riemann zeta-function. On the other hand, Lu Yaming [6] studied the solutions of an equation involving the F.Smarandache function S(n), and proved that for any positive integer k ≥2, the equation S(m +m +···+m )=S(m )+S(m )+···+S(m ) 1 2 k 1 2 k has infinite groups positive integer solutions (m ,m ,··· ,m ). 1 2 k Jozsef Sandor [7] proved for any positive integer k ≥2, there exist infinite groups positive integer solutions (m , m , ··· , m ) satisfied the following inequality: 1 2 k S(m +m +···+m )>S(m )+S(m )+···+S(m ). 1 2 k 1 2 k Also, there exist infinite groups of positive integer solutions (m ,m ,··· ,m ) such that 1 2 k S(m +m +···+m )<S(m )+S(m )+···+S(m ). 1 2 k 1 2 k Themainpurposeofthispaperisusingtheelementaryandanalyticmethodstostudythe mean value properties of [S(n)−S(S(n))]2, and give an interesting mean value formula for it. That is, we shall prove the following conclusion: Theorem. Let k be any fixed positive integer. Then for any real number x>2, we have the asymptotic formula (cid:181) (cid:182) (cid:195) (cid:33) (cid:88)[S(n)−S(S(n))]2 = 2 ·ζ 3 ·x32 ·(cid:88)k ci +O x32 , 3 2 lnix lnk+1x n≤x i=1 where ζ(s) is the Riemann zeta-function, c (i = 1, 2,··· , k) are computable constants and i c =1. 1 §2. Proof of the Theorem In this section, we shall prove our theorem directly. In fact for any positive integer n>1, let n=pα1pα2···pαs be the factorization of n into prime powers, then from [3] we know that 1 2 s S(n)=max{S(pα1), S(pα2), ··· , S(pαs)}≡S(pα). (2) 1 2 s Now we consider the summation (cid:88) (cid:88) (cid:88) [S(n)−S(S(n))]2 = [S(n)−S(S(n))]2+ [S(n)−S(S(n))]2, (3) n≤x n∈A n∈B 106 ZhongtianLv No. 2 where A and B denote the subsets of all positive integer in the interval [1, x]. A denotes the set involving all integers n ∈ [1, x] such that S(n) = S(p2) for some prime p; B denotes the set involving all integers n ∈ [1, x] such that S(n) = S(pα) with α = 1 or α ≥ 3. If n ∈ A, then n = p2m with P(m) < 2p, where P(m) denotes the largest prime factor of m. So from the definition of S(n) we have S(n)=S(mp2)=S(p2)=2p and S(S(n))=S(2p)=p if p>2. From (2) and the definition of A we have (cid:88) [S(n)−S(S(n))]2 n∈A (cid:88) (cid:163) (cid:164) (cid:88) (cid:163) (cid:164) = S(p2)−S(S(p2)) 2+ S(p2)−S(S(p2)) 2 n≤x n≤x √ √ p2(cid:107)n, n<p2 p2(cid:107)n, p2≤ n (cid:88) (cid:163) (cid:164) (cid:88) (cid:163) (cid:164) = S(p2)−S(S(p2)) 2+ S(p2)−S(S(p2)) 2 p2n≤x p2n≤x n<p2, (p, n)=1 p2≤n, (p, n)=1 (cid:88) (cid:88) = p2+ p2+O(1) p2n≤x p2n≤x n<p2, (p, n)=1 n≥p2, (p, n)=1 = (cid:88) (cid:88) p2+O (cid:88) (cid:88) p2+O (cid:88) (cid:88) p2 √ n≤ xn<p2≤nx m≤x41 p≤(x)13 p≤x41 p2≤n≤px2 (cid:195) (cid:33) m = (cid:88) (cid:88) p2+O x54 , (4) √ √ lnx n≤ xp≤ x n where p2(cid:107)n denotes p2|n and p3†n. By the Abel’s summation formula (See Theorem 4.2 of [8]) and the Prime Theorem (See Theorem 3.2 of [9]): (cid:181) (cid:182) (cid:88)k a ·x x π(x)= i +O , lnix lnk+1x i=1 where a (i=1, 2,··· ,k) are computable constants and a =1. i 1 We have √ (cid:181)(cid:114) (cid:182) (cid:90) (cid:88) p2 = x ·π x − nx 2y·π(y)dy √ n n 3 p≤ x 2 n (cid:195) (cid:33) = 1 · x32 ·(cid:88)k b(cid:112)i +O x32 , (5) 3 n32 i=1 lni nx n32 ·lnk+1x √ where we have used the estimate n≤ x, and all b are computable constants and b =1. i 1 (cid:181) (cid:182) (cid:88)∞ 1 3 (cid:88)∞ lnin Note that =ζ , and is convergent for all i=1, 2, 3, ···, k. So from n=1n32 2 n=1 n32 Vol. 3 OntheF.Smarandachefunctionanditsmeanvalue 107 (4) and (5) we have (cid:88) [S(n)−S(S(n))]2 n∈A (cid:34) (cid:195) (cid:33)(cid:35) (cid:195) (cid:33) = (cid:88) 1 · x32 ·(cid:88)k b(cid:112)i +O x32 +O x54 n≤√x 3 n32 i=1 lni nx n32 ·lnk+1x lnx (cid:181) (cid:182) (cid:195) (cid:33) = 2 ·ζ 3 ·x32 ·(cid:88)k ci +O x23 , (6) 3 2 lnix lnk+1x i=1 where c (i=1,2, 3,··· ,k) are computable constants and c =1. i 1 NowweestimatethesummationinsetB. Foranypositiveintegern∈B,ifS(n)=S(p)= p, then [S(n)−S(S(n))]2 =[S(p)−S(S(p))]2 =0; If S(n)=S(pα) with α≥3, then [S(n)−S(S(n))]2 =[S(pα)−S(S(pα))]2 ≤α2p2 and α≤lnx. So that we have (cid:88) (cid:88) [S(n)−S(S(n))]2 (cid:191) α2·p2 (cid:191)x·ln2x. (7) n∈B npα≤x α≥3 Combining (3), (6) and (7) we may immediately deduce the asymptotic formula (cid:181) (cid:182) (cid:195) (cid:33) (cid:88)[S(n)−S(S(n))]2 = 2 ·ζ 3 ·x32 ·(cid:88)k ci +O x32 , 3 2 lnix lnk+1x n≤x i=1 where c (i=1,2, 3,··· ,k) are computable constants and c =1. i 1 This completes the proof of Theorem. References [1] F. Smarandache, Only Problems, Not Solutions, Chicago, Xiquan Publishing House, 1993. [2] Farris Mark and Mitchell Patrick, Bounding the Smarandache function. Smarandache Nations Journal, 13(2002), 37-42. [3] A. Murthy, Some notions on least common multiples, Smarandache Notions Journal, 12(2001), 307-309. [4] Le Maohua, An equation concerning the Smarandache LCM function, Smarandache Notions Journal, 14(2004), 186-188. [5] Xu Zhefeng, The value distribution property of the Smarandache function, Acta Math- ematica Sinica, Chinese Series, 49(2006), No.5, 1009-1012. 108 ZhongtianLv No. 2 [6] Lu Yaming, On the solutions of an equation involving the Smarandache function, Sci- entia Magna, 2(2006), No.1, 76-79. [7] Jozsef Sandor, On certain inequalities involving the Smarandache function, Scientia Magna, 2(2006), No. 3, 78-80. [8]TomM.Apostol,IntroductiontoAnalyticNumberTheory,NewYork,Springer-Verlag, 1976. [9] Pan Chengdong and Pan Chengbiao, The elementary proof of the Prime theorem, Shanghai, Shanghai Science and Technology Press, 1988.